172 Chapter 4 Measurement: Area and Volume Student Learning Opportunities 1 (C) Your students have accepted the formula for the volume of a simple solid. They are now curious to know how you can show that if you have the solid shown in Figure 4.72 h B Figure 4.72 where each cross section has area B and height is h, the volume of the solid is Bh. How do you show it? 2* Many times strokes are caused by a buildup of plaque in the arteries. Imagine the plaque buildup in an artery to be a region between two concentric circular cylinders of length L, and assume that the inner radius is 0.3cm and the outer radius is 0.307 cm. Estimate the volume of the artery blocked by plaque. (See Figure 4.73.) .307 .3 L . .307 .3 Figure 4.73 3* The French physiologist Jean Poisseuille discovered the law that the volume of blood flowing through an artery per minute is given by V = kR4 where k is a constant, and R is the inner radius. When a person’s arteries are blocked, a procedure called angioplasty is done. Here a balloon is inserted into the artery and the artery is expanded. Suppose that under angioplasty, an artery has its radius increased 5%. Estimate the change in the volume of blood flow that results. 4* (C) A student asks why can’t you use formula (4.30) to find R xxdx. What is your answer? 5* Let R be the region bounded by y = x2, the x-axis, the lines x = 1 and x = 5. (a) Suppose S is a solid whose base is R and whose cross sections perpendicular to the x-axis are semicircles. What is the volume of the solid, S? (b) Suppose S is a solid whose base is R and whose cross sections perpendicular to the x-axis are equilateral triangles. What is the volume of the solid, S?
4.6 Introduction to Volume 173 6 (C) A student asks how you show that the volume of a sphere with radius R is 4 pR3: How do you 3 explain it by using integrals? 7 If a plane perpendicular to a diameter at a distance a from the center of a sphere chops the sphere into two parts, the smaller part is called a spherical cap. Find the volume of a spherical cap in terms of a and R where R is the radius of the sphere. 8 (C) A student asks you how to show that the volume of a pyramid with square base whose area is B is 1 Bh: Using Figure 4.74 and the hint, fill in the details. 3 origin O x SQ h PR Area of the base is B x-axis Figure 4.74 [Hint: Put the pyramid so that its apex is at the origin. We have also shown a typical cross section at a distance x from the origin (the square with the letters S and Q). Observe that tri- angles SOQ and POR are similar. Thus, SQ = hx. But SQ is half the side of the square containing PR SQ, and PR is half the side of the square containing PR. Thus, the sides of the squares containing SQ and PR are double SQ and PR, and the ratio of the areas of the squares containing these lines is ð2SQÞ2 or just ðSQÞ2 : Recalling that the area of the square base is B and calling the area of the ð2PRÞ2 ðPRÞ2 cross section at distance x from the origin A(x), we have AðxÞ ¼ ðSQÞ2 ¼ x 2 . Thus, AðxÞ ¼ B ðPRÞ2 h Bx2 : Now finish it. The key step in the proof was to show that AðxÞ ¼ Bx2 : This is true for h2 h2 any pyramid with a vertical axis, regardless of the shape of the base.]
CHAPTER 5 THE TRIANGLE: ITS STUDY AND CONSEQUENCES 5.1 Introduction If you ask adults what theorem they remember from their study of mathematics, they will most probably say, the Pythagorean Theorem. Why should this theorem, usually studied in secondary school, make such a lasting impression? As will be demonstrated in this chapter, this one theorem concerning the relationship of the sides of a right triangle can be extended to the study of (a) all types of triangles, (b) relationships concerning circles, (c) key trigonometric relationships, and (d) concepts of area. It is really quite amazing! To begin, we need only remember a few basic definitions: In a right triangle with acute angle, A, sin A ¼ length of side opposite A ; length of hypotenuse cos A ¼ length of side adjacent to A; length of hypotenuse tan A ¼ length of side opposite to A : length of side adjacent to A Also, we easily see that tan A ¼ sin AA: cos We begin by discussing how the Pythagorean Theorem can be extended to generate the Law of Cosines and then follow it with the study of the Law of Sines. What is different about this chapter is that these will then be used to prove all the congruence laws, all the main similarity laws, facts about circles, and a host of other relationships in triangles. While people normally think that trigonometry follows geometry, we have demonstrated that we can do things in reverse by showing how the theorems of geometry can be derived from theorems in trigonometry. We will use the following notational conventions throughout: When we say that ∡A = ∡B, we mean that the measures of the two angles are equal. An alternate way of saying this is that ∡A is congruent to ∡B. AB will represent the length of AB: When we say that AB = CD we are saying that the measures of AB and CD are equal, or equivalently, that AB is congruent to CD.
176 Chapter 5 The Triangle: Its Study and Consequences 5.2 The Law of Cosines and Surprising Consequences LAUNCH Draw a large triangle on a clean sheet of paper. Then measure the length of each of the sides of the triangle you have drawn. Using these same three lengths, try to draw another triangle that is NOT congruent to the first one you drew. Could you do it? Why or why not? We hope that the launch question helped you to recall some of the work you did in secondary school regarding congruent triangles. You probably remember the theorem that one of the ways to prove that two triangles are congruent is to show that the three sides of one are congruent to the three sides of another (often represented as SSS = SSS). But, if a student asked you if this was an axiom or a theorem, would you know what to say? Don’t feel badly if you wouldn’t, since many secondary school textbooks have listed it in different ways. It will probably surprise you to know that it can indeed be proven, by applying the Law of Cosines. In this section, we will demonstrate how this and other similar congruence results can be shown. The Pythagorean Theorem tells us that, in a right triangle with legs a and b and hypotenuse c, a2 + b2 = c2. What happens if the triangle is not a right triangle? The following theorem answers this. Theorem 5.1 (Law of Cosines): In any triangle ABC, c2 ¼ a2 þ b2 À 2ab cos C: Proof. Notice that “c2 = a2 + b2” is part of the theorem. How interesting! It seems very likely that we will be using the Pythagorean Theorem in this proof. We begin with triangle ABC and draw altitude AD and call its height h. We will prove the theorem in the case when the altitude is inside the triangle (that is, when the triangle has all its angles less than 90°). In Student Learning Opportunity 4 you will prove it for the case when the altitude is outside of the triangle. A bh c C xD a −x B Figure 5.1
5.2 The Law of Cosines and Surprising Consequences 177 Using the Pythagorean Theorem on triangle ABD we have that (a À x)2 + h2 = c2, which when expanded gives us a2 À 2ax þ x2 þ h2 ¼ c2: ð5:1Þ Using the Pythagorean Theorem on triangle ACD we have x2 þ h2 ¼ b2: ð5:2Þ Substituting equation (5.2) in equation (5.1) we have a2 À 2ax þ b2 ¼ c2 ð5:3Þ Now, from triangle ACD, cos C = x from which it follows that x = b cos C. Substituting this in b equation (5.3) for x we get a2 À 2ab cos C þ b2 ¼ c2 ð5:4Þ which can be rearranged to read ð5:5Þ c2 ¼ a2 þ b2 À 2ab cos C and we are done. & ð5:6Þ There are two other versions of the law of cosines: a2 ¼ b2 þ c2 À 2bc cos A and b2 ¼ a2 þ c2 À 2ac cos B ð5:7Þ and they are proved in exactly the same way, only we draw altitudes to the other sides of the tri- angle. You will prove one of the versions in the Student Learning Opportunities. Taking the theorems further, if we solve for cos C in equation (5.5) we get that a2 þ b2 À c2 ð5:8Þ cos C ¼ 2ab : Similarly, we can solve for cos A and cos B in equations (5.6) and (5.7) respectively, to get cos A ¼ b2 þ c2 À a2 ð5:9Þ 2bc and cos B ¼ a2 þ c2 À b2 ; respectively: ð5:10Þ 2ac What equations (5.9), (5.10), and (5.8) tell us, respectively, is, if we know the lengths a, b, and c of the three sides of a triangle ABC, then we immediately know cos A, cos B, and cos C and hence angles A, B, and C. This brings us to the topic of congruence. 5.2.1 Congruence Recall that, in geometry, two triangles ABC and DEF are congruent if their corresponding sides and corresponding angles are congruent. That is, they have the same measure. So, if ABC is congruent to DEF, then AB = DE, BC = EF, AC = DF, and ∡A = ∡D, ∡B = ∡E, and ∡C = ∡F. (See Figure 5.2.)
