Question

$△ABC$ and $△DBC$ are two iosceles triangle on the same base $BC$ and vertices $A$ and $D$ are on the same sides of $BC$. If $AD$ is exerted to intersect $BC$ at $P$, show that
$△ABP\cong triangleACP$

Answer 1

Given:$△ABC$ is isosceles, $AB=AC$ ......$\left(1\right)$

Also,$△DBC$ is isosceles, $DB=DC$ ......$\left(2\right)$

In $△ABC$ and $△DBC$, we have

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

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

$AD=AD$(common)

$△ABD\cong △ACD$(SSS congruence rule)

So, $\angle BAP=\angle PAC$ by CPCT ......$\left(1\right)$

In $△ABP$ and $△ACP$,

$AB=AC$(given)

$\angle BAP=\angle PAC$ from $\left(1\right)$

$AP=AP$(common)

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