Question

The side $BC$ of $\Delta ABC$ is produced to a point $D.$ The bisectors of $\angle ABC$ and $\angle ACD$ meet at point $E.$ If $\angle BAC={60}^{\circ },$ then $\angle BEC$ is :

• A. ${15}^{\circ }$
• B. ${30}^{\circ }$
• C. ${60}^{\circ }$
• D. ${120}^{\circ }$

Answer 1

The correct option is B ${30}^{\circ }$

REF.Image

$\angle ABE=\angle EBC$ { BE bisects $\angle ABC$}

Let $\angle ABE=x$

$\varphi \angle ACE=y$

$\angle ACE=\angle ECD$ { CE bisects $\angle ACD$}

In $\Delta BEC$,

$\angle EBC=\angle BEC=\angle ECD$ { exterior angle is equal to sum of opposite interior angle}

$x+\angle BEC=y$

$\angle BEC=y-x$ \angle _________ (1)

In $\Delta ABC$

$\angle ABC+\angle BAC=\angle ACD$ { from above statement}

$2x+\angle BAC=2y$

$\angle BAC=2y-2x$ __________ (2)

$2\times eq\left(1\right)-eq\left(2\right)$

$2\angle BEC=\angle BAC=0$

$\angle BAC=2\angle BEC$

$\angle BEC=\frac{1}{2}\angle BAC$ $\angle BAC={60}^{\circ }$

$\angle BEC=\frac{1}{2}\times 60={30}^{\circ }$