178 Chapter 5 The Triangle: Its Study and Consequences B E ca fd A bC D e F Figure 5.2 Under this correspondence, angles A and D are called corresponding angles, as are the angles B and E, and C and F. Sides AB and DE are called corresponding sides, as are the sides BC and EF, and the sides AC and DF. By definition of congruent triangles, corresponding parts have the same measure, which means corresponding sides have the same length and corresponding angles have the same degree measure. Notice that the order in which we write the letters tells us the angle correspon- dence and side correspondence. Had we written that triangle ACB was congruent to FDE, then it would mean that ∡A = ∡F, ∡C = ∡D, and ∡B = ∡E, and that AC = FD, CB = DE, and BA = EF. The first result we talk about is something we are all familiar with: If three sides of one triangle have the same lengths as three sides of another triangle, then the triangles are congruent. That is, all their corresponding parts match! This is quite remarkable since we have said nothing about the angles of these triangles. Yet, this follows immediately from the Law of Cosines. Theorem 5.2 (SSS = SSS) If the three sides of triangle ABC are congruent to the three sides of triangle DEF, then the triangles ABC and DEF are congruent. Proof. Let us assume that the sides that match are a and d, b and e, and c and f. (Refer to Figure 5.2.) So a = d, b = e, and c = f. From equation (5.8) we have that a2 þ b2 À c2 ð5:11Þ cos C ¼ 2ab and using the same law in triangle DEF with the corresponding sides, we have cos F ¼ d2 þ e2 À f 2 ð5:12Þ 2de : Since a = d, b = e, and c = f we can substitute them in equation (5.11) to get d2 þ e2 À f 2 2de cos C ¼ ð5:13Þ and we see from equations (5.12) and (5.13) that cos C ¼ cos F: ð5:14Þ It follows that ∡C = ∡F. In a similar manner using the other versions of the Law of Cosines, equations (5.9) and (5.10), we can show that ∡A = ∡D and ∡B = ∡E. Thus, if three sides of one triangle are equal to three sides of another triangle, then the corre- sponding angles match, and so the triangles are congruent. & Note: In the proof of Theorem 5.2 (refer to equation (5.14)), we used the fact that, if cos C = cos F, then ∡C = ∡F. While you may have accepted this, much more is involved in this statement
5.2 The Law of Cosines and Surprising Consequences 179 than meets the eye. For now, we will use this fact continually and ask you to accept it. But, we will examine the reason behind it in a later section on technical issues. We now turn to a corollary of our SSS congruence theorem. Corollary 5.3 (HL = HL) Two right triangles are congruent if the hypotenuse and leg of one triangle are congruent to the hypotenuse and leg of the other triangle. Proof. In Figure 5.3 we see two right triangles where the hypotenuse and leg of one have the same lengths as the hypotenuse and leg of the other. H ab H L L Figure 5.3 We will show the third sides of the triangles have the same length. By the Pythagorean Theorem, a2 þ L2 ¼ H2 ð5:15Þ and ð5:16Þ b2 þ L2 ¼ H2: From equations (5.15) and (5.16) we have a2 þ L2 ¼ b2 þ L2: Subtracting L2 from both sides, we get that a2 ¼ b2 and therefore a = b. Thus, the lengths of three sides of one triangle are equal to the lengths of three sides of the other triangle and the two triangles are congruent by SSS = SSS. & Corollary 5.4 (SAS = SAS) If two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then the triangles are congruent. Proof. Suppose we have triangles ABC and DEF and suppose that a = d, b = e, and ∡C = ∡F. (See Figure 5.4.) BE cf ad A bC Fe D Figure 5.4
180 Chapter 5 The Triangle: Its Study and Consequences Then, by the Law of Cosines, applied to triangle ABC, ð5:17Þ c2 ¼ a2 þ b2 À 2ab cos C: By the Law of Cosines, applied to triangle DEF, we have ð5:18Þ f 2 ¼ d2 þ e2 À 2de cos F: But we know that a = d, b = e, and ∡C = ∡F, and if we substitute these into equation (5.18) we get f 2 ¼ a2 þ b2 À 2ab cos C: ð5:19Þ Since the right hand sides of equations (5.17) and (5.19) are the same, so are the left sides. That is, c2 ¼ f 2: From this we get that c = f (since both c and f are positive). Thus, the three sides of the first triangle are equal to the three sides of the second triangle, and so the triangles are congruent by Theorem 5.2. & Suppose we have one triangle and we only know the measures of two of its angles and one side. Would we be able to use the Law of Cosines to determine information about the other two sides? Well, the answer is, “No.” Since the Law of Cosines requires us to know two sides and one angle, we do not have enough information to use it. To find the missing information about the triangle, we need another law which we will discuss in the next section: the Law of Sines. Student Learning Opportunities 1* Given that the sides of a triangle are a = 3, b = 5, and c = 7, find all three angles. 2* If the sides of a parallelogram are 3 and 4 and the angle between them is 30 degrees, how long is each diagonal? 3* A surveyor needs to estimate the distance across a lake from point A to point B. Standing at point C, 4.6 miles from A and 7.3 miles from B, he measures the angle shown in Figure 5.5 to be 80 degrees. Estimate the distance AB. A B 4.6 miles 7.3 miles 80 Figure 5.5 C 4 We will point out in the chapter on trigonometry that cos(180° À x) = Àcos x. Use this fact to prove the Law of Cosines when the altitude AD in Figure 5.1 is outside of the triangle. 5 (C) A student asks how you prove the other two versions of the Law of Cosines found in equa- tions (5.9) and (5.10). How do you do it?
5.3 The Law of Sines 181 6* Your students are intrigued by how the Pythagorean Theorem was used to prove the Law of Cosines. They wonder because of the similar structure of the theorems, if one can go in reverse. That is, can one use the Law of Cosines to prove that if c2 = a2 + b2 holds in a triangle, then the triangle is right? What is your answer and how do you show it? 7 (C) A student asks how you can prove that if two angles and any side of one triangle are equal to two angles and the corresponding side of another triangle, the triangles are congruent (i.e., AAS = AAS). How do you prove it? 5.3 The Law of Sines LAUNCH Give an example (draw it) where two angles and a side of one triangle are equal to two angles and a side of another triangle, but the triangles are not congruent. [Hint: make sure the sides are not corresponding.] We hope that you were able to construct two different shaped triangles and that if you hadn’t realized it before, you realize now, the importance of always specifying that corresponding parts be congruent in congruence proofs. You most likely remember the theorem, which says that one of the ways to prove that two triangles are congruent is to show that they have two angles and a cor- responding side that are congruent. What you probably did not know is why this is true. In this section you will be surprised to see how instrumental the Law of Sines can be in verifying the proof of this relationship. Theorem 5.5 (Law of Sines): In any triangle ABC a A ¼ b B ¼ c C : sin sin sin Proof. We may use Figure 5.1 (from earlier in the chapter), which we copy here for convenience. A bh c C xD a −x B Figure 5.1
182 Chapter 5 The Triangle: Its Study and Consequences In triangle ADB, sin B ¼ h : Hence, c h ¼ c sin B: ð5:20Þ ð5:21Þ In triangle ADC, sin C ¼ h : Thus, ð5:22Þ b h ¼ b sin C: Setting the two expressions equal for h in equations (5.20) and (5.21) we have c sin B ¼ b sin C; and dividing both sides of equation (5.22) by (sin B) (sin C) we get c ¼ b B : ð5:23Þ sin C sin That is the first half of our theorem. In the Student Learning Opportunities you will draw a dif- ferent altitude and show that a ¼ c C : ð5:24Þ sin A sin And the two relationships, equations (5.23) and (5.24) together tell us that a A ¼ b ¼ c C : & sin sin B sin Since this law involves several angles, we can now find the missing parts of a triangle in which we are given two angles and a side or two sides and an angle. Corollary 5.6 (ASA = ASA) If two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, the triangles are congruent. Proof. Suppose that we have two triangles ABC and DEF from Figure 5.2 earlier in the chapter, which we copy here, BE ca fd A bC D e F Figure 5.2 and we are given that ∡A = ∡D, ∡B = ∡E, and c = f. Since ∡A = ∡D and ∡B = ∡E we also have ∡C = ∡F, because the sum of the angles of a triangle is 180 degrees. Using the Law of Sines in triangle ABC we get that a ¼ c C : ð5:25Þ sin A sin Using the Law of Sines in triangle DEF we have d D ¼ f F : ð5:26Þ sin sin
5.3 The Law of Sines 183 Since ∡A = ∡D, ∡C = ∡F, and c = f, we can substitute these values into equation (5.25) to get a ¼ f ð5:27Þ sin D sin F and since the right sides of equations (5.26) and (5.27) are the same, we see that a ¼ d D : ð5:28Þ sin D sin Multiplying both sides of equation (5.28) by sin D we get that a = d. Since we were given c = f and we showed that a = d, and we were given that ∡B = ∡E (see Figure 5.2), by Corollary 5.4 (SAS = SAS), we have that the two triangles ABC and DEF are congruent. & Student Learning Opportunities 1* (C) Your students tell you that they are confused about when they should use the Law of Cosines and when they should use the Law of Sines when trying to find missing parts of a tri- angle. What do you tell them? 2* In triangle ABC, ∡A = 37°, ∡B = 64°, and c = 12. Find the lengths of all three sides. 3* In triangle ABC, AC = 56, AB = 80, ∡C = 64°. Find ∡B to the nearest degree. 4* Find ∡ACD to the nearest degree in Figure 5.6. C 16 10 10 30° A DB Figure 5.6 5* A satellite orbiting the earth is being tracked. The observation stations are 300 miles apart in two different towns A and B. When the satellite is visible from both towns, the angles of eleva- tion of the satellite are recorded and found to be 63 and 72 degrees respectively, as shown in Figure 5.7. Satellite 63° 300 miles 72° A Figure 5.7 B How far is the satellite from station A? from station B?
184 Chapter 5 The Triangle: Its Study and Consequences 6 (C) The proof of the Law of Sines we gave was for acute triangles only. A student is curious to know how the proof would have to be modified to show that it is still true if h is outside the triangle. How would you show it? [Hint: It is a fact that sin(180° À x) = sin x.] 7 The following interesting result also follows from the Law of Sines: If ABC is any triangle and BD is the angle bisector of angle B, then AB ¼ AD : (The angle bisector divides the opposite side into seg- BC DC ments having the same ratio as the sides.) Prove this. You will need the fact that sin(180 À p)° = sin p. [Hint: Use Figure 5.8.] B aa pq A xD y C Figure 5.8 8* In triangle BAC, AC is the shortest side. Angle bisector BD is drawn to AC dividing it into seg- ments AD and BD. If AC = 14 and the ratio of the sides of the triangle is 2 : 3 : 4, how long is the shorter of the segments AD and BD? Explain. 9 (C) Your students want to know if there is a way to find the area of a triangle without knowing its altitude. Show them that there is by proving that the area of a triangle ABC is a2 sin B sin C: 2 sin A [Hint: The area of a triangle is 1 base × height. Take the base to be a.] 2 5.4 Similarity LAUNCH 1 Using a ruler, a pen, and a piece of paper, draw a triangle. Label the vertices A, B, and C. Start- ing at vertex A, extend side AB its own length to point D (side AD will now be twice as long as side AB). Now, from vertex A, extend side AC its own length to point E (side AE will now be twice as long as side AC). 2 Measure the length of side BC. Based on what you have measured, what do you predict is the length of side DE? Are you correct? 3 What can you say about the shapes of triangles ABC and AED? Why do you believe this is true?
