MATH 3 part 2

Mathematics III PART 2

Module 3 Similarity What this module is about This module is about similarities on right triangles. As you go over theexercises you will develop skills in applying similarity on right triangles and solvefor the missing lengths of sides using the famous Pythagorean theorem. What you are expected to learn 1. Apply AA similarity on Right triangles 2. In a right triangle, the altitude to the hypotenuse separates the triangle into two triangles each similar to the given triangle and similar to each other. 3. On a right triangle, the altitude to the hypotenuse is the geometric mean of the segments in which it divides, each leg is the geometric mean of the hypotenuse and the segment of the hypotenuse adjacent to it. 4. Pythagorean Theorem and its application to special right triangles How much do you know Use the figure to answer each of the following: C AB D 1. What is the hypotenuse of rt. ∆ABC?

2. If ∠ C is the right angle of ∆ABC and CD ⊥ AB then ∆ABC ∼ ∆BDC ∼ ___.3. Complete the proportion: AD = ? CD BD4. In rt. ∆PRO, ∠ R is a right angle OR = 24 and PO = 26,find PR: O 26 24 PR5. In a 30°-60°-90° triangle the length of the hypotenuse is 14. Find the length of the longer leg.6. In a 30°-60°-90° Triangle, the length of the hypotenuse is 11 1 . Find the 2 length of the shorter leg.7. In a 45° - 45° - 90° triangle, the length of the hypotenuse is 16. Find the length of a leg.8. Find the length of the altitude of an equilateral triangle if the length of a side is 6.9. Find the length of the diagonal of a square if the length o a side is 10 cm.10. ΔBAC is a right triangle ∠C is right angle CD ⊥ AB. Find CD if AD = 14, DB = 6 C AB 14 D 6

What will you do Lesson 1 Similarity on Right Triangle Let us recall the AA Similarity Theorem. Given a correspondence between the vertices of two triangles. If two pairsof corresponding angles are congruent, then the triangles are similar. B S RT AC From the theorem, if ABC ↔ RST and ∠A = ∠R, ∠B = ∠S then ΔABC ∼ΔRST. We can apply this theorem to prove another theorem, this time in a righttriangle. Theorem: In a right triangle, the altitude to the hypotenuse separates the triangle into to two triangles each similar to the given triangle and similar to each other. C BP A __Given: Right ΔABC with altitude CPProve: ΔACP ∼ ΔCBP ∼ ΔABCTo prove this theorem, we apply the AA Similarity Theorem

Examples: If you are given ΔPRT a right triangle and RM an altitude to the hypotenusethen we can have three pairs of similar triangles.ΔRMP ∼ ΔTRP RΔTMR ∼ ΔTRP MΔRMP ∼ ΔTMR P TTry this outA. Use the figure to answer each of the following:1. Name the right triangle of ΔABC C B2. What is the altitude to the hypotenuse of ΔABC? D3. Name the hypotenuse of ΔABC4. Two segments of the hypotenuse Are AD and ____.5. The hypotenuse of ΔBCD if A CD ⊥ AB is ____.6. Name the right angle of ΔACD7. Name the hypotenuse of right ΔBCD8. ΔADC ∼ _____9. ΔABC ∼ _____10.ΔABC ∼ _____B. Name the pairs of right triangles that are similar. T 1. S 2. 3. R O

M4. S5.6. R OC. Use the figure at the right. A1. Name all the right triangles. D2. In ∆ABC, name the altitude to the hypotenuse.3. Name the hypotenuse in ∆ADC. C B4. Name the hypotenuse of ∆ACB.One of the segments shown is an altitude to the hypotenuse of a righttriangle. Name the segment.5. 6. 7. B E ID H JA CG FK D OName the three pairs of similar triangles:8. M9. N10. P

Lesson 2 Geometric Mean in Similar Right TrianglesThe previous theorem states that: In a right triangle, the altitude to the hypotenuse divides the triangle intosimilar triangles, each similar to the given triangle.If : ∆ACB is right with /C, the right angle C CD is the altitude to the hypotenuse AB.Then : ∆ADC ∼ ∆ ACB ∆ CDB ∼ ∆ ACB ∆ ADC ∼ ∆CDB AB DCorollary: 1. In a right triangle, the altitude to the hypotenuse is the geometric mean of the segments into which divides the hypotenuseIn the figure: C AD = CD CD DB A DBCorollary 2: In a right triangle, each leg is the geometric mean of the hypotenuse and the segment of the hypotenuse adjacent to it.In the figure: A AB = AC D AC ADAB = C B C BC B AD

