Honors Geometry Practice: Proofs Ch. 4
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1 Honors Geometry Practice: Proofs h. 4 omplete each proof. Number your steps. Name: Hour: ANSWERS PAY LOSE ATTENTION TO THE ORER OF THE VERTIES AN BE AREFUL WITH USING THE ORRET NOTATION.. GIVEN: RO MP, MO OP MRO PRO. RO MP, MO OP. ROP and ROM are right angles. efinition of perpendicular 3. ROP ROM 3. Right Angles ongruence Theorem 4. RO RO 4. Reflexive Property of ongruence 5. MRO PRO 5. SAS. GIVEN: SV bisects TSB VS bisects TVB TV BV 3 4. SV bisects TSB.. efinition of angle bisector 3. VS bisects TVB 3. Given efinition of angle bisector 5. SV SV 5. Reflexive Property of ongruence 6. STV SBV 6. ASA 7. TV BV 7. PT 4; PR TS ; 3. GIVEN: NP NT NR NS N. PR TS; NP NT. 3. Base Angles Theorem 3. NPR NTS 3. SAS 4. NR NS 4. PT 3 4 P R S T NOTE: There is more than one way to do this proof. This particular way does not use 4 and is not needed anywhere in the proof.
2 4. GIVEN: 6; B E AB E. B E Vertical Angles ongruence Theorem A Given n B E 6 m 4. and are a linear pair 5 and 6 are a linear pair 4. efinition of Linear Pair 5. 5 and 6 are supplementary 5. Linear Pair Postulate 6. 5 and are supplementary 6. Substitution (step 3 and step 5) 7. and are supplementary 7. Linear Pair Postulate ongruent Supplements Theorem 9. AB E 9. ASA 5. GIVEN: S and T trisect RV R V; BST BTS BRS BVT. S and T trisect RV ; R V; BST BTS. RS ST VT. efinition of trisect 3. BR BV 3. onverse of Base Angles Theorem 4. BRS BVT 4. SAS NOTE: You will not have to know the definition of "trisect" for the chapter 4 test. 6. GIVEN: HGJ KJG; KGJ HJG G HG KJ K. HGJ KJG; KGJ HJG. GJ JG. Reflexive Property of ongruence 3. HJG KGJ 3. ASA 4. HG KJ 4. PT H J
3 NOTE: The definition of a circle is not something that we have yet defined. The definition of a circle talks about equidistant not congruent. That s why it shows up in these proofs as length = length (not segment segment). You will not have to know the definition of circle for the chapter 4 test, but please note that in future proofs, we will use the definition of circle as shown here. 7. GIVEN: circle with center O RO MP MR PR. circle with center O; RO MP. ROM and ROP are right angles. efinition of perpendicular 3. ROM ROP 3. Right Angles ongruence Theorem 4. OM = OP 4. efinition of circle 5. OM OP 5. efinition of congruent segments 6. RO RO 6. Reflexive Prop. 7. ROM ROP 7. SAS 8. MR PR 8. PT 8. GIVEN: B Y is the midpoint of BY AB ZY. B Y; is the midpoint of BY. B Y. efinition of midpoint Vertical Angles ongruence Theorem 4. AB ZY 4. ASA 5. AB ZY 5. PT NOTE: A corresponds to Z and B corresponds to Y. This is a typing error in this proof. 9. GIVEN: circle with center O E O OE. circle with center O; E. O = EO. efinition of circle 3. O EO 3. efinition of congruent segments 4. O O 4. Reflexive Property of ongruence 5. O OE 5. SSS 6. O OE 6. PT NOTE: The given and what you are to prove suggests that the triangles are rotations of each other, but the triangles can also be viewed as reflections of each other because O = O = OE.
4 0. GIVEN: 5 6, JHG O GH MO J P. 5 6, JHG O, GH MO. 5 and 7 are a linear pair. efinition of linear pair 6 and 8 are a linear pair 3. 5 and 7 are supplementary 3. Linear Pair Postulate 4. 6 and 7 are supplementary 4. Substitution (steps, 3) 5. 6 and 8 are supplementary 5. Linear Pair Postulate ongruent Supplements Theorem 7. GHJ MOP 7. ASA 8. J P 8. PT. GIVEN: 7 8; ZY WX W Y. 7 8; ZY WX Vertical Angles ongruence Theorem Transitive Property of ongruence 7 Z 9 Y Vertical Angles ongruence Theorem Transitive Property of ongruence W 0 X 8 6. XZ ZX 6. Reflexive Property of ongruence 7. WZX YXZ 7. SAS 8. W Y 8. PT. GIVEN: AE F; FB E FB AE F E A B. AE F; FB E. AE = F. efinition of congruent segments 3. AE + EF = AF 3. Segment Addition Postulate 4. F + EF = AF 4. Substitution 5. EF + F = E 5. Segment Addition Postulate 6. AF = E 6. Transitive Property of Equality 7. AF E 7. efinition of congruent segments 8. FB AE 8. Given 9. AE and E are a linear pair 9. efinition of linear pair BFA and FB are a linear pair 0. AE and E are supplementary 0. Linear Pair Postulate. FB and E are supplementary. Substitution (step 8, 0). BFA and FB are supplementary. Linear Pair Postulate 3. BFA E 3. ongruent Supplements Theorem 4. ABF E 4. SAS PT
5 3. GIVEN: ircle with center O PQ QR QO bisects PQR. ircle with center O; PQ QR. PO = RO. efinition of circle 3. PO RO 3. efinition of congruent segments 4. OQ OQ 4. Reflexive Property of ongruence 5. POQ ROQ 5. SSS 6. POQ ROQ 6. PT 7. QO bisects PQR 7. efinition of angle bisector 4. GIVEN: 4 6; 3; 4 5 R T. 3. m = m 3. efinition of congruent angles 3. m PQR = m + m 3. Angle Addition Postulate 4. m PQR = m 3 + m 4. Substitution 5. m SQT = m 3 + m 5. Angle Addition Postulate 6. m PQR = m SQT 6. Transitive Property of Equality 7. PQR SQT 7. efinition of congruent angles Given 9. QP QS 9. onverse of Base Angles Theorem Given. SQT PQR. ASA. R T. PT 5. GIVEN: ; 7 8; 5 6. ; 7 8. and 3 are a linear pair and 4 are a linear pair. efinition of linear pair 3. and 3 are supplementary 3. Linear Pair Postulate 4. and 3 are supplementary 4. Substitution 5. and 4 are supplementary 5. Linear Pair Postulate ongruent Supplements Theorem 7. BE BE 7. Reflexive Property of ongruence 8. ABE BE 8. ASA 9. AE E 9. PT Base Angles Theorem
6 6. GIVEN: l // m and AB B A. l // m and AB A 3 B 4 m l.. Alternate Interior Angles Theorem 3. A A 3. Reflexive Property of ongruence 4. AB A 4. SAS 5. B A 5. PT 7. GIVEN: UQ // VR and UP RP UQP RVP Q 4 3 P U R V. UQ // VR and UP RP Vertical Angles ongruence Theorem 3. U R 3. Alternate Interior Angles Theorem 4. UQP RVP 4. ASA ALTERNATE PROOF: 3. Q V 3. Alternate Interior Angles Theorem 4. UQP RVP 4. AAS 8. GIVEN: E is the midpoint of B E is the midpoint of A AB // A 3 B. E is the midpoint of B. BE E. efinition of midpoint 3. E is the midpoint of A 3. Given E 4. AE E 4. efinition of midpoint Vertical Angles ongruence Theorem 4 6. ABE E 6. SAS PT 8. AB // 8. Alternate Interior Angles onverse Theorem
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