Homework Problems Set 24
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1 Homework Problems Set 24 Proof (1) ACB & ADB both intercept arc Assumption for Conditional Proof AB in O (2) m ACB = ½ m(arc AB) By (1), Theorem 6.9 (3) m ADB = ½ m(arc AB) By (1), Theorem 6.9 (4) m ACB = m ADB By Transitivity of Equality, (2), (3) (5) ACB ADB By (4), Theorem 5.6 (6) ACB & ADB both intercept arc Conclusion of Conditional Proof AB in O ACB ADB
2 Proof (1) ABC is inscribed in a circle & AC Assumption for Conditional Proof is the diameter (2) AOC is a linear angle By (1), Definition of Linear Angles (3) m ACB = ½ m(arc AC) By (1), Theorem 6.9 (4) m ACB = ½ m AOC By (3), Definition 6.4 (5) 2(m ACB) = m AOC By (4), Multiplication Property of Equality (6) m ACB + m ACB = m AOC By (5), Definition of Multiplication (7) ACB ACB AOC By (6), Theorem 5.6 (8) ACB & ACB are supplementary By (2), (7), Definition 4.14 (9) ACB is a right angle By (8), Definition 4.15 (10) ABC is inscribed in a circle & AC Conclusion of Conditional Proof is the diameter ACB is a right angle
3 Proof (1) Quadrilateral ABCD is inscribed in Assumption for Conditional Proof O (2) m ABC=½ m(arc ADC) By Theorem 6.9 (3) m BCD=½ m(arc DAB) By Theorem 6.9 (4) m CDA=½ m(arc ABC) By Theorem 6.9 (5) m DAB=½ m(arc BCD) By Theorem 6.9 (6) Arc ADC Arc ABC = O Given (by Construction) (7) m(arc ADC) + m(arc ABC) = m( O) By (6), Theorem 5.4 (8) Arc DAB Arc BCD = O Given (by Construction) (9) m(arc DAB) + m(arc BCD) = m( O) By (8), Theorem 5.4 (10) m ABC + m CDA = ½ (m(arc ADC)+ By (2), (4) m(abc)) (11) m ABC + m CDA = ½ m( O) By Axiom C11, (7), (10) (12) m( EF) = ½ m( O) By Definition 5.2 (13) m ABC + m CDA = m( EF) By Axiom C11, (11), (12) (14) m BCD + m DAB = ½ (m(arc DAB)+ By (3), (5) m(bcd)) (15) m BCD + m DAB = ½ m( O) By Axiom C11, (9), (14) (16) m BCD + m DAB = m( EF) By Axiom C11, (12), (15) (17) m( O) = 360 o By Degrees in a Circle (18) m( EF) = 180 o By (12), (17), Substitution Property of Equality (19) m ABC + m CDA = 180 o By (13), (18), Axiom C11 (20) m BCD + m DAB = 180 o By (16), (18), Axiom C11 (21) ( ABC & CDA are opposite angles) By Definition 4.21 & ( BCD & DAB are opposite angles)
4 (22) Opposite angles are supplementary By (19), (20), (21), Definition 4.14 (23) Quadrilateral ABCD is inscribed in Assumption for Conditional Proof O Opposite angles are supplementary (1) ATB is formed from Tangent TB Assumption for Conditional Proof & Secant TA in O Subproof A: Cases Case 1 (2) Secant TA passes through the Assumption for Case 1 center of the circle (3) ATB intercepts Arc ACT By Definition 6.5 (4) TA is a diameter of O By (2), Definition 6.1 (5) AOT is a linear angle By Definition of Linear Angles (6) OT TB By Theorem 6.5 (7) OTB is a right angle By (6), Definition 4.17 (8) OTB is supplementary to itself By (7), Definition 4.15 (9) OTB OTB A Linear Angle By (8), Definition 4.14 (10) 2( OTB) A Linear Angle By (9), Definition of Multiplication (11) 2( OTB) AOT By (5), (10), Axiom C11 (12) OTB ½ ( AOT) By Division Property of
5 Congruence (13) OTB= ATB By Definition 3.11 (14) ATB ½ ( AOT) By (12), (13), Axiom C11 (15) m ATB = ½ m(arc ACT) By Theorem 6.9 (16) Secant TA passes through the Conclusion of Case 1 center of the circle m ATB = ½ m(arc ACT) Case 2 (17) The center of the circle is in Assumption for Case 2 the exterior of ATB (18) CT is a diameter of O By Construction, Definition 6.1 (19) COT is a linear angle By Definition of Linear Angles (20) OT TB By Theorem 6.5 (21) OTB is a right angle By (20), Definition 4.17 (22) OTB is supplementary to itself By (21), Definition 4.15 (23) OTB OTB A Linear Angle By (22), Definition 4.14 (24) 2( OTB) A Linear Angle By (23), Definition of Multiplication (25) 2( OTB) COT By (19), (24), Axiom C11 (26) OTB ½ ( COT) By Division Property of Congruence (27) OTB= CTB By Definition 3.11 (28) CTB ½ ( COT) By (26), (27), Axiom C11 (29) m CTB = ½ m(arc CAT) By Theorem 6.9 (30) CTB CTA ATB By Construction (31) Arc CAT Arc CA Arc AT By Construction (32) m CTA = ½ m(arc CA) By Theorem 6.9 (33) ½ m(arc CAT) ½ m(arc CA) By Substitution Property of Equality, = m ATB (29), (30), (32) (34) m(arc CAT) m(arc CA) = By (33), Multiplication Property of 2 m ATB Equality
6 (35) m(arc AT) = 2 m ATB By (31), (34), Substitution Property of Equality (36) m ATB = ½ m(arc AT) By (35), Division Property of Equality (37) The center of the circle is in Conclusion of Case 2 the exterior of ATB m ATB = ½ m(arc AT) Case 3 (38) The center of the circle is in Assumption for Case 3 the interior of ATB (39) CT is a diameter of O By Construction, Definition 6.1 (40) COT is a linear angle By Definition of Linear Angles (41) OT TB By Theorem 6.5 (42) OTB is a right angle By (41), Definition 4.17 (43) OTB is supplementary to itself By (42), Definition 4.15 (44) OTB OTB A Linear Angle By (42), Definition 4.14 (45) 2( OTB) A Linear Angle By (44), Definition of Multiplication (46) 2( OTB) COT By (40), (45), Axiom C11 (47) OTB ½ ( COT) By Division Property of Congruence (48) OTB= CTB By Definition 3.11 (49) CTB ½ ( COT) By (47), (48), Axiom C11 (50) m CTB = ½ m(arc CDT) By Theorem 6.9 (51) CTB CTA ATB By Construction (52) Arc CDT Arc CA Arc ADT By Construction (53) m CTA = ½ m(arc CA) By Theorem 6.9 (54) ½ m(arc CDT) + ½ m(arc CA) By Substitution Property of Equality, = m ATB (50), (51), (53) (55) m(arc CDT) + m(arc CA) = By (54), Multiplication Property of
7 2 m ATB Equality (56) m(arc ADT) = 2 m ATB By (52), (55), Substitution Property of Equality (57) m ATB = ½ m(arc ADT) By (56), Division Property of Equality (58) The center of the circle is in Assumption for Case 3 the interior of ATB m ATB = ½ m(arc ADT) (59) The measure of ATB is equal to By (1), Subproof A (by cases) half the measure of the intercepted arc (60) ATB is formed from Tangent TB Conclusion of Conditional Proof & Secant TA in O The measure of ATB is equal to half the measure of the intercepted arc
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