Examples:1. How long is the altitude of a right triangle that separates the hypotenuse into lengths 4 and 20?4= a a 20a 20 4a2 = 80a = 80a = 16 ⋅ 5a =4 52. Use the figure at the right to solve for x and y. 2=x xy x8 x2 = 16 2 x = 16 8 x =4Solve for y 6=y y8 y2 = 48 y = 48 y = 16 ⋅ 3 y =4 3

Try this outA. Supply the missing terms:1. RW = RT T RT ?2. WS = TS ? RS3. ? = TS RW S TS RS4. RS = RT RT ?Give the indicated proportions. F O5. The altitude is the geometric mean S6. The horizontal leg is the geometric mean7. The vertical leg is the geometric mean R PFind: 2 B8. BS9. RS 810. ST STB. Solve for x and y1. 4 x= 7 y= x y2. 10 x= 20 y= yx

3. y x= x y= 4 10 x= x y=4. 5 y x= y= 105. x 6 yC. OP M RGiven: Right ∆POM OR ⊥ PM, the hypotenuse:1. PR = 5, RM = 10, OR =2. OR = 6, RM = 9, PR =3. PR = 4, PM =12, PO =4. RM = 8, PM =12, OM =5. PO = 9, PR = 3, PM =6. PR = 6, RM = 8, PO =

7. PR = 4, RM = 12, OM =8. PR = 4, PO = 6, RM =9. PR = 8, OR = 12, RM =10. PM =15, OM = 12, RM = Lesson 3 The Pythagorean TheoremIn a right triangle, the square of the length of the hypotenuse is equal to thesum of the squares of the lengths of the legs. BIn the figure: ac A∆BCA is right with leg lengths, C Aa and b and hypotenuse length, c. bThe Pythagorean Theorem in symbol: c2 = a2 + b2 Pythagorean Theorem is named after Pythagoras, a Greek Mathematician ofthe sixth century BC. This theorem can be used to find a missing side length in aright triangle.Examples:1. In the figure c = 13, b = 12Find a: c2 = a2 + b2 a=? c = 13132 = a2 + 122 b = 12 a2 =132 - 122 a2 =169 – 144 a2 = 25 a = 25 a =5

2. Find c if a = 16 b = 12 c2 = a2 + b2 c=? a = 16 c2 = 162 + 122 c2 = 256 + 144 c = 400 c = 20 b = 123. c2 = a2 + b2 82 = 42 + b2 c=8 b2 = 82 – 42 a4 b2 = 64 –16 b2 = 48 ? b = 48 b b = 16 ⋅ 3 b=4 3Try this out:A. State whether the equation is correct or not1. c a b a2 + b2 = c22. r t sr2 = s2 + t23. e d

4. f a e2 = f2 – d2 c b a2 = c2 – b25. l k m k2 = l2 – m26. r p q p2 = r2 + q27. x 3 4 x2 = 32 + 428. 10 7 x

x2 = 102 – 72 x9. 4 6 x2 = 42 + 6210. x 7 5 x2 = 72 – 52B. Write the equation you would use to find the value of x.1. x 7 52. 6 x 5

3. 10 7 x x4. 5 6Classify each statement as true or false 5. 32 + 42 = 52 6. 102 – 62 = 82 7. 12 + 12 = 22 8. 22 + 22 = 42 9. 72 – 52 = 52 10. 92 + 122 = 152C. Given the lengths of two sides of a right triangle. Find the length of the third side B 1. a = 6, b = 8, c =2. a = 5, b = 12, c = c3. a = 12, c = 15, b = a4. b = 4, c = 5, b = A bC5. a = 24, c = 26, b =6. b = 16, c = 20

a b c 24 ?7. 7 6 ? 9 ?8. 4 ? 129. 710 63 Lesson 4 Special Right TriangleIsosceles Right Triangle or 45° – 45° – 90° Theorem In a 45° – 45° – 90° triangle, the length of the hypotenuse is equal to thelength of a leg times 2 . CIn the figure: s sIf ∆ABC, a 45° – 45° – 90° triangle 45° 45°when AC = BC = s then AB = s 2 . A C30° - 60° - 90° TheoremIn a 30° - 60° - 90° triangle, the length of the hypotenuse is twice thelength of the shorter leg, and the length of the longer leg is 3 times the lengthof the shorter leg. P 60°In the figure:If ∆PRT, is right where /R isA right angle, /T = 30°Then: T 30° R a. PT = 2PR b. RT = PR 3Examples:

