Question

$△ABC$ and $△DBC$ are two isosceles triangles on the same base $BC$ and vertices $A$ and $D$ are on the same side of $BC$ as in figure.If $AD$ is extended to intersect $BC$ at $P,$Show that $△ABD\cong △ACD$

Answer 1

Given:$△ABC$ is isosceles,

$AB=AC$ .....$\left(1\right)$

Also,$△DBC$ is isosceles,

$DB=DC$ .....$\left(2\right)$

Proof:In $△ABD$ and $△ACD$, we have

$AB=AC$ from $\left(1\right)$

$BD=DC$ from $\left(2\right)$

$AD=AD$ (common)

$△ABD\cong △ACD$ by SSS congruence rule.