5.4 Similarity 185 After having done the launch question, you are probably beginning to recall some of the basic properties of similar triangles. Did you ever question how the similarity theorems you learned in secondary school could be proven? In this section we will surprise you with how they can be done. We have used the Law of Sines and the Law of Cosines to derive all the congruence theorems that are taught in geometry. But now let us turn to another set of results that are also critically important, those that deal with similarity. Applications of similarity range from the mundane to the surprising. For example, similarity is used on a daily basis by engineers who use scale drawings to create a model of a building that is going to be constructed. When you take a picture of a person, the picture you get is similar to the person. Similarity is also used (surprisingly) in radiation therapy for cancer patients for accuracy in focusing the beam. To find out more about this, visit the website: www.learner.org/courses/learningmath/geometry/session8/video.html and watch the video on “Similar Triangles and Radiation Therapy.” We will now show how the Law of Sines and the Law of Cosines can be used to develop ALL the main results about similarity. The versatility of these laws is quite remarkable. Recall that two tri- angles, ABC and DEF are called similar if they have the same shape, but not necessarily the same size. This means that one is a scaled version of the other. The formal definition of similarity between triangles ABC and DEF is that the angles of triangle ABC are congruent to those of triangle DEF, and that the sides opposite the corresponding congruent angles (that is, the corresponding sides) are proportional. When one writes that two triangles are similar, the order in which the letters are written reveals which angles are congruent. Thus, when we write that triangle ABC is similar to triangle DEF, it follows that ∡A = ∡D, ∡B = ∡E, and ∡C = ∡F. Saying that the sides of the triangles are proportional means that there is a number k such that a = kd, b = ke, and c = kf, where a, b, c and d, e, f are respectively, the corresponding sides of the triangles. See Figure 5.9. E B fd ca A b CD e F Figure 5.9 Another way of expressing that the sides of two triangles ABC and DEF are proportional is to write a ¼ b ¼ c ¼ k: d e f Here is our first theorem on similarity. Theorem 5.7 If the corresponding sides of two triangles are proportional, then the corresponding angles of the triangle are equal. Hence, the two triangles are similar. Proof. Suppose that our triangles are ABC and DEF as in Figure 5.9, and suppose that we are given that a = kd, b = ke, and c = kf. Using (5.8) we have that in triangle ABC cos C ¼ a2 þ b2 À c2: ð5:29Þ 2ab
186 Chapter 5 The Triangle: Its Study and Consequences In triangle DEF we have the similar result that cos F ¼ d2 þ e2 À f 2 ð5:30Þ 2de : But we know that a = kd, b = ke, and c = kf. Substituting these in equation (5.29) we have cos C ¼ ðkdÞ2 þ ðkeÞ2 À ðkf Þ2 2kdke ¼ k2ðd2 þ e2 À f 2Þ 2k2de ¼ d2 þ e2 À f 2 2de ¼ cos F ðFrom equation ð5:30ÞÞ: Since cos C = cos F, C = F. In a similar manner we can show that the other angles are equal. Thus, since the corresponding sides were in proportion and as a result of this, we showed the corresponding angles were equal, the triangles ABC and DEF are similar. & Now we prove the converse. Theorem 5.8 If the three angles of one triangle are equal to three angles of another triangle, then the corresponding sides of the triangles are in proportion. Hence, the two triangles are similar. Proof. Suppose that we have triangles ABC and DEF where angle ∡A = ∡D, ∡B = ∡E and ∡C = ∡F. (See the figure from the previous theorem.) Then by the Law of Sines, applied to triangle ABC we have a A ¼ b sin sin B which can be written as a ¼ sin A : ð5:31Þ b sin B Using the Law of Sines in triangle DEF we have in a similar manner that d ¼ sin D : ð5:32Þ e sin E But since ∡A = ∡D and ∡B = ∡E, we can substitute ∡A and ∡B for ∡D and ∡E in equation (5.32) and we get d ¼ sin A : ð5:33Þ e sin B Since the right sides of equations (5.31) and (5.33) are the same, the left sides are also, so we have that a ¼ de: b
5.4 Similarity 187 Therefore, ae = bd by cross multiplying, and dividing both sides by de we get that a ¼ be: d In the Student Learning Opportunities you will show in a similar manner that a ¼ fc: So, with your d work and ours, we get a ¼ b ¼ c d e f which says the sides are in proportion. & The following result is probably the most familiar to you. Corollary 5.9 (AA = AA) If two angles of one triangle are equal to two angles of another triangle, the triangles are similar. Proof. The third angles of the triangles will also be equal since the sum of the angles of a triangle is 180°. The result now follows from Theorem 5.8. & What Theorems (5.7) and (5.8) are saying is that to show that two triangles are similar, we need to show that either the corresponding sides are in proportion or that the corresponding angles are equal. One automatically implies the other. There is one other result about similar triangles that is useful, but less well known. Theorem 5.10 If two sides of one triangle are proportional to two sides of another triangle, and the angle between the proportional sides of these triangles is the same, then the two triangles are similar. Proof. Using Figure 5.9 we may suppose that the sides that are in proportion are b and e, and c and f, and that the angle A between sides b and c is equal to the angle D that is between e and f. Saying that the sides b, e, c, and f are in proportion means that b ¼ c ¼ k: ð5:34Þ e f To prove the triangles are similar, we will show that the third sides are also in proportion. That is, we will show a is also k. Once we have that all three sides are in proportion, we know by Theorem d 5.7 that the triangles are similar. Now, using the Law of Cosines in triangle ABC we have that a2 ¼ b2 þ c2 À 2bc cos A: ð5:35Þ Using the Law of Cosines in triangle DEF we have ð5:36Þ d2 ¼ e2 þ f 2 À 2ef cos D:
188 Chapter 5 The Triangle: Its Study and Consequences From equations (5.34) we have b = ke and c = kf, and we were given that ∡A = ∡D. Replacing b by ke, c by kf and A by D in equation (5.35) we get a2 ¼ ðkeÞ2 þ ðkf Þ2 À 2ðkeÞðkf Þ cos D ¼ k2ðe2 þ f 2 À 2ef cos DÞ ¼ k2d2 ðUsing equation ð5:36ÞÞ: This string of equalities shows that a2 = k2d2. Hence, a = kd. It follows that a is also k and using equa- d tion (5.34) we see that a ¼ b ¼ c ¼ k: d e f So, we have shown that all three sides are in proportion, and hence by Theorem 5.7, triangles ABC and DEF are similar. & Student Learning Opportunities 1 If triangle ABC is similar to triangle DEF, and the following facts about the sides are given, find the remaining sides and angles to the nearest degree. (a)* AB = 6, DE = 18, EF = 12, CA = 8 (b) BC = 12, EF = 6, DF = 18, DE = 8 2 (C) A student wants to know if in similar triangles corresponding altitudes are in the same ratio as corresponding sides. Are they? If so, how can you show it? 3 (C) A student wants to know if you are given two similar triangles, does it mean that any pair of their corresponding medians are in the same proportion as the sides? How do you respond? What is your explanation? 4 (C) A student wants to know if it is true that the ratio of areas of similar triangles is the same as the ratio of the corresponding sides. Is it? If not, what is true about the ratio of the areas of similar triangles and the ratio of corresponding sides and how can you show it? [Hint: Take two specific similar right triangles, like the 3-4-5 right triangle and the 6-8-10 right triangle, and answer these questions before answering the general question.] 5 Using Theorem 5.10, show that the length of the line segment connecting the midpoints of two sides of a triangle is 1 the length of the third side. 2 6 (C) Your students notice that when you draw the line segment connecting the midpoints of two sides of a triangle it appears to always be parallel to the third side. They want to know if this is always true, and if it is, how can it be proven. How do you respond and how do you prove it? [Hint: Show that you have a pair of corresponding angles equal.] 7 (C) You have encouraged your students to use some dynamic geometric software to make some conjectures about the shape of the quadrilateral created by connecting the midpoints of the adjacent sides of any quadrilateral. (See Figure 5.10.) They have been surprised to see that regardless of the size or shape of the exterior quadrilateral, it seems that the interior quad- rilateral they have created always looks like a parallelogram. They want to know if it really is,
5.4 Similarity 189 and if it is, how it can be proven. How would you prove it? [Hint: Use the results from Student Learning Opportunity 6 for your proof that the students are correct.] B AC D Figure 5.10 8 Finish the proof of Theorem 5.8 by proving a ¼ cf : d 9 (C) You have had your students use some dynamic geometric software to create a right triangle and draw an altitude to the hypotenuse. After dragging the points of the right triangle they have noticed that the two smaller triangles that are formed within the larger right triangle appear to always be similar to each other, and more surprisingly, seem to always be similar to the big triangle. They want to know if this is always true, and if it is, how can it be proven. How do you prove it? 10 Give another proof of the Pythagorean Theorem using similar triangles. [Hint: Refer to Á ÁFigure 5.11. Show that (AC)2 = AB AD and that (BC)2 = AB DB. Now add the equations. You might want to use the result of the previous Student Learning Opportunity.] C AD B Figure 5.11 11* Find the value of x in Figure 5.12 if DE is parallel to AC: B 8 x E D 2x x+3 A C Figure 5.12
190 Chapter 5 The Triangle: Its Study and Consequences 12 Two poles of heights 5 and 10 are separated by a distance of 20 feet. A wire is placed tautly from the top of each pole to the bottom of the other pole and they overlap at a point P shown in Figure 5.13. D A 10 P 5 BE C 20 Figure 5.13 (a)* How high above the ground is P? [Hint: Triangles ABC and PEC are similar, as are triangles BPE and BDC. Let BE = x and EC = 20 À x and PE = h. From the first set of similar triangles, 5 ¼ 20 h x : Find a similar relationship for the second pair of triangles and work from there.] 20 À (b) Do the same problem as in (a), only now assume that the distance between the poles is 100 feet instead of 20 feet. Show that the overlap of the wires is at the same height above the ground as it was in part (a). Does this surprise you? Why or why not? (c) Suppose the distance between the poles is d. Show that the distance h, that P is above ground does not depend on d. (d) Generalize the solution. Suppose that the poles are at heights a and b and that the distance between them is d. Show that h ¼ a ab b : þ 5.5 Sin(A + B) LAUNCH 1 Most students believe that sin(A + B) = sin(A) + sin(B). (a) Is this always true? If you said “yes,” then justify why. If you said “No,” then support your answer by giving a counterexample. (b) Is this ever true? If you said “Yes,” then give one example when it is true and state how often you think it is true. If you said “No” then justify why you believe it is never true.