1.Find the length of the hypotenuse of an isosceles right triangle with a leg 7 2 cm long. 7245° 45° Hypotenuse = leg ⋅ 2 . =7 2⋅ 2 = 7·2 = 14 45° 2. Find the length of each leg of a 45°-45°-90°12cm triangle with a hypotenuse 12 cm long. 45° Leg = hypotenuse √2 = 12 = 12 . √2 = 12√2 = 6√2 cm √2 √2 √2 √2 3.Find the length of the longer leg and the length of the hypotenuse.30° Longer leg = shorter leg . 3 30 m = 30 . 3 = 30 3 m hypotenuse = shorter leg . 2 = 30 . 2 = 60 mTry this outA. Use the figure to answer the following: C 60° 30° 30° D 60° A B1. The hypotenuse of a rt. ∆ABC is ___________.2. The shorter leg of rt. ∆ABC is ___________.

3. The shorter leg of rt. ∆ADC is ___________.4. The longest side of rt. ∆ADC is _________.5. The altitude to the hypotenuse of ∆ACD is ________.6. The longer leg of rt. ∆ACB is _________.7. The longer leg of rt. ∆ADC is ________.8. When CD = 2 then ____ = 4.9. When CB = 6 then _____ = 6 310. When CB = 6 then _____ = 3B. Find the value of x1. 2. 3. x 60° 60° 60° 30° 6 10 x 16 30° 30° x4. 5. 26 6. x 60° 60° 30° 45° 45°18 x 7 x 30°7. 8. 9. 60° 24 60° 45° 30° x 12 30 x 45 ° 30°10. x 45° 10 45°

xC. Find the missing lengths x and y1. 45° 2. 45° x y x y 45° 5 45° 3/2 3. y 4. x x y45° 45° 45° 45° 3/2 3 x 6. 45° 5. y y5 1.5 x7. x 8. 60° y 30° 7 y x 60° 30° 10 9. 10. 60° x 10 3 y x

30° y .5 3Beyond the Pythagorean Theorem In symbol c2 = a2 + b2, where c is the hypotenuse and a and b are the legs ofa right triangle.Figure shows acute trianglesFigure shows obtuse trianglesFigure shown right triangle

Activity: This activity will help you extend your understanding of the relationship of thesides of a triangle.Materials: Strips of paper cut in measured lengths of 2,3, 4, 5, 6 and 8 units.Procedure: 1. Form triangles with strips indicated by the number triplets below. 2. Draw the triangle formed for each number triple. 3. Fill out the table: Number triplets What kind of Compute c2 Compute a2 + b2 triangle 52 = 25 32 + 42 = 251. 3 4 5 Right2. 2 3 43. 2 4 54. 5 4 85. 6 5 86. 4 5 67. 2 3 38. 3 3 4After the computation, the completed table will look like thisNumber triplets Kind of c2 a2 + b2 Comparison of c with triangle (a2 + b2 )1. 3 4 5 Right 25 25 Equal to2. 2 3 4 Obtuse 16 133. 2 4 5 Obtuse 25 20 Greater than4. 5 4 8 Obtuse 64 41 Greater than5. 6 5 8 Obtuse 64 61 Greater than6. 4 5 6 Acute 36 41 Greater than7. 2 3 3 Acute 9 13 Smaller than8. 3 3 4 Acute 16 18 Smaller than Smaller than1. What kind of ∆ did you get from triplet no. 1?2. In triplet no. 1, what is the relation between c2 and (a2 + b2)?3. Which triplets showed obtuse triangle?4. For each obtuse triangle compare the result from c2 and (a2 + b2).

