answer to ch 6 proofs

Practice Proofs – Chapter 6U1. Given: ∡ 1 and ∡ 2 are complementary;∡ 2 and ∡ 3 are complementary.Prove: GU || HSG1 23SHSTATEMENTSREASONS1. ∠ 1 and ∠ 2 are complementary; 1. Given∠ 2 and ∠ 3 are complementary2. ∠ 1 = ∠ 3 2. ∠ ’s complementary to same ∠ =3. ∠ 1, ∠ 3 are alt. int ∠ ’s (transversal SG) 3. def. alternate interior angles4. GU || HS 4. Alt. interior ∠ ’s = parallel lines2. Given: CH = HA; ∡ 2 = ∡ 3;CT is a transversal of CH and AN .HNProve: CH AN1 2 3C A TSTATEMENTSREASONS1. CH = HA 1. Given2. ∠ 1 = ∠ 2 2. = sides in ∆ opposite ∠ ’s =3. ∠ 2 = ∠ 3 3. Given4. ∠ 1 = ∠ 3 4. Transitive Prop. = /substitution (#2, 3)5. CT is a transversal of CH and AN . 5. Given6. ∠ 1, ∠ 3 are corr. ∠ ’s (transversal CT) 6. def. corresponding angles7. CH || AN 7. Corresponding ∠ ’s = parallel lines

3. Given: KA = KL;NE || KL .NKProve: ∆ ANE is isoscelesSTATEMENTSAREASONSEL1. KA = KL 1. Given2. ∠ A = ∠ L 2. = sides in ∆ opposite ∠ ’s =3. NE || KL 3. Given4. ∠ NEA, ∠ L are corresponding ∠ ’s 4. def. corresponding angles5. ∠ NEA = ∠ L 5. Parallel lines corresponding ∠ ’s =6. ∠ NEA = ∠ A 6. Substitution/transitive (2 5)7. NA = NE 7. = ∠ ’s in ∆ opposite sides =8. ∆ ANE is isosceles 8. Equal sides isosceles ∆S4. Given: ∆ ASE and ∆ AGE are right triangleswith right angles S and G; AS = GEProve: SE || AGAEGSTATEMENTSREASONS1. ∆ ASE and ∆ AGE are right triangles 1. Givenwith right angles S and G; AS = GE2. AE = AE 2. Reflexive Prop. =3. ∆ ASE ≅ ∆ EGA 3. HL (#1, 1, 2)4. ∠ SEA = ∠ GAE 4. CPCTE5. ∠ SEA, ∠ GAE are alt. int ∠ ’s 5. def. alternate interior angles(transversal AE )6. SE || AG 6. Alt. interior ∠ ’s = parallel lines

L5. Given: ∡ 1 and ∡ 2 are vertical angles;B is the midpoint of LW ; EL WO .Prove: EL = OWE1B2WOSTATEMENTSREASONS1. ∠ 1 and ∠ 2 are vertical angles 1. Given2. ∠ 1 = ∠ 2 2. Vertical angles are =3. EL || WO 3. Given4. ∠ L, ∠ W ( ∠E, ∠O) are alt. int ∠ ’s 4. def. alternate interior angles5. ∠L = ∠ W (∠E = ∠ O) 5. Parallel lines alt. interior ∠ ’s =6. B is the midpoint of LW 6. Given7. BL =BW 7. Midpoint = parts/segments8. ∆ LBE ≅ ∆ WBO 8. ASA (#2, 5, 7) (AAS – same)9. EL = OW 9. CPCTE6. Given: R-A-N and A-N-C; RA = NC;FN || AE; RF || ECProve: ∆ FRN ≅ ∆ ECARAF• •NCESTATEMENTSREASONS1. R-A-N; A-N-C; RA = NC 1. Given2. RA + AN = AN + NC 2. Addition POE3. RA + AN = RN; AN + NC = AC 3. B.O.P.4. RN = AC 4. Substitution (3 2)5. FN || AE; RF || EC 5. Given6. ∠ RNF, ∠ CAE, ∠ R, ∠ C are alt. int ∠ ’s 6. def. alternate interior angles7. ∠RNF = ∠CAE; ∠R = ∠C 7. || lines alternate interior ∠’s ≅8. ∆ FRN ≅ ∆ ECA 8. ASA (#4, 7, 7)

7. Given: BA ||SL with transversal BI ;∡ A = ∡ I; BS = SLALProve: ∆ BAS ≅ ∆ SILBSISTATEMENTSREASONS1. BA ||SL with transversal BI 1. Given2. ∠ B, ∠ ISL are corresonding ∠ ’s 2. def. corresponding angles3. ∠ B = ∠ ISL 3. Parallel lines corresponding ∠ ’s =4. ∠ A = ∠ I; BS = SL 4. Given5. ∆ BAS ≅ ∆ SIL 5. AAS (#3, 4, 4)