5.5 Sin(A + B) 191 After having responded to the launch question, you are now most likely curious about the behaviors of the sine of the sum of two angles. You must be wondering if the trigonometric rela- tionships share the same distributive property that algebraic relationships have. Wouldn’t it be nice if they did? In this section we will pursue this question further to arrive at the true relation- ships in a manner that will most likely surprise you. We have used the concept of area to prove the Pythagorean Theorem and applied that theorem to prove the Law of Sines and the Law of Cosines, which yielded the main theorems about congru- ence and similarity. We now take the concept of area and use it to prove theorems in trigonometry. It is hard not to appreciate how powerful and fundamental the concept of area is. It is a common misconception among secondary school students that sin(A + B) = sin(A) + sin(B). This is not true, which is easy to see by just taking the counterexample A = 30° and B = 60°. Using pffiffiffi 12; 23, and sin 90° = 1, we see that a calculator, or the well-known values of sin 30 ¼ sin 60 = pffiffiffi 1 3 sin(A + B) = 1 but sin A + sin B ¼ 2 þ 2 % 1: 366: They are quite far apart. However, it is true that sin(A + B) = sin A cos B + cos A sin B, and you can verify it for the angles given earlier. Of course, a proof of this relationship is needed, which we will give now using areas. Our proof is only valid for triangles with interior altitudes. But the theorem is true in general and will follow from results we set forth in the trigonometry chapter. See Theorem 12.8. First we need a preliminary result which is commonly taught in secondary school. Theorem 5.11 The area of a triangle is 1 ab sin C where a and b are two sides of a triangle and C is the 2 included angle. Proof. Using Figure 5.14 with altitude h and base b. B ha AD C Figure 5.14 we have that the area of the triangle is 1 bh: ð5:37Þ 2 But from right triangle BDC; sin C ¼ opposite ¼ ha, so h = a sin C. Substituting for h in (5.37) we hypotenuse get that the area of a triangle is 1 ab sin C: & 2 In what follows we give a proof of the formula for the sine of the sum of two angles. This proof is valid for angles α and β where α + β < 180°, though it is true for all angles and we will prove it, in Chapter 12. (Ray Siegrest from SUNY Oneonta showed us this proof.)
192 Chapter 5 The Triangle: Its Study and Consequences Theorem 5.12 If α and β are angles, then sin(α + β) = sin α cos β + cos α sin β. Proof. We place the two angles next to each other and form a triangle ABC shown in Figure 5.15. We drop a perpendicular, AD, to the base BC: Suppose that AD = h. Figure 5.15 Now, the area of triangle ABC, by the previous theorem, is 1 bc sin ða þ bÞ, while the areas of tri- 2 angles ACD and ADB are 1 bh sin ðaÞ and 1 ch sin ðbÞ: Since the area of triangle ABC is the sum of the 2 2 areas of of triangles ACD and ADB, we have 1 bc sin ða þ bÞ ¼ 1 bh sin a þ 1 ch sin b: 2 2 2 Multiplying by 2 and dividing both sides by bc, we get sin ða þ bÞ ¼ h sin a þ h sin b: ð5:38Þ c b But from triangle ADB, h ¼ cos b ð5:39Þ c and from triangle ACD, h ¼ cos a: ð5:40Þ b Substituting equations (5.39) and (5.40) into equation (5.38) we get sin ða þ bÞ ¼ cos b sin a þ cos a sin b ¼ sin a cos b þ cos a sin b: & Notice how this theorem ties together algebraic concepts with the geometric concept of area, and the trigonometric concepts of sines and cosines. What a nice interplay of ideas! Later in this chapter, we will get this result in a different and quite unexpected manner. (See Ptolemy’s Theorem.) Theorem 5.12 is attributed to the Persian mathematician and astronomer Abul Wafa and dates back to the 10th century.
5.5 Sin(A + B) 193 Student Learning Opportunities 1 (C) One of your very clever students asks: “If the two adjacent sides of a triangle are 6 and 8 and the angle between them varies, different triangles will be formed and their areas will be different. What would the measure of the included angle have to be to make a triangle of largest area?” How would you respond and does this same angle work regardless of the lengths of the adjacent sides? [Hint: Use Theorem 5.11.] 2 Verify that sin(30 + 60) = sin 30 cos 60 + cos 30 sin 60, using the known values of these trigno- metric functions. 3* A rectangle ABCD has sides 3 and 6. If diagonal AC is split into three equal parts by points E and F, find the area of triangle BEF. 4 We have shown that the area of a triangle is 21ab sin C: We can also compute the area two other ways: 12bc sin A and 21ac sin B: Of course the area that we get is the same no matter which formula we use. Show that from this observation, we can get another proof of the Law of Sines. 5* (C) Despite learning the formula for sin(A + B) several of your students still maintain that sin2θ = sin θ + sin θ. How can you prove to them that they are incorrect and how would you illustrate graphically that in fact sin 2θ = 2 sin θ cos θ? pffiffiffi 22, and that the exact values of 6* Using the fact that the exact values of sin 45° and cos 45° are pffiffiffi 1 23, find the exact value of sin 75°. sin 30 and cos 30 are respectively, 2 and 7 Using Figure 5.16, do a proof similar to the one given in Theorem 5.12, using areas, to find a formula for sin(α À β) when β < α < 180°. In this figure, the measure of angle ABC is α, and the measure of angle DBC is β. Figure 5.16
194 Chapter 5 The Triangle: Its Study and Consequences 5.6 The Circle Revisited LAUNCH In a circle with center O, draw two adjacent central angles of 120 degrees. Let the intersection of the sides of these angles with the circle be A, B, and C as shown in Figure 5.17: B 120° 120° AC Figure 5.17 Connect A to B, B to C, and C to A, and you should now have an equilateral triangle. Pick any point P on the circle and draw line segments PA, PB, and PC. Do you notice any relationship between the lengths of the two shorter segments and the length of the longer segment? What do you notice? Pick another point P0 and do the same thing. Do you notice the same relationship? What is it? Do you think this will always happen? Why or why not? After having completed the launch, you are probably beginning to realize that the circle is a most fascinating figure, especially when you begin to inscribe other geometric figures within it. In this section, you will learn more about the most interesting relationships that exist within a circle. You will revisit the launch problem at the end of the section in the Student Learning Oppor- tunities, after you become familiar with Ptolemy’s Theorem. 5.6.1 Inscribed and Central Angles We hope you have appreciated seeing how we have used the Law of Sines and the Law of Cosines to develop all the congruence theorems, similarity theorems, and related laws of trigonometry that are part of the secondary school curriculum. Recall that the Law of Cosines essentially depended on the Pythagorean Theorem which, in turn, depended on the concept of area. Thus, it seems that area is the driving concept in the secondary school curriculum. This is why our chapter on areas preceded this one. Let us now investigate the circle and see what other relationships we can prove. Other than congruence and similarity, the main theorems in a geometry course are those con- cerning circles, their chords, their tangents, and their secants. We now prove some of the main the- orems about these, and guide you through many of the others in the Student Learning Opportunities. These tasks will not only review the theorems, but show, yet again, how all the
5.6 The Circle Revisited 195 concepts connect. After this, we will continue to investigate further applications of the Pythago- rean Theorem and the concept of area. To begin, we need to recall a fact from geometry. A central angle is one whose vertex is at the center of the circle. Thus, in the picture, θ is a central angle and arc AB (denoted by A_B) is the arc subtended by the central angle or the intercepted arc. We define the measure of the arc sub- tended by a central angle of θ, to be θ also. That is, each central angle has the same measure as the arc subtended by it and vice versa. This is a definition. (See Figure 5.18.) A Degree measure of the arc AB is the same as the central angle. q B Figure 5.18 When you think about this definition, it makes sense. We know that a complete rotation is 360 degrees. So, if we drew adjacent central angles each of 1 degree, we would need 360 of them to fill the circle. But this divides the circle into 360 parts. Thus, the number of central angles each of degree 1, and the number of parts of the circle are both 360. Therefore, it seems reasonable that each arc associated with a 1 degree central angle should be called a 1 degree arc. It follows from this that any arc will have the same number of degrees as its central angle. Here is our first theorem. Recall that an inscribed angle is one whose vertex is on the circle and whose sides are chords of the circle, as is shown in Figure 5.19. Arc AC is called the intercepted arc. A B C Figure 5.19 Angle ABC is an inscribed angle. Theorem 5.13 An angle inscribed in a circle is measured by 1 of its intercepted arc. 2 Proof. We give one half of the proof leaving the other half to you. We begin with the central angle AOB. By definition, this has the same degree measure as _ . Pick a point P on the circle. We suppose AB that our picture is as given in Figure 5.20. (This simplifies the first part of the proof. There is another picture as you will see.)
196 Chapter 5 The Triangle: Its Study and Consequences A x x Oα Py β y B Figure 5.20 Because the radii of a circle are equal, triangles AOP and BOP are isosceles, and their base angles are equal, as indicated in the diagram. Since the exterior angle of a triangle is the sum of the remote interior angles (Chapter 1, Section 2, Student Learning Opportunity 1) we have that α = 2x and in the same way, β = 2y. Thus, α + β = 2x + 2y. This means that x þ y ¼ a þ b: ð5:41Þ 2 But α + β is the measure of arc AB and x + y is the measure of the inscribed angle P. Thus, equation (5.41) says that ∡P ¼ 1 ∡AOB: ð5:42Þ 2 You might think the proof is complete. It isn’t. There is another possible picture (see Figure 5.21). In the Student Learning Opportunities, you will be asked to prove the theorem for this case. & PA O B Figure 5.21 Corollary 5.14 If an inscribed angle and a central angle intercept the same arc, the central angle is twice the inscribed angle. Proof. In Figure 5.21, ∡P is measured by 1 A_B, while angle O is measured by A_B : So, the measure of 2 the central angle is twice the measure of the inscribed angle. & There is a very surprising corollary of Theorem 5.13—an extended version of the Law of Sines! Corollary 5.15 Given any triangle ABC; a ¼ b ¼ c C ¼ d where d is the diameter of the sin A sin B sin circumscribed circle.