5. For acute triangles how will you compare the result of c2 and (a2 + b2)Fill in the blanks with <, =, >: 6. In a right triangle, c2 ____ a2 + b2 7. In an obtuse triangle, c2 ______ a2 + b2 8. In an acute triangle, c2 ____a2 + b2 Let’s Summarize Theorem: In a right triangle, the altitude to the hypotenuse separates the triangle into two triangles each similar to the given triangle and similar to each other. Corollary 1: In a right triangle, the altitude to the hypotenuse is the geometric mean of the segments into which it divides the hypotenuse. Corollary 2: In a right triangle, each leg is the geometric mean of the hypotenuse and the segment of the hypotenuse adjacent to it. Pythagorean Theorem: The square of the length of the hypotenuse is equal to the sum of the squares of the legs. 45°-45°-90° Theorem: In a 45°-45°-90° triangle, the length of the hypotenuse is equal to the length of a leg times 2. 30°- 60° - 90° Theorem: In a 30°- 60°- 90° triangle, the length of the hypotenuse is twice the length of the shorter leg, and the length of the leg is 3 times the length of the shorter leg.

What have you LearnedFill in the blanks:1. The _______ to the hypotenuse of a right triangle forms two triangles each similar to the given triangle & to each other.2. The lengths of the ________ to the hypotenuse is the geometric mean of the lengths of the segments of the hypotenuse.3. In the figure M AB = MA for nos. 3 & 4 MA ?4. If BP = 8AB = 4Find PM ___ P A B5. If in a right triangle the lengths of the legs are 8 and 15, the length of thehypotenuse is _______6. Find the length of an altitude of an equilateral triangle if the length of a side is 10.7. In a 30° – 60° – 90° triangle, the length of the hypotenuse is 8. Find the length of the shorter leg.8. – 9. ∆ACB is an isosceles right triangle. CD bisects ∠C, the right angle.Find AB and CB. C 3 AB D10. What is the height of the Flag Pole? 2m 8m

Answer key 6. 5 ¾ 7. 8 2How much do you know 8. 3 3 9. 10 2 1. AB 10. 2 21 2. ∆ACB 3. CD B. ∆ROS ∼ ∆RST 4. 10 1. ∆TOS ∼ ∆RST 5. 7 3 2. ∆ROS ∼ ∆TOS 3.Lesson 1: 4. ∆MST ∼ ∆MORA. 5. ∆RSO ∼ ∆MOR 1. ∠C or ∠ACB 6. ∆MST ∼ ∆RSO 2. CD 3. AB ___ B. C. 4. BD or DB 1. x = 2 11 1. 5 2 5. BC 2. 4 6. ∠ADC y=2 7 7. BC 8. ∆BDC 9. ∆ADC 10. ∆BDCC. 1. ∆ADC, ∆BDC, ∆ACB 2. CD 3. AC 4. AB 5. BD 6. GH 7. OK 8. ∆MNR ∼ ∆MPO 9. ∆ONP ∼ ∆MPO 10. ∆MNP ∼ ∆ONP Lesson 2A. 1. WS 2. TS

3. WS 2. x = 10 3 3. 4 3 4. RW y = 30 4. 4√6 5. SF = OS 5. 27 3. x = 2√14 6. 4 3 OS SP y = 2√35 6. SP = PO 7. 8 3 4. x = 20 8. 9 PO PF y = 5√5 7. FS = OF 5. x = 100 or 16 2 9. 18 OF PF 63 10. 9.6 8. 4 y = 10 9. 2√5 10. 4√5 B. C. 1. x2 = 52 + 72 1. 10Lesson 3 2. x2 = 62 – 52 2. 13 3. x2 = 102 – 72 3. 9A. 4. x2 = 62 – 52 4. 31. correct 5. true 5. 102. Correct 6. true 6. 123. not 7. false 7. 254. correct 8. 2√135. correct 8. false 9. √1306. not 10. 67. correct 9. false8. correct 10. true9. not10. not B. C. y 1. 12 xLesson 4A. 2. 10 3 1. 5 3 5 3. 8 1. AB 2. 3 3 √2 2. BD 4. 18 3 3. CD 5. 13 2 2 4. AC 3. 3 3 5. CD 6. 7 2 6. AC 4. 3√2 3√2 7. AD 7. 12 3 8. AC 8. 12√2 22 9. AC 9. 15 3 10. DB 10. 10√3 5. 5√2 5√2 3 22 6. 1.5√2 1.5√2 22 7. 14 73

8. 5 53 9. 5 3 15 10. 5 10What have you learned 1. altitude 2. altitude 3. AP 4. 4 2 5. 17 6. 5 3 7. 4 8. 3 2 9. 3 10. 2 3



