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 $△ABP\cong △ACP$

Answer 1

Given:$△ABC$ is isosceles,

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

Also,$△DBC$ is isosceles,

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

Proof:In $△ABD$ and $△DBC$, we have

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

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

$AD=AD$ (common)

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

So,$\angle BAP=\angle PAC$ by CPCT .........$\left(3\right)$

In $△ABP$ and $△ACP,$

$AB=AC$(given)

$\angle BAP=\angle PAC$ by CPCT (from$\left(3\right)$)

$AP=AP$(common)

$△ABP\cong △ACP$ by SAS congruence rule.