5.6 The Circle Revisited 197 Proof. According to the previous corollary, the central angle intercepting the same arc as an angle of the triangle whose vertex lies on the circle will have twice the measure of the inscribed angle. Look at Figure 5.22 which shows an acute triangle and its circumscribed circle. C a D B r aa F A Figure 5.22 If the triangle is acute, then the center of the circle is always inside the triangle and we can draw altitude DF to isosceles triangle ADB and it will bisect angle ADB as well as the base. Now, in triangle ADF, sin a ¼ AF ¼ AF where d is the diameter of the circle. Inverting the d and multiplying, we get r d 2 2 that sin a ¼ 2AF ¼ AdB: Rewriting this as d AB ¼ d ð5:43Þ sin a and realizing that sin α = sin C and AB is side c in triangle ABC, equation (5.43) becomes c C ¼ d: ð5:44Þ sin That is, the ratio of c to sin C is the diameter. Since, there was nothing special about angle C, a similar proof shows that a and b are both d. So, a ¼ b ¼ c C ¼ d and this gives sin A sin B sin A sin B sin us not only the law of sines, but tells us exactly what the common ratios are, the diameter of the circumscribed circle! & There is yet another corollary of this which may seem obscure now, but will be put to good use later in section 5.6.2. Corollary 5.16 In a circle with diameter 1, if we have an inscribed angle of measure α, intercepting arc AB, then the length of the chord joining the points A and B has length sin α. That is, sin α = AB. Proof. We simply let d = 1 in equation (5.43) and the result follows immediately by cross multiply- ing. &
198 Chapter 5 The Triangle: Its Study and Consequences 5.6.2 Secants and Tangents In this section we deal with some of the main theorems concerning tangents and secants. Recall that a secant line to a circle is a line drawn from an external point which cuts through the circle and stops at the other side. We have drawn a picture of a secant in Figure 5.23. AP is the secant. AB P Figure 5.23 Theorem 5.17 If two secants, with lengths s1 and s2 are drawn to a circle from an external point, and the parts of them which are external to the circle are e1 and e2, then s1e1 = s2e2. Proof. Suppose that the secants hit the circle at points A, B, C and D as shown in Figure 5.24. AB P D C Figure 5.24 We are calling AP = s1, CP = s2, BP = e1 and DP = e2. Draw CB and AD as shown in Figure 5.25. AB P D C Figure 5.25 Then we have that ∡A = ∡C since both angles A and C are inscribed angles and both are half the degree measure of arc BD. Of course, angle P is common to both triangles ADP and CBP. Thus, by Corollary 5.9 (AA = AA), triangles ADP and CBP are similar. It follows that the corresponding sides are in proportion. Thus, AP ¼ CP DP BP Cross multiplying, we get AP Á BP ¼ CP Á DP
5.6 The Circle Revisited 199 which says nothing more than s1e1 ¼ s2e2: & Theorem 5.18 If a tangent line with length t and a secant line with length s and external segment e are both drawn to a circle from an external point, then t2 = se. Proof. Although we could give a purely geometric proof of this, we prefer instead to show you a different approach which ties together concepts you learned in calculus with geometry. Imagine a group of additional secants with lengths s1, s2, and so on being drawn to the circle, and that these secants approach the tangent line. See Figure 5.26. The bold parts of the diagram are the original secant line with length s and external segment of length e, and the original tangent e3 e2 t P e1 e s Figure 5.26 whose length is t. Now we know from the previous theorem that se = s1e1 = s2e2 and so on. This says that the sequence of numbers {snen} is constant and every term is equal to the constant se. Now we know that the limit of a constant sequence of numbers is the constant. That is lim snen ¼ se: ð5:45Þ n!1 Furthermore, since the secant lines approach the tangent line, we see that the lengths of the secant lines approach the length of the tangent, and the lengths of the external segments of the secant lines also approach the length t of the tangent line. In symbols: lim sn ¼ t and lim en ¼ t: Now, n!1 n!1 using (5.45) we have se ¼ lim snen n!1 ¼ lim sn Á lim en n!1 n!1 ¼tÁt ¼ t2 and we are done. & 5.6.3 Ptolemy’s Theorem One result relating to circles that is done in some secondary school courses is Ptolemy’s Theorem, which we present next. Ptolemy lived in the second century and was one of Greece’s most influen- tial astronomers. He propounded the theory that the Earth was the center of the solar system,
200 Chapter 5 The Triangle: Its Study and Consequences which was believed until about 1543, when Copernicus showed otherwise. Ptolemy is credited with making the first tables of sines, cosines, and tangents and applying these to problems in astronomy. The proof of Ptolemy’s Theorem uses similar triangles. Its consequences are quite unexpected and powerful and relate directly to the secondary school curriculum. Here is Ptolemy’s Theorem. Theorem 5.19 Suppose that quadrilateral PQRS is inscribed in a circle as shown in Figure 5.27: Q R T P S Figure 5.27 If we multiply the lengths of the opposite sides of the quadrilateral and sum the results, we get the product of the diagonals. That is, QR Á PS þ PQ Á SR ¼ QS Á PR: Proof. Here is a plan for the proof. We will pick a point T on diagonal PR such that ∡RQT = ∡SQP. We will then show that triangle RQT is similar to SQP, and triangle RQS is similar to triangle TQP and thereby establish proportions that will lead to our theorem. So, begin by picking a point T on diagonal PR such that ∡RQT = ∡SQP. Thus, by construction, we have one angle of triangle RQT equal to one angle of triangle PQS. Since ∡QRP and ∡QSP subtend the same arc, PQ, it follows that ∡QRP = ∡QSP. This gives us a second pair of equal angles in triangles RQT and SQP. Since two angles of triangle RQT are equal to two angles of triangle SQP, triangles RQT and SQP are similar by Corollary 5.9 (AA = AA). It follows that QR ¼ QPSS: Cross multiplying, we get TR QS Á TR ¼ QR Á PS: ð5:46Þ Now we show that triangle RQS is similar to triangle TQP. We know that, by construction ∡RQT = ∡PQS. If we add ∡SQT to both of these angles, we get that ∡PQT = ∡RQS, providing us with one pair of equal angles in triangles RQS and TQP. Also, since ∡QPR and ∡QSR both subtend arc QR, we have that ∡QPR = ∡QSR. So now triangles RQS and TQP are similar by AA = AA. So their sides are in proportion. That is, QP ¼ QSRS: Cross multiplying, we have PT PT Á QS ¼ QP Á SR: ð5:47Þ Now, if we add equations (5.46) and (5.47) we get QS Á TR þ PT Á QS ¼ QR Á PS þ QP Á SR
5.6 The Circle Revisited 201 and if we factor out QS we get QS Á ðPT þ TRÞ ¼ QR Á PS þ PQ Á SR and since PT + TR = PR, this simplifies to QS Á PR ¼ QR Á PS þ PQ Á SR: & The tendency is to say “So what?” Let’s see how you feel after the next example. Example 5.20 Show using Ptolemy’s Theorem that sin(α + β) = sin α cos β + cos α sin β. Solution: Work in a circle of diameter 1 shown in Figure 5.28. Q R P ab S Figure 5.28 Here we have drawn a quadrilateral, one of whose sides, QS, is a diameter. Thus, QS = 1. Now since QPS is inscribed in a semicircle, it is a right angle. (It is measured by half its intercepted arc which is 180 degrees.) So cos a ¼ adjacent ¼ PS ¼ PS ¼ PS ð5:48Þ hypotenuse QS 1 Similarly, cos b ¼ SR; ð5:49Þ sin a ¼ PQ , and ð5:50Þ sin b ¼ QR: ð5:51Þ (Verify!) Since the diameter of the circle is 1, we have by Corollary (5.16) that ð5:52Þ PR ¼ sin ða þ bÞ: Now we are ready to proceed. By Ptolemy’s Theorem, ð5:53Þ PQ Á SR þ PS Á QR ¼ PR Á QS: Now we know that QS = 1, and substituting the values for PQ, SR, PS, QR and PR obtained in (5.49)– (5.52), in (5.53) we get sin a Á cos b þ cos a Á sin b ¼ sin ða þ bÞ Á 1
202 Chapter 5 The Triangle: Its Study and Consequences which was our goal! So we have seen yet another surprising and corroborating proof of the formula for sin(α + β) when α + β < 180°. We can get other trigonometric identities from Ptolemy’s Theorem, some of which we leave for the Student Learning Opportunities. Student Learning Opportunities 1* Triangle ABC is inscribed in a circle. The smaller arcs, A_B ; B_C , and C_A are respectively 2x, 3x, and 5x degrees. What are the angles of the triangle? 2 (C) Your students want you to explain why an angle inscribed in a semicircle is a right angle. Show how you would explain this by using theorems from this chapter. 3 (C) You have encouraged your students to use some dynamic geometric software to examine what happens if you create an isosceles triangle and draw an altitude from the vertex angle to the base. They have noticed that regardless of the size and shape of the isosceles triangle they make, the altitude seems to always bisect the vertex angle as well as the base. Your students want to know if this is always true and if so, how can you prove it. What is your response and how do you prove it? [Note that this relationship was used in Theorem 5.15.] 4 (C) Through the use of dynamic geometric software, your students have become convinced that every triangle ABC can be inscribed in a circle. (That circle is called the circumscribed circle.) They ask you how you can construct the circumscribed circle for any triangle. How do you do it? [Hint: Proceed as follows. Draw the perpendicular bisectors of two sides, say AB and BC of the triangle. They will intersect at some point P. Prove that AP = PB using congru- ent triangles. Now, in a similar manner PB = PC. Finish it.] 5 (C) Through the use of dynamic geometric software, your students have noticed that whenever they inscribe a right triangle in a circle, the hypotenuse always seems to be the diameter of the circle. They want to know if this is always the case, and if so, how to prove it. How will you do it? They then want to know why the median to the hypotenuse of a right triangle has length half the hypotenuse. How can you show it based on what you have already done? 6 Use Ptolemy’s Theorem to show that, if a rectangle with legs a and b and diagonal c is inscribed in a circle, that a2 + b2 = c2. Thus, we have yet another proof of the Pythagorean Theorem. 7 Use Ptolemy’s Theorem to prove that sin(α À β) = sin α cos β À cos α sin β. Use Figure 5.29 where we take the diameter AD = 1 and use Corollary 5.16. [Hint: AB = cos α, BC = sin(α À β) by Corollary 5.14.] C B ab D A 1 Figure 5.29
5.6 The Circle Revisited 203 8 Prove the following theorem which answers the launch question: If an equilateral triangle ABC is inscribed in a circle, and P is any point on the circle, then the shorter two of the three segments, PA, PB, and PC adds up to the third. [Hint: Call the side of the equilateral triangle s, and connect P to the three vertices of the triangle to form a quadrilateral. Then use Ptolemy’s Theorem.] Note: We have been using Ptolemy’s Theorem to derive geometric and trigonometric results. In fact, there is a very famous and exceedingly useful law in optics that says that, when light traveling with velocity v1 in a medium, say air, enters a medium, say water, entering at an angle of θ1 relative to the vertical, the light is refracted, that is, bent, and travels at an angle θ2 to the vertical. Furthermore, its velocity in the medium it enters changes to v2. The relation- ship is known as Snell’s Law and says that sin y1 ¼ v1: sin y2 v2 A big surprise: Ptolemy’s Theorem can be used to show that Snell’s Law as well as Fermat’s prin- ciple—which states that light travels in such a way so as to minimize its travel time from point A to point B lead to the same geometry of refraction. 9* Find the area of the pentagon shown in Figure 5.30. Angle A is a right angle. C B 10 10 D 12 10 A 16 E Figure 5.30 10 (C) You have encouraged your students to use some dynamic geometric software to make a conjecture regarding the angle a tangent line drawn to a circle makes with a radius drawn to that tangent line at the point of tangency. They have all discovered that these two lines always seem to be perpendicular to one another. They want to know why this happens. How can you help them to discover this? [Hint: Draw the radius of the circle to the point of tangency. So, its length is r, the radius of the circle. Then draw a line to any other point P on the tangent line and show that its length is more than r. How does this show it?] 11 (C) Your students have been using dynamic geometric software to investigate the relationship of two tangent lines drawn to a circle from a common external point. They have made the con- jecture that the two tangent lines have the same length and now they need some guidance on how to prove it. How can you help them? 12 (C) You have encouraged your students to use some dynamic geometric software to make some conjectures. They have observed over and over that, if a line passing through the center of a circle is drawn perpendicular to a chord, AB, it bisects the chord. They ask you for a proof. How do you prove it? (Drawing radii to the endpoints of the chord will help.)
