Question

In $△ABC,AB=AC$. Points $D$ and $E$ are on the sides $BC$ and $AC$ respectively such that $AD=AE$. If $\angle BAD={30}^{\circ }$, then the measure of $\angle EDC$ is:

• A. ${10}^{\circ }$
• B. ${15}^{\circ }$
• C. ${20}^{\circ }$
• D. ${25}^{\circ }$

Answer 1

The correct option is B ${15}^{\circ }$

Given, $AB=AC$
$\Rightarrow \angle B=\angle C=x^{\circ }$ (Isosceles triangle Property)
Given, $AD=AE$
$\Rightarrow \angle D=\angle E=y^{\circ }$ (Isosceles triangle Property)
Sum of angles $={180}^o$
$\angle DAC+\angle ADE+\angle AED={180}^o$
$\angle DAC={180}^o-2y$
In $\Delta DEC,\angle y$ is external angle
$\angle DEA=\angle EDC+\angle ECD$ (Exterior angle property)
$\Rightarrow y^{\circ }=a+x^{\circ }\Rightarrow a^{\circ }=y-x^{\circ }$
In $\Delta ABC,$
Sum of angles $={180}^o$
$\angle ABC+\angle ACB+\angle BAC={180}^o$
$2x^{\circ }+{30}^{\circ }+{180}^{\circ }-2y^{\circ }={180}^{\circ }$
$2(y-x)={30}^{\circ }$
$y-x={15}^{\circ }=a^{\circ }$
$\therefore \angle EDC={15}^{\circ }$
Hence, option $B$ is correct.