Answer Key To Extra Problem
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1. Triangle AMB is right at M since it is inscribed in a semi-circle. . Therefore, triangle MAB is a semi-equilateral.
ˆ = 60 MBA
o
2.
ˆ = 90o = BMK ˆ KOB Therefore, K, M, B and O belong to the same circle since KMBO is inscribed (opposite angles are supplementary) in a circle of diameter [KB]. 3. Since K belongs to the perpendicular bisector of [AB], then KA=KB And EK=KB (by symmetry). Therefore, KA= EK=KB (by substitution) Then, triangle EAB is right at A. (OK) is perpendicular to (AB) and (EA) and perpendicular to (AB), therefore, (OK) is parallel (EA) (two lines perpendicular to the same line are parallel) Thus, (AE) tangent at A. 4. a.
» = 2 × ABF · AF = 2 × 60 = 120o b. KE=KA and
thus AKE is equilateral. Thus AE=3 and
· EAK = 60o OK=1.5
ˆ = 60 MBA
o
2.
ˆ = 90o = BMK ˆ KOB Therefore, K, M, B and O belong to the same circle since KMBO is inscribed (opposite angles are supplementary) in a circle of diameter [KB]. 3. Since K belongs to the perpendicular bisector of [AB], then KA=KB And EK=KB (by symmetry). Therefore, KA= EK=KB (by substitution) Then, triangle EAB is right at A. (OK) is perpendicular to (AB) and (EA) and perpendicular to (AB), therefore, (OK) is parallel (EA) (two lines perpendicular to the same line are parallel) Thus, (AE) tangent at A. 4. a.
» = 2 × ABF · AF = 2 × 60 = 120o b. KE=KA and
thus AKE is equilateral. Thus AE=3 and
· EAK = 60o OK=1.5
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