Will be a property of an Isosceles Triangle?

Let $\angle ABC=\angle ACB=2\alpha$ . Let $E$ be a point on $BC$ such that $BE=BD$ .

Then, $EC=BC-BE=BC-BD=AD$ .

Since $BD$ is the internal angle bisector of $\angle ABC$ , $\frac{AD}{AB}=\frac{DC}{BC}$ $\Rightarrow$ $\frac{EC}{AB}=\frac{DC}{BC}$

Observe that, in $\bigtriangleup ECD$ & $\bigtriangleup ABC$ , $\frac{EC}{AB}=\frac{DC}{BC}$ & $\angle ECD=\angle BCA=\angle ABC$ ; That is, both the triangles have two of their sides in proportion with two sides of the other triangle and the angle between these two sides is equal in both of them. Thus, $\bigtriangleup ECD$ $\sim$ $\bigtriangleup ABC$ .

$\Rightarrow$ $\angle CDE=2\alpha$ & $\angle DEB= 2\alpha+2\alpha=4\alpha$ .

Also, observe that, $\bigtriangleup BED$ is isosceles ( $BD=BE$ ). $\Rightarrow$ $\angle EDB=\angle DEB=4\alpha$

In $\bigtriangleup BED$ , $\angle EBD=\alpha$ & $\angle EDB=\angle DEB=4\alpha$ . Thus, $\alpha+4\alpha+4\alpha=180$ $\Rightarrow$ $\alpha=20^{\circ}$

Therefore, $\angle BAC=180-4\alpha=180-4\cdot20=\boxed {100^{\circ}}$ .