Question

In the figure below lines A'A" and C'C" are parallel. AB is the bisector of $\angle$CAA" and BC is the bisector of $\angle$ ACC". What is the measure of $\angle$ABC?

• A. 60°
• B. 70°
• C. 80°
• D. 90°

Answer 1

The correct option is D 90°


Given: A'A" || C'C"
AB is the angle bisector of $\angle CAA"$ and CB is the angle bisector of $\angle ACC"$

Let $\angle ACC"=x$
$\angle ACC"+\angle CAA"={180}^{\circ }$
(Since co-interior angles are supplementary)
$\angle CAA"={180}^{\circ }-x$

Since BC is the angle bisector of $\angle ACC"$,
$\angle BCC"=\frac{x}{2}$
$\angle BCA=\frac{x}{2}$

Similary, since AB is the angle bisector of $\angle CAA"$,
$\angle CAB=\frac{{180}^{\circ }-x}{2}={90}^{\circ }-\frac{x}{2}$
$\angle BAA"=\frac{{180}^{\circ }-x}{2}={90}^{\circ }-\frac{x}{2}$

In $△ABC$,
$\frac{x}{2}+({90}^{\circ }-\frac{x}{2})+\angle ABC={180}^{\circ }$
$\Rightarrow \angle ABC={90}^{\circ }$