Module 3 Triangle Congruence What this module is about This module is about using triangle congruence to prove congruentsegments and angles. You will understand that a correspondence between twotriangles is a congruence if the corresponding angle and corresponding sides arecongruent. What you are expected to learnThis module is designed for you to:1. prove congruent segments and angles using the conditions for triangle congruence.2. solve routine and non-routine problems.How much do you know1. Given: HS bisects ∠ THE ∠ HTS ≅ ∠ HES TProve: TS ≅ ES H S E

2. Given: PM // NS; PM ≅ SN M Prove: ML ≅ LN P L NS3. Given: NO ≅ XS; OE ≅ SI; NI ≅ XE Prove: ∠ O ≅ ∠ S ONI EX S4. Given: ∆ PNC and ∆ TNC are isosceles triangles with common base NC. PProve: ∠ PNT ≅ ∠ PCT NC T 2

5. Given: AE is the ⊥ bisector of BC AProve: ∠ B ≅ ∠ C B C E6. Given: CH ≅ EH, FH ≅ GHProve: ∠ C ≅ ∠ E D FG H EC7. Given: ∠ A and ∠ B are right angles and AP ≅ BP. P Prove: AQ ≅ BQ AB Q 3

8. Given: ∠ 1 ≅ ∠ 2 O is the midpoint of SP Prove: MO ≅ NO M1 PSO 2 N9. Given: LO ≅ LN ∆ JLO and ∆ MLN are right ∆ s Prove: ∠ J ≅ ∠ K J NOL10. Given: PQ ≅ RS M ∠ QRP ≅ ∠ SRP S Prove: SP ≅ QR PQR 4

What you will do LessonCongruent Segments and Congruent Angles To prove two segments or two angles are congruent, you must show thatthey are corresponding parts of congruent triangles. For triangle congruence, you have the following: • SSS congruence • SAS congruence • ASA congruence • SAA congruence For right triangle congruence, you have the following: • LL congruence • LA congruence • HyL congruence • HyA congruenceExamples: 3 D Formal Proofs: 2 1. Given: AB // DC, AD // BC Prove: AB ≅ DC AB 14 C 5

Proof: Reason Statement 1. Given 2. If two parallel lines are cut by a transversal,1. AB // DC, AD // BC2. ∠ 1 ≅ ∠ 2, ∠ 3 ≅ ∠ 4 then the alternate interior angles are ≅ 3. Reflexivity3. BD ≅ BD 4. ASA4. ∆ ABD ≅ ∆ CDB 5. Corresponding parts of ≅ triangles are ≅5. ∴AB ≅ DC or CPCTC2. Given: LS bisects ∠ TLE ∠ LTS ≅ ∠ LES Prove: TS ≅ ES T SLProof: E Statement Reason1. LS bisects ∠ TLE 1. Given2. ∠ TLS ≅ ∠ ELS 2. Definition of angle bisector3. LS ≅ LS 3. Reflexivity4. ∠ LTS ≅ ∠ LES 4. Given5. ∆ TSL ≅ ∆ ESL 5. SAA congruence6. TS ≅ ES 6. Corresponding parts of ≅ ∆ s are ≅ or CPCTC 6

3. Given: MO ≅ YS, OB ≅ SP, MP ≅ YBProve: ∠ O ≅ ∠ S OMP BYProof: S Statement Reason 1. Given1. MO ≅ YS OB ≅ SP 2. Reflexivity MP ≅ YB 3. By addition 4. SSS congruence2. PB ≅ PB 5. CPCTC3. MB ≅ PY4. ∆ MOB ≅ ∆ YSP5. ∠ O ≅ ∠ S4. Given: AR is the ⊥ bisector of BX. Prove: ∠ B ≅ ∠ X AB X R 7

Proof: Reason Statement 1. Given 2. Reflexivity1. AR is the ⊥ bisector of BX 3. Definition of ⊥ bisector2. AR ≅ AR 4. LL congruence3. BR ≅ XR 5. CPCTC4. ∆ ARB ≅ ∆ ARX5. ∠ B ≅ ∠ X5. Given: ∠ 1 ≅ ∠ 2 O is the midpoint of SPProve: DO ≅ SO D1 PSO 2Proof: S Statement Reason1. ∠ 1 ≅ ∠ 2 1. Given2. ∠ DSO ≅ ∠ SPO 2. Supplements of ≅ ∠ s are also ≅3. O is the midpoint of SP 3. Given4. SO ≅ PO 4. Definition of midpoint5. ∠ DOS ≅ ∠ SOP 5. Vertical ∠ s are ≅6. ∆ SDO ≅ ∆ PSO 6. ASA congruence7. DO ≅ SO 7. CPCTC 8

