Question

$AB$ is a line segment and $P$ is its mid-point. $D$ and $E$ are points on the same side of
$AB$ such that $\angle BAD=\angle ABE\text{and}\angle EPA=\angle DPB$ (See the given figure). Show that
$\Delta DAP\cong \Delta EBP$

Answer 1

Given:

$AB$ is a line segment and $P$ is its mid-point.

$\angle BAD=\angle ABE\text{and}\angle EPA=\angle DPB$

To prove: $\Delta DAP\cong \Delta EBP$

Proof:

In figure,

$\angle EPA=\angle DPB$
$\Rightarrow \angle EPA+\angle EPD=\angle DPB+\angle EPD$

[Adding~$\angle EPD$ on both sides]

$\angle APD=\angle BPE\dots \left(i\right)$

In $\Delta DAP$ and $\Delta EBP$

$AP=BP$ [Since, $P$ is a midpoint of $AB$]

$\angle APD=\angle BPE$ [Since, (From~(i)]

$\angle BAD=\angle ABE$ [Given]

By ASA congruence~rule
$\Delta DAP\cong \Delta EBP$

Hence, proved.