Question

Triangle$A$BC is isosceles with $AB=AC$. The measure of $\angle BAD$ is ${30}^{\circ }$ and $AD=AE$. The measure of $\angle EDC$ is

• A. $5^{\circ }$
• B. ${10}^{\circ }$
• C. ${15}^{\circ }$
• D. ${20}^{\circ }$

Answer 1

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

In $△ABC$
$AB=AC$
$\angle B=\angle C$ (isosceles triangle property) (1)
$\angle ADC=\angle B+30$ (Exterior angle property) (2)
Also, $\angle AED=\angle EDC+\angle C$ (Exterior angle property)
$\angle AED=\angle EDC+\angle B$ (From 1)
Now, given, $AD=AE$
hence, $\angle ADE=\angle AED$ (Isosceles triangle property)
Thus, $\angle ADE=\angle EDC+\angle B$
$\angle ADE+\angle EDC=2\angle EDC+\angle B$
$\angle ADC=2\angle +\angle B$
$\angle B+30=2\angle EDC+\angle B$ (From 2)
$\angle EDC={15}^{\circ }$