Try this out K M 1. Given: ∠ SLO and ∠ KMO are right angles LO ≅ MO Y Prove: ∠ S ≅ ∠ K S O L 2. Given: XY ≅ RS ∠ YXR ∠ SRX Prove: SX ≅ YR XSR3. Given: ∆ PNU and ∆ TNU are isosceles triangles with common base NUProve: ∠ PNT ≅ ∠ PUT P ONu T 9

4. Given: CD ≅ ED, FC ≅ FE Prove: ∠ C ≅ ∠ E CDF E5. Given: ∠ A ≅ ∠ C are right angles C AK ≅ MC Prove: MA ≅ KC M A K6. Given: MN ⊥ NR, PR ⊥ NR, MR ≅ PN P Prove: MN ≅ NR MNR 10

7. Given: AI ≅ BN, BI ≅ AN B Prove: ∠ I ≅ ∠ N A IN8. Given: In the figure, JV and NC bisect each other at O Prove: ∠ J ≅ ∠ V JC O NV9. Given: MQ and PN bisect each other at O Prove: ∠ P ≅ ∠ N N M QO P 11

10. Given: GQ ≅ RS Q ∠ QGR ≅ ∠ SRG Prove: SG ≅ QR G SR Let’s summarize To prove two segments or two angles are congruent, you must show thatthey are corresponding parts of congruent triangles. For triangle congruence, you have the following: • SSS congruence • SAS congruence • ASA congruence • SAA congruence For right triangle congruence, you have the following: • LL congruence • LA congruence • HyL congruence • HyA congruence 12

What have you learned T1. Given: TS bisects ∠ MTE ∠ TMS ≅ ∠ TES Prove: TM ≅ TE M SE2. Given: PQ // NS; PQ ≅ SN Q Prove: QL ≅ LN P LNS 13

3. Given: XO ≅ YS; OE ≅ SI; XI ≅ YE Prove: ∠ O ≅ ∠ S OXI EY S4. Given: ∆ PNQ and ∆ TNQ are isosceles triangles with common base NQ.Prove: ∠ PNQ ≅ ∠ TNQ P NQ5. Given: HE is the ⊥ bisector of BC T Prove: ∠ B ≅ ∠ C H B C E 14

6. Given: CG ≅ EG, FG ≅ HG D Prove: ∠ C ≅ ∠ E F H G E HC7. Given: ∠ R and ∠ S are right angles and RP ≅ SP. PProve: RQ ≅ SQ RS Q8. Given: ∠ 1 ≅ ∠ 2 O is the midpoint of SPProve: RO ≅ TO R 1 P SO 2 T 15

9. Given: ∠ JLO and ∠ KMO are right ∠ s K LO and MO M Prove: ∠ J ≅ ∠ K Q J O L10. Given: PQ ≅ RS ∠ QPR ≅ ∠ SRP Prove: SP ≅ QR PSR16

Answer key ReasonHow much do you know 1. Given 1. Proof 2. Definition of ∠ bisector 3. Reflexivity Statement 4. SAA congruence 5. CPCTC 1. HS bisects ∠ THE ∠ HTS ≅ ∠ HES Reason 2. ∠ THS ≅ ∠ EHS 1. Given 3. HS ≅ HS 2. If 2 // lines cut by a transversal, 4. ∆ THS ≅ ∆ EHS 5. TS ≅ ES the alternate interior ∠ s are ≅ 3. ASA congruence 2. Proof: 4. CPCTC Statement Reason 1. PM // NS, PM ≅ SN 1. Given 2. ∠ P ≅ ∠ S, ∠ M ≅ ∠ N 2. Reflexivity 3. ∆ PLM ≅ ∆ SLN 3. By addition 4. ML ≅ LN 4. SSS congruence 5. CPCTC 3. Proof: Reason Statement 1. Given 1. NO ≅ XS, OE ≅ SI NI ≅ XE 2. Definition of isosceles 3. Reflexivity 2. IE ≅ IE 4. SSS congruence 3. NE ≅ IX 5. CPCTC 4. ∆ NOE ≅ ∆ XSI 5. ∠ O ≅ ∠ S 4. Proof: Statement 1. ∆ PNC and ∆ TNC are isosceles ∆ 2. PN ≅ PC, TN ≅ TC 3. PT ≅ PT 4. ∆ PNT ≅ ∆ PCT 5. ∠ PNT ≅ ∠ PCT 17

