Question

A triangle $ABC$ is inscribed in a circle. The internal bisectors of angles $A,B$ and $C$ meet the circumference in $X,Y$ and $Z$ respectively, $\angle YXZ$ is equal to

• A. ${90}^o-\frac{A}{2}$
• B. ${90}^o+\frac{A}{2}$
• C. ${90}^o+A$
• D. ${180}^o-A$

Answer 1

The correct option is A ${90}^o-\frac{A}{2}$

The point where all angle bisector meets = center of circle

Since $O$ is center of circle

$\angle BOC=2\angle A$

$\therefore \angle BOX=\angle COX=A$

Similarly

$\angle AOY=\angle COY=B$

$\angle AOZ=\angle BOZ=C$

Also,

$\angle ZXY=\frac{1}{2}\angle ZOY$

$=\frac{1}{2}(B+C)$ $(A+B+C={180}^{\circ })$

$=\frac{1}{2}(180-A)$

$=90-\frac{A}{2}$