204 Chapter 5 The Triangle: Its Study and Consequences 13* Suppose we have two circles with the same center and that the area of the shaded region between them is 25π. A chord is drawn in the larger circle which is tangent to the smaller circle. (See Figure 5.31.) B A Figure 5.31 What is the length of the chord? [Hint: Draw a radius to the point of tangency and another to B.] 14 (C) Using dynamic geometric software, your students have noticed that if they draw two chords that are the same distance from the center of a circle, they always have the same length. They want to know how to prove it. How can you prove it? [Hint: Draw the lines that give the distance from the center. Also draw radii to one endpoint of each chord.] 15 (C) Using dynamic geometric software, your students have noticed that if in one circle they draw two chords that have the same length, then they are the same distance from the center of the circle. They want to know how to prove it. Show them. 16 (C) Using dynamic geometric software, your students have noticed that, if in one circle they draw chords of equal length, they always seem to subtend minor arcs that have the same angle measure. They want to know if this is always the case, and if it is, how can it be proven. What do you say and how can you prove it? 17 Show that, if a circle of radius r is inscribed in a triangle with perimeter P and area A, then P ¼ A 2r : [Hint: Draw radii to the sides and connect the center of the circle to the vertices. Then sum the areas of the three triangles formed. Each has height r.] 18 (C) Your students have learned that if two chords intersect within a circle, then the product of the segments of one chord is equal to the product of the segments of the other chord. That is, in the following diagram, ab = cd. Some of your more curious students want to know how that is proven. How do you do it? [Hint: Draw the dotted lines as shown in Figure 5.32, and try to get similar triangles.] P Q c a T b d RS Figure 5.32
5.7 Is the Cosine Well Defined? 205 19 Prove Theorem 5.13 for Figure 5.21. 20 Using Figure 5.33, find the missing piece of information in parts (a)–(d). B E C D A Figure 5.33 (a)* BE = 12, AE = 9, DE = 1, CE = ? (b)* BC = 5, CE = 4, DE = 9, AD = ? (c)* BC = 6, AD = 4, DE = 2, CE = ? (d) CE = 4, DE = 3, AD = 4, BC = ? 5.7 Is the Cosine Well Defined? LAUNCH Historically, the cosine of an acute angle A in a right triangle was defined as the ratio of the length of the side adjacent to A to the length of the hypotenuse. Unless this ratio is the same for all right tri- angles having an acute angle equal to A, this definition is unusable. But, how can we be sure that the cosine of an angle is the same regardless of the right triangle in which it occurs? We hope that, after thinking about this, you realized that similar triangles are used to show this. The usual argument given is: “If A is an angle in two right triangles, then the triangles must be similar, since they both have angle A and a right angle. Since they are similar, their sides are in pro- portion. That means that the ratio of the adjacent side to the hypotenuse is the same in both tri- angles.” That is, the cosine of A is independent of the right triangle in which it resides. Our goal in this chapter was to prove the theorems about similarity using the fact that sines and cosines are well defined, and we cannot use theorems about similar triangles to assume that the sine and cosine are well defined, or else we would be engaging in what is known as circular reasoning (using theorems about similar triangles to prove the same theorems about similar triangles). So, to avoid this circular reasoning, we will now give an independent proof of the fact that the sine of an angle is independent of the triangle in which it resides. Surprisingly, we can use areas of triangles to prove this, which we will now do. But first we need the following preliminary theorem. Theorem 5.21 Suppose that, in a given triangle ABC, a line is drawn from B to AC intersecting AC at D. Then, the ratio of the areas of triangle ABD to triangle CDB is the same as the ratio of AD to DC.
206 Chapter 5 The Triangle: Its Study and Consequences Proof. We refer to Figure 5.34. B h AD C Figure 5.34 We see that both triangles ABD and DBC have the same height h. Thus, Area of triangle ABD ¼ 1 AD Á h ¼ AD : & Area of triangle DBC 2 DC Á h DC 1 2 Corollary 5.22 Given a right triangle ABC with right angle at C, draw a line DE parallel to BC where AC ¼ AADE : That is, cos A is the same D is any point on AC and E is where this line intersects AB. Then AB whether we take the ratio of the opposite side to hypotenuse in right triangle AED or right triangle ABC. Proof. We begin with our picture, Figure 5.35. B E A DC Figure 5.35 First draw EC, dividing the triangle into triangles I and II as shown in Figure 5.36: B E I II A DC Figure 5.36
5.7 Is the Cosine Well Defined? 207 Now, from Theorem 5.21, applied to triangle AEC we have Area of I ¼ DADC: ð5:54Þ Area of II Now, draw DB, yielding Figure 5.37. E B III C I AD Figure 5.37 Again, by Theorem 5.21, only applying it to triangle ABD, we have Area of I ¼ AEBE: ð5:55Þ Area of III Finally, we note that both triangles II and III have base ED and height DC (since ED is parallel to BC and hence are the same distance from each other everywhere). Thus, the areas of II and III are the same. Replacing Area III by Area II in (5.55) we get that Area of I ¼ AEBE: ð5:56Þ Area of II Comparing equations (5.54) and (5.56) we see that AD ¼ AEBE: ð5:57Þ DC We will now add the number 1 to both sides of equation (5.57). (You will soon see what this accom- plishes.) We get AD þ1 ¼ AE þ 1: DC EB Combining each side into a single fraction we get AD þ DC ¼ AEEþBEB: ð5:58Þ DC Dividing equation (5.58) by equation (5.57) we get that AD þ DC AE þ EB DC ¼ EB AD AE DC EB which simplifies to AD þ DC ¼ AE þ EB AD AE
208 Chapter 5 The Triangle: Its Study and Consequences and since AD + DC = AC and AE + EB = AB, this is just AC ¼ AABE: ð5:59Þ AD From this proportion it follows that AC ¼ AD ð5:60Þ AB AE and we are done. & The importance of this theorem cannot be underestimated. It says that, if A is any angle, cos A is unique without the assumption of similar triangles. So, if we have two right triangles, ABC and AED, each containing an acute angle A, then we can just overlap them together so that we get Figure 5.38 as shown: B E A DC Figure 5.38 and then from equation (5.60) we see that the cosine of A is independent of the triangle we are using. We can use a similar proof to show that sin A is independent of the triangle in which it resides, or we can prove it using trigonometric identities. (See Student Learning Opportunity 1.) Now that we know that sin A and cos A are well defined, and don’t change depending on which triangle we are working in, we can use these facts freely. There is one other result we need for our work to be complete. Theorem 5.23 As an acute angle A in a right triangle increases, its cosine decreases. As angle A increases, so does its sine. Proof. Since cos A is independent of the triangle in which it resides, we will consider only triangles with hypotenuse = 1. Such a triangle will look as shown in Figure 5.39. B 1 A xC Figure 5.39
5.8 Ceva’s Theorem 209 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi From the Pythagorean Theorem, x2 + (BC)2 = 1. It follows that x ¼ 1 À ðBCÞ2: From this relation- ship we see that, as BC increases, x decreases. Now, if A increases, BC also increases and hence x decreases. But from the triangle, x = cos A. So as angle A gets bigger, cos A gets smaller. The proof of the second part is similar. & We used the following theorem throughout this chapter (see for example Theorem 5.2). Now it can be justified. Corollary 5.24 If A and B are acute angles and if cos A = cos B, then ∡A = ∡B. Proof. From the theorem, as an angle increases, its cosine decreases. Hence, it is not possible for two different acute angles to have the same cosine. Thus, it must be that ∡A = ∡B. & Student Learning Opportunities 1* Draw a right triangle ABC with right angle C. Starting with the Pythagorean Theorem and dividing both sides by c2, show that sin2 A + cos2 A = 1. From this, show that sin A ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 À cos 2A: (We don’t use the ± square root since the cosine of an angle in a triangle can’t be negative. It is the ratio of the lengths of two sides.) Use this to show that sin A is indepen- dent of the triangle in which it resides. 2 (C) Your students have been investigating triangles using dynamic geometric software. Under your direction they have noticed that if they create any triangle ABC and they draw a line par- allel to AC, intersecting AB at D and BC at E, then AD ¼ CEBE: They want to know how to prove DB this. How do you respond? 3* Show that if A and B are acute angles, and if sin A = sin B, it follows that ∡A = ∡B. 4 In the proof of Corollary 5.22 we said that equation (5.60) follows from equation (5.59). Show it. 5.8 Ceva’s Theorem LAUNCH 1 On a piece of standard loose-leaf paper, draw a triangle at the top half of the paper. Using a ruler, or by folding the segments, locate the midpoints of each segment and then connect these midpoints to the opposite vertices. You should have just drawn three medians. What do you notice? Do the medians intersect? Do they all intersect at the same point? Do you think this will this always happen? 2 On the bottom half of your paper, draw an entirely different type of triangle. As before, draw the three medians. Did the same thing happen? Do you think this will always happen? Explain.
