Question

In the given figure, $△ABC$ is inscribed in a circle. The bisector of $\angle BAC$ meets BC at D and the circle at E. If EC is joined and $\angle ECD$ $={30}^o$, find $\angle BAC$.

• A. ${30}^o$
• B. ${40}^o$
• C. ${50}^o$
• D. ${60}^o$

Answer 1

The correct option is D ${60}^o$

Given, $AD$ is perpendicular bisector of $BC$

and $\angle ECD={30}^o$

In $△DCE$
$\angle DEC={180}^{\circ }-(\angle DCE+\angle CDE)\angle DEC={180}^{\circ }-(30+90{)}^{\circ }\angle DEC={60}^o$

Also, $\angle ABC=\angle AEC={60}^{\circ }$ (half angle property)

In $△ABD$

$\angle ADB={90}^o,\angle ABD={60}^o$

Hence,

$\angle BAD={180}^{\circ }-(90+60{)}^{\circ }={30}^o$

$\angle BAC=2\angle BAD={60}^o$ (angular bisector)