Question

In $\Delta ABC$, AB = AC and D is a point on side AC such that AD = BD = BC. Show that :
$\angle ADB={108}^{\circ }$.

Answer 1

In $△ABD,$

$\angle ADB+\angle A+\angle ABD={180}^o$

$\Rightarrow \angle ADB+36+36={180}^o$

$\Rightarrow \angle ADB=180-72={108}^o$

In an isosceles triangle , opposite to equal sides are equal. In a triangle, one side is extended,

the exterior angle is equal to sum of interior opposite angles