5. Proof: Reason Statement 1. Given 3. Definition of ⊥ bisector 1. AE is the ⊥ bisector of BC 2. Reflexivity 2. BE ≅ CE 4. LL congruence 3. AE ≅ AE 5. CPCTC 4. ∆ AEB ≅ ∆ AEC 5. ∠ B ≅ ∠ C6. Proof: Reason Statement 1. Given 2. Vertical ∠ s are ≅ 1. CH ≅ EH, FH ≅ GH 3. SAS 2. ∠ FHC ≅ ∠ GHE 4. CPCTC 3. ∆ FHC ≅ ∆ GHE 4. ∠ C ≅ ∠ E7. Proof: Reason Statement 1. Given 1. ∠ A and ∠ B are rt. ∠ s 2. Reflexivity AP ≅ BP 3. HyL congruence 4. CPCTC 2. PQ ≅ PQ 3. ∆ PAQ ≅ ∆ PBQ 4. AQ ≅ BQ8. Proof: Reason Statement 1. Given 2. Supplements of ≅ ∠ s are also ≅1. ∠ 1 ≅ ∠ 2 3. Given2. ∠ MSO ≅ ∠ NPO 4. Definition of midpoint3. O is the midpoint of SP 5. Vertical ∠ s are ≅4. SO ≅ PO 6. ASA congruence5. ∠ MOS ≅ ∠ NOP 7. CPCTC6. ∆ MDO ≅ ∆ NPO7. MO ≅ NO 18

9. Proof: Reason 1. Given Statement 2. Vertical ∠ s are ≅ 3. LA congruence 1. LO ≅ LN 4. CPCTC 2. ∠ JLO ≅ ∠ MLN 3. ∆ JLO ≅ ∆ MLN Reason 4. ∠ J ≅ ∠ K 1. Given 2. Reflexivity 10. Proof: 3. SAS congruence 4. CPCTC Statement Reason 1. PQ ≅ RS 1. Given ∠ QRP ≅ ∠ SRP 2. Vertical ∠ s are ≅ 3. LA congruence 2. PR ≅ PR 4. CPCTC 3. ∆ PSR ≅ ∆ RQP 4. SP ≅ QR Reason 1. GivenTry this out 2. Reflexivity 3. SAS congruence Lesson 4. CPCTC 1. Proof: Statement 1. LO ≅ MO 2. ∠ SOL ≅ ∠ KOM 3. ∆ SLO ≅ ∆ KMO 4. ∠ S ≅ ∠ K 2. Proof: Statement 1. XY ≅ RS ∠ YXR ≅ ∠ SRX 2. RX ≅ RX 3. ∆ XYR ≅ ∆ RSX 4. SX ≅ YR 19

3. Proof: Reason 1. Given Statement 2. Definition of isosceles 3. Reflexivity 1. ∆ PNU and ∆ TNU 4. SSS congruence are isosceles ∆ 5. CPCTC 2. PN ≅ PU, TN ≅ TU Reason 3. PT ≅ PT 1. Given 4. ∆ PNT ≅ ∆ PUT 2. Reflexivity 5. ∠ PNT ≅ ∠ PUT 3. SSS congruence 4. CPCTC4. Proof: Reason Statement 1. Given 2. Reflexivity 1. CD ≅ ED, FC ≅ FE 3. HyL congruence 2. DF ≅ DF 4. CPCTC 3. ∆ DCF ≅ ∆ DEF 4. ∠ C ≅ ∠ E Reason 1. Given5. Proof: 2. Reflexivity 3. Hyl congruence Statement 4. CPCTC 1. ∠ A ≅ ∠ C AK ≅ MC 2. MK ≅ MK 3. ∆ MCK ≅ ∆ KAM 4. MA ≅ KC6. Proof: Statement 1. MR ≅ PN 2. NR ≅ NR 3. ∆ MNR ≅ ∆ PRN 4. MN ≅ PR 20

