Question

In the given fig $AD$ divides $\angle BAC$ in the ratio $1:3$ and $AD=DB$. Determine the value of $x.$

Answer 1

Given, AD bisects $\angle BAC$
$\frac{\angle BAD}{\angle DAC}=\frac{1}{3}$

Let $\angle BAD=k$ and $\angle DAC=3k$
Now, $\angle BAC+\angle EAC=180$ (liner pair)
$\angle BAD+\angle DAC+\angle EAC=180$
$3k+k+108=180$
$4k=72$
$k=18$
In $△BAD$
Since, $AD=BD$, $\angle DBA=\angle DAB=18$ (Isosceles triangle property)
In $△ABC$,
$\angle EAC=\angle ABC+\angle ACB$ (Exterior angle property)
$108=18+x^{\circ }$
$x={90}^{\circ }$