210 Chapter 5 The Triangle: Its Study and Consequences Now that you have completed the launch question, we hope you are marveling at the mysteries of the triangle. The truth is that there are many interesting results related to the triangle that you will be reading more about in this section. For example, in addition to the medians meeting at a point, the three altitudes meet at a point, and the three angle bisectors meet at a point, although the points at which they meet are usually different. All of these theorems, as well as many others, follow from one remarkable result called Ceva’s Theorem (after the mathematician Giovanni Ceva (1647–1674) who we know little about, except that he was a professor of mathematics in Mantua, Italy, and published one of the first works in mathematical economics). We begin with a definition. A cevian in a triangle is a line drawn from a vertex to the opposite side of a triangle (or its extension). Thus, altitudes, medians and angle bisectors are all cevians. In this section we will once again use areas of triangles in a completely unexpected way. But first we need an interesting lemma about fractions. Lemma 5.25 If a ¼ c then a ¼ c ¼ a À c : Thus, if we subtract numerators and denominators of two b d, b d b À d equal fractions, we get an equivalent fraction. To illustrate: 2 ¼ 6 ¼ 2À6 which is true! 8 24 8 À 24 Here is the proof: Proof. Let t be the common value of the fractions a and c : So, a ¼ t and c ¼ t: Multiplying these b d b d a À c equations by b and d respectively, we get a = bt and c = dt. Using these values in the fraction b À d we get a À c ¼ bt À dt = tðb À dÞ ¼ t: Thus, a À c has the same value, t, as the other fractions have. So all b À d b À d bÀd b À d three fractions are equivalent. & Theorem 5.26 (Ceva’s Theorem): In triangle ABC, if AE; BF, and CD are cevians that meet at G inside the triangle, then BD Á AF Á CE ¼ 1. (See Figure 5.40.) DA FC EB B DE G AF C Figure 5.40 Proof. Before getting into the heart of proof, we just remind you of Theorem 5.21 which says that if two triangles have the same height, then the ratio of their areas is the same as the ratio of their bases. Now, since triangles ABF and CBF have the same height, area ðABFÞ ¼ AF ð5:61Þ area ðCBFÞ FC Similarly, since triangles AGF and FGC have the same height, area ðAGFÞ ¼ AF , ð5:62Þ area ðCGFÞ FC
5.8 Ceva’s Theorem 211 From equations (5.61) and (5.62), area ðABFÞ ¼ area ðAGFÞ ¼ AF ð5:63Þ area ðCBFÞ area ðCGFÞ FC and so by the previous lemma, with a, b, c, and d replaced by the appropriate numerators and denominators of equations (5.63), we have area ðABFÞ À area ðAGFÞ ¼ AF : ð5:64Þ area ðCBFÞ À area ðCGFÞ FC But area (ABF) À area (AGF) = area(AGB) and area (CBF) À area (CGF) = area(CGB). (Look at the figure to confirm!) Substituting into equation (5.64) we have areaðAGBÞ ¼ AF ð5:65Þ areaðCGBÞ FC We now use the cevian AE in a similar manner to get that CE ¼ areaðAGCÞ ð5:66Þ EB areaðAGBÞ And then again use cevian CD in a similar manner to get BD ¼ areaðCGBÞ ð5:67Þ DA areaðAGCÞ Using equations (5.65), (5.66), and (5.67) and multiplying we get AF Á CE Á BD ¼ areaðAGBÞ Á areaðAGCÞ Á areaðCGBÞ ¼ 1 ½Dividing common areas: & FC EB DA areaðCGBÞ areaðAGBÞ areaðAGCÞ The converse of Theorem 5.26 is also true. Theorem 5.27 If in triangle ABC we have three cevians, AE; BF, and CD, and AF Á CE Á BD ¼ 1, then the FC EB DA cevians AE; BF, and CD meet at a point. Proof. Suppose that AE and DC meet at G. Draw BG and let it intersect AC at F0 as shown in Figure 5.41. B DE G A F F’ C Figure 5.41
212 Chapter 5 The Triangle: Its Study and Consequences Then by Ceva’s Theorem, AF0 Á CE Á BD ¼ 1 ð5:68Þ F0 C EB DA But we are given that AF Á CE Á BD ¼ 1: ð5:69Þ FC EB DA From equations (5.68) and (5.69) we have ð5:70Þ AF0 CE BD AF CE BD F0 C Á BE Á DA ¼ FC Á BE Á DA : Dividing both sides of (5.70) by CE Á DBDA, we get BE AF0 AF F0 C ¼ FC From which it follows that F = F 0. Thus, AF0 really is AF and the three cevians go through the same point, G. & Student Learning Opportunities 1* (C) Using their dynamic geometric software your students have been investigating what happens in a triangle when they construct medians from all three vertices. No matter how they drag their figures and change the sizes and shapes of their triangles, it always seems to be that the medians meet at one point and that the six triangles formed by the medians have the same area (which they have programmed their software to calculate). They ask you if these relationships can be proven, and if so, how it is done. Use Ceva’s Theorem to prove this. 2 Prove that if a circle is inscribed in a triangle (see Figure 5.42), B DE A C F Figure 5.42 then the cevians drawn from each vertex to the points where the circle is tangent meet in a point. (You will need to use the fact that the tangents drawn to a circle from an external point are equal. That is, BD = BE, and so on.) This point where they meet is known as the Ger- gonne point and is usually not the center of the circle.
5.8 Ceva’s Theorem 213 3* We only proved Ceva’s Theorem for the case where the cevians met inside of the triangle. But the cevians can meet outside the triangle. Consider Figure 5.43 where we begin with triangle ABC and draw cevians AE; CD, and BF . Go through the proof of Theorem 5.27 line by line to see if the proof of the theorem works in this case. D B G E AC F Figure 5.43 4 (C) Your students have been using their dynamic geometric software to explore what happens when you create a triangle and draw the angle bisectors from each of the three vertices. They notice that the three angle bisectors always meet at a point. They ask you how to prove that this will always happen. How can you prove it to your students, using Ceva’s Theorem? [Hint: Student Learning Opportunity 7 from Section 5.3 may help.] 5 (C) Your students have continued their exploration of triangles using their dynamic geometric software and have now drawn a triangle and constructed the altitudes from all three of the sides. Much to their surprise, they notice that no matter the size or shape of their triangle, the altitudes meet at one point. They are eager to know if this always happens and why. How does Ceva’s Theorem help you to prove it? [Hint: In Figure 5.44 AD = AB cos(ffBAC). Get similar relationships for the other segments.] B FE AD C Figure 5.44 6* In the proof of the converse of Ceva’s Theorem we made the statement that if AF 0 ¼ AF F0C FC then F = F 0. Verify that this is true.
214 Chapter 5 The Triangle: Its Study and Consequences 5.9 Pythagorean Triples LAUNCH 1 Pick any two positive integers m and n where m is greater than n. Now compute the values for a, b, and c where a = m2 À n2, b = 2mn, and c = m2 + n2. Examine your values for a, b, and c. Specifically, check if a2 + b2 = c2. What does this mean? What did you just find? 2 Pick two different positive integers m and n where m is greater than n. Follow the directions again. Did the same relationship exist between a, b, and c? 3 Do you think this will always work? What does this mean? Now that you have completed this launch question, you may be thinking that you have found a way to generate three numbers that can serve as sides of a right triangle. Pretty cool, isn’t it? We will investigate this further in this section. Thus far, we have demonstrated how the Pythagorean Theorem and its consequences can be used to develop some very important relationships. In this section we wish to study the Pythago- rean Theorem a bit further. As you know, there are many “special” right triangles that are included in the secondary school curriculum. For example, there is the 3–4–5 right triangle, the 5–12–13 right triangle, and so on. Such sets of 3 positive integers that can serve as the sides of the same right triangle are called Pythagorean triples. You may be thinking that the method you used in the launch question can be used to generate all Pythagorean triples. Surprisingly, that is true. In this section, we will show how this can be done, and in the process we will connect the material we studied in Chapter 2 on divisibility with geom- etry and algebra. The connections alone make the journey worthwhile. Here is our first theorem: Theorem 5.28 Suppose we have a triangle ABC with sides a, b, and c, and suppose that there are pos- itive integers m and n such that a = m2 À n2, b = 2mn, and c = m2 + n2. Then, the triangle ABC will automatically be a right triangle. Proof. All we have to do is show that a2 + b2 = c2. Since a = m2 À n2, ð5:71Þ a2 ¼ m4 À 2m2n2 þ n4 ð5:72Þ ð5:73Þ as you can easily verify by multiplying a by itself. Since b = 2mn, b2 ¼ 4m2n2: And, since c = m2 + n2, c2 ¼ m4 þ 2m2n2 þ n4:
5.9 Pythagorean Triples 215 However, notice that if you add equations (5.71) and (5.72), you get equation (5.73). Thus, a2 + b2 = c2, and by Theorem 4.6 from the previous chapter, the triangle ABC is a right triangle regardless of the values of m and n. & What is surprising is that the converse of Theorem 5.28 is true; namely, if we start with a right triangle with legs a and b, and hypotenuse c, where a, b, and c have no common factors (other than 1), then there must be positive integers m and n such that a = m2 À n2, b = 2mn, and c = m2 + n2. This result will be the main theorem of this section. We reiterate, we are assuming from the outset that a, b, and c have no common factors. What this means is that the greatest common divisor of a, b, and c is 1. We first observe that, if a number is squared, then all its prime factors are raised to even powers. Furthermore, if all the prime factors of a number (when it is factored completely into primes) are raised to even powers, then the number is a perfect square. Let us give a numerical example to demonstrate. If N = 2356, then N2 = 26512 and all exponents are even. Conversely, if P = 3674, then P = T2, where T = 3372. That is, if all powers of the primes in the factorization of a number are even, then the number is a square. Armed with this fact, we can prove our first lemma. Lemma 5.29 If s and t are positive integers with no common factors and st is a perfect square, then both s and t are perfect squares. Proof. Let us factor st into primes. Since it is a square, all primes in the factorization are raised to even powers. Since s and t have no common factors, each prime raised to its power goes with either s or with t. You cannot have a prime going with s and with t because that would mean that s and t will have a common factor. Since all the powers of the primes are even, those that go with s have even powers and all the primes that go with t also have even powers. Thus, s and t are squares. & Lemma 5.30 If a, b, and c are positive integers with no common factor and if a2 + b2 = c2, then one of a or b is even, and the other is odd. Proof. If both a and b are even then c2 being the sum of two even numbers is even, and hence c is even. That means that each of a, b, and c are even. This contradicts that they have no common factor. So a and b cannot both be even. If both a and b are odd, then so are their squares. And since c2 = a2 + b2, c2 must be even, being the sum of two odd numbers. Hence, c must be even. Since a and b are odd and we now know that c is even, we can write a = 2k + 1 and b = 2l + 1 and c = 2r where k, l, and r are integers. Substituting this into a2 + b2 = c2, we get that ð2k þ 1Þ2 þ ð2l þ 1Þ2 ¼ ð2rÞ2 or, after squaring and simplifying, that 4k2 þ 4k þ 4l2 þ 4l þ 2 ¼ 4r2; which in turn, can be written as 4r2 À ð4k2 þ 4k þ 4l2 þ 4lÞ ¼ 2:
