question_answer

A triangle ABC is inscribed in a circle, the bisectors of whose angles meet the circumference at X, Y and Z. Determine the angles X, Y and Z respectively.

A)  \[{{90}^{o}}-\frac{A}{2},{{90}^{o}}-\frac{B}{2},{{90}^{o}}-\frac{C}{2}\]

B)  \[{{90}^{o}},{{60}^{o}},{{30}^{o}}\]

C)  \[\frac{A}{2},\frac{B}{2},\frac{C}{2}\]

D)  \[\frac{B}{2},\frac{A}{2},\frac{A}{2}-\frac{B}{2}\]

Correct Answer: A

Solution :

Clearly, \[\angle BYX=\angle BAX=\frac{\angle A}{2}.\] Also \[\angle ZYB=\angle ZCB=\frac{\angle C}{2}.\](Angles in the same segment.) Hence, \[\angle ZYX=\frac{\angle C}{2}+\frac{\angle A}{2}=\frac{\angle A+\angle C}{2}\] \[\Rightarrow \]\[\angle ZYX=\frac{\angle C}{2}+\frac{\angle A}{2}=\frac{\angle A+\angle C}{2}\] \[=\frac{{{180}^{o}}-\angle B}{2}={{90}^{o}}-\frac{\angle B}{2}\] Similarly, the other angles are \[{{90}^{o}}-\frac{A}{2}\]and \[{{90}^{o}}-\frac{C}{2}.\]