7. Proof: Reason 1. Given Statement 2. Reflexivity 3. SSS congruence 1. AI ≅ BN 4. CPCTC BI ≅ AN Reason 2. IN ≅ IN 1. Given 3. ∆ AIN ≅ ∆ BNI 2. Definition of Segment Bisector 4. ∠ I ≅ ∠ N 3. Vertical ∠ s are ≅ 4. SAS8. Proof: 5. CPCTC Statement Reason 1. Given 1. JV and NC bisect 2. Definition of Segment Bisector each other at O 3. Vertical ∠ s are ≅ 4. SAS 2. JO ≅ VO 5. CPCTC CO ≅ NO Reason 3. ∠ JOC ≅ ∠ VON 1. Given 4. ∆ JOC ≅ ∆ VON 2. Reflexivity 5. ∠ J ≅ ∠ V 3. SAS congruence 4. CPCTC9. Proof: 21 Statement 1. MQ and PN bisect each other at O 2. MO ≅ QO PO ≅ NO 3. ∠ POM ≅ ∠ NOQ 4. ∆ POM ≅ ∆ VOQ 5. ∠ P ≅ ∠ N10. Proof: Statement 1. GQ ≅ RS ∠ QGR ≅ ∠ SRG 2. RG ≅ RG 3. ∆ QGR ≅ ∆ SRG 4. SG ≅ QR

What have you learned Reason 1. Given 1. Proof 2. Definition of ∠ bisector 3. Reflexivity Statement 4. SAA congruence 5. CPCTC 1. TS bisects ∠ MHE ∠ TMS ≅ ∠ TES Reason 1. Given 2. ∠ MTS ≅ ∠ ETS 2. If 2 // lines cut by a transversal, 3. TS ≅ TS 4. ∆ SMT ≅ ∆ SET the alternate interior ∠ s are ≅ 5. TM ≅ TE 3. ASA congruence 4. CPCTC 2. Proof: Reason Statement 1. Given 2. Reflexivity 1. PQ // SN 3. By addition 2. ∠ QPL ≅ ∠ NSL 4. SSS congruence 5. CPCTC ∠ PQL ≅ ∠ SNL 3. ∆ PQL ≅ ∆ SNL Reason 4. QL ≅ LN 1. Given 2. Definition of isosceles 3. Proof: 3. Reflexivity 4. SSS congruence Statement 5. CPCTC 1. XO ≅ YS, OE ≅ SI 22 XI ≅ YE 2. IE ≅ IE 3. XE ≅ YI 4. ∆ XOE ≅ ∆ YSI 5. ∠ O ≅ ∠ S 4. Proof: Statement 1. ∆ PNQ and ∆ TNQ are isosceles ∆ 2. PN ≅ PQ, TN ≅ TQ 3. PT ≅ PT 4. ∆ PNT ≅ ∆ PQT 5. ∠ PNQ ≅ ∠ TNQ

5. Proof:Statement Reason1. HE is the ⊥ bisector of BC 1. Given2. BE ≅ CE 3. Definition of ⊥ bisector3. HE ≅ HE 2. Reflexivity4. ∆ HEB ≅ ∆ HEC 4. LL congruence5. ∠ B ≅ ∠ C 5. CPCTC6. Proof: Reason Statement 1. Given 2. Vertical ∠ s are ≅ 1. CG ≅ EG, FG ≅ HG 3. SAS 2. ∠ FGC ≅ ∠ HGE 4. CPCTC 3. ∆ FGC ≅ ∆ HGE 4. ∠ C ≅ ∠ E7. Proof: Reason Statement 1. Given 2. Reflexivity 1. RP ≅ SP 3. HyL congruence 2. PQ ≅ PQ 4. CPCTC 3. ∆ PRQ ≅ ∆ PSQ 4. RQ ≅ SQ8. Proof: Reason Statement 1. Given 2. Supplements of ≅ ∠ s are also ≅1. ∠ 1 ≅ ∠ 2 3. Given2. ∠ RSP ≅ ∠ TPO 4. Definition of midpoint3. O is the midpoint of SP 5. Vertical ∠ s are ≅4. SO ≅ PO 6. ASA congruence5. ∠ SOR ≅ ∠ POT 7. CPCTC6. ∆ SOR ≅ ∆ POT7. RO ≅ TO 23