216 Chapter 5 The Triangle: Its Study and Consequences Now, since 4 can be factored out of the left side of this equation, the left side is divisible by 4. But the right side isn’t. This can’t be. Thus, it can’t be that both a and b are odd. We have dispensed of the case where both a and b are even and where a and b are odd. The only case left is that one of a and b must be even and the other odd. & Lemma 5.31 If a is odd and c is odd, and if a and c have no common factor (other than 1), then c þ a 2 c À a and 2 have no common factor (other than 1). Proof. First, we know that c þ a and c À a are integers, since c + a and c À a are even. Now, if c þ a 2 2 2 c À a c þ a c À a and 2 had a common factor, say d, then since d divides both 2 and 2 , and d must divide their sum and difference. That is, d must divide cþa þ c À a ¼ c and d must divide cþa À c À a ¼ 2 2 2 2 a: But c and a have no common factors other than 1. So d must be 1. & We are now ready to prove the main theorem about generating our Pythagorean triples. Theorem 5.32 Given that a, b, and c are positive integers with no common factors, and a is odd and b is even, then if a2 + b2 = c2, there are positive integers m and n such that a ¼ m2 þ n2 b ¼ 2mn c ¼ m2 þ n2: Proof. Write a2 + b2 = c2 as b2 ¼ c2 À a2 Dividing both sides by 4 we have ð5:74Þ b2 c2 À a2 ¼; 44 which in turn can be written as b2 ¼ ðc þ aÞ Á ðc À aÞ : 2 22 So, on the left we have the square of an integer, and on the right, we have the product of two integers ðc þ aÞ and ðc À aÞ with no common factor other than 1. So, by Lemma 5.30 we have 2 2 that both ðc þ aÞ and ðc À aÞ are perfect squares. That is, 2 2 ðc þ aÞ ¼ m2 and ð5:75Þ 2 ðc À aÞ ¼ n2: ð5:76Þ 2
5.9 Pythagorean Triples 217 Substituting these values from equations (5.75) and (5.76) into equation (5.74) we get b2 ¼ ðc þ aÞ Á ðc À aÞ ¼ m2n2 , 2 2 2 from which it follows that that b ¼ mn. Hence, b = 2mn. Also, subtracting equation (5.76) from 2 equation (5.75) we get a = m2 À n2, and adding equations (5.76) and (5.75) we get c = m2 + n2. Thus, we have shown that a, b, and c satisfy the formulas we gave in the theorem. & This theorem is telling us that all right triangles are generated in the same way using different values for m and n. This is a rather interesting result, don’t you think? As we hope you are aware, throughout this text we have been trying to point out the connec- tions that exist between different areas of the mathematical content studied in secondary school. We now take the opportunity to make connections between number theory and geometry by pffiffiffi showing another proof that 2 is irrational (though this is not as elegant as the proof in Chapter 1, which required minimal knowledge). Begin with a right triangle whose legs are 1 and 1, and suppose that the hypotenuse, which is pffiffiffi is rational and equal to pq, where p and q are positive integers with no common factor. Then, the 2, sides of our triangle are 1, 1, and p and, if we multiply all the sides by q, we get a similar right tri- q angle with legs a = q, b = q, and c = p. But, by Lemma 5.30, one of the legs, say a, of the right triangle has to be odd and the other, b, has to be even. But a and b are the same! How could this be? We have a contradiction! pffiffiffi pffiffiffi Our contradiction arose from assuming that 2 was rational. Thus, 2 is irrational. Student Learning Opportunities 1* What is the Pythagorean triple generated by m = 2 and n = 1? 2* Generate a Pythagorean triple which gives us a value of 12 for one of the legs of the triangle. Is there more than one triple with this property? 3* Find the sides of a right triangle that has all integer sides, and for which one leg is 9. Can you find a Pythagorean triple one of whose sides is k where k is any odd integer? Explain. 4* Find the sides of a right triangle with all integer sides where the hypotenuse is 25. 5* Find a Pythagorean triple where the even leg is the smallest side. 6 (C) In trying to generate Pythagorean triples some of your students have noticed that if you start with the Pythagorean triple, 3–4–5, and multiply each number by 2, you get 6–8–10, which is another Pythagorean triple. If they multiply each number in the original triple by 3, they get 9–12–15, which is again a Pythagorean triple. They are wondering if it is always true that if each entry in a Pythagorean triple is multiplied by a positive integer k, the result will be a Pythagorean triple. How do you respond and how do you prove it? 7 (C) One of your students has made the observation that in all of the Pythagorean triples she has seen, at least one member is divisible by 3. For example, in the triple 6–8–10, the first member
218 Chapter 5 The Triangle: Its Study and Consequences 6 is divisible by 3. In the triple 5–12–13, the second member 12 is divisible by 3. She has made a conjecture that every Pythagorean triple has a member that is divisible by 3. Is it true, and if so, how do you prove it? [Hint: What can m and n be congruent to mod 3?] 5.10 Other Interesting Results About Areas LAUNCH In Figure 5.45, A CB Figure 5.45 the sides of the triangle A, B, C, where C is the right angle and B is the vertex at the base of the tri- angle, are of length a = 3 units, b = 4 units, and c = 5 units. Let’s see how many different ways you can calculate the area of this triangle. 1 First use the formula A ¼ 1 b  h: What did you get? 2 2 Now use the formula A ¼ 1 ab sin C : What did you get? 2 3 Now, examine the grid, noting that the triangle is one half of a rectangle whose sides are 3 units and 4 units. What did you get for the area? 4 Next, you will do something that is probably unfamiliar to you. (a) Figure out the perimeter of the triangle and let s equal 1 the perimeter. 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (b) Now, use the strange formula A ¼ sðs À aÞðs À bÞðs À cÞ: Did you get the same area as before? 5 Next, use Pick’s Theorem from the previous chapter. 6 Compare and contrast the different formulas you used to calculate the area of the triangle. When would you use one over another? Having completed the launch question you are probably still wondering about the formulas you used in question 4. Does it always give the area of a triangle and if so, where did it come from? This section will explain.
5.10 Other Interesting Results About Areas 219 5.10.1 Heron’s Theorem Thus far, when finding the area of a triangle, we have used either A ¼ 1 b  h or 1 ab sin C: While 2 2 these are most useful formulas, they necessitate knowing either the height or an angle of the trian- gle. But, what if this information is not available? It would be most helpful if we could derive a formula for the area of a triangle that only requires information about the sides. Actually, given the relationships we have established thus far, we can do just that. We will use the Pythagorean Theorem, the fact that we can factor m2 À n2 into (m À n)(m + n), and the formula for the area of a triangle, 1 ab sin C: Let us proceed. 2 Theorem 5.33 (Heron’s formula) The area, A, of a triangle with sides a, b, and c is given by pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi A ¼ sðs À aÞðs À bÞðs À cÞ where s is half the perimeter of the triangle; that is, where s ¼ aþbþc : 2 Proof. We know from Theorem 5.11 that the area of a triangle is A ¼ 1ab sin C: Let us square both 2 sides to get A2 ¼ 1a2b2 sin 2C: Since sin2 C is 1 À cos2 C we can write this last statement as 4 A2 ¼ 1 a2b2ð1 À cos2CÞ 4 ¼ 1 a2b2ð1 þ cos CÞð1 À cos CÞ: ð5:77Þ 4 Now, from equation (5.29) we have that cos C ¼ a2 þ b2 À c2 : Thus, 1 þ cos C ¼ 1 þ 2ab a2 þ b2 À c2 ¼ a2 þ 2ab þ b2 À c2 ¼ ða þ bÞ2 À c2 ¼ ða þ b þ cÞða þ b À cÞ. Similarly, 1 À cos C = 2ab 2ab 2ab 2ab ðc þ a À bÞðc À a þ bÞ: Substituting these expressions for 1 + cos C and 1 À cos C into equation 2ab (5.77) we get A2 ¼ 1 a2b2ð1 þ cos CÞð1 À cos CÞ 4 ¼ 1 a2b2 ða þ b þ cÞða þ b À cÞ Á ðc þ a À bÞðc À a þ bÞ 4 2ab 2ab ¼ ða þ b þ cÞða þ b À cÞðc þ a À bÞðc À a þ bÞ 16 ¼ ða þ b þ cÞ Á ða þ b À cÞ Á ðc þ a À bÞ Á ðc À a þ bÞ : ð5:78Þ 2 2 2 2 Since s ¼ a þ b þ c our first factor in equation (5.78) is s. Since sÀc ¼ aþbþc Àc ¼ a þ b À c, our 2 2 2 second factor in equation (5.78) is s À c. In a similar manner, the third factor in line (5.78) is s À b, and the fourth factor in line (5.78) is s À a. Thus, our previous string of equalities now simplifies to A2 ¼ sðs À cÞðs À bÞðs À aÞ:
220 Chapter 5 The Triangle: Its Study and Consequences Taking the square root of both sides and rearranging the terms, we have pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi A ¼ sðs À aÞðs À bÞðs À cÞ: & Student Learning Opportunities 1 (C) Your students have learned that there are multiple strategies for solving a problem and some methods come easier to them than others. Given the problem of finding the area of the triangle in Figure 5.46, Figure 5.46 do the following: (a) List at least four different formulas they could use to find the area. (b)* Calculate the area using each of the formulas you have listed. (c) Which of the formulas might be helpful for your more visual learners? Explain. 2* Find the area of a triangle whose sides are 10, 12, and 15. 3* Find to the nearest integer, the area of a quadrilateral ABCD if AB = 5, BC = 6, CD = 7, and DA = 8, if ∡B = 100 degrees. 4 In Figure 5.47, the circles of radii 8, 10, and 12 are tangent to one another. Find the area of the region between the circles to the nearest tenth. 10 8 12 Figure 5.47 5* A person is to pay a one-time tax of 10 dollars per square meter on the area of his backyard. The shape and dimensions of the parts of the backyard are given in Figure 5.48.
5.10 Other Interesting Results About Areas 221 R 30 meters Q S 130° 40 meters 15 meters 80° P Figure 5.48 Estimate the area of the land to the nearest square foot and the tax paid on this land. [Hint: Draw RP and find its length as well as the measure of angle RPQ.]
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