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 $\angle BAD=\angle ABE$ and $\angle EPA=\angle DPB$(see in the figure).Show that $\left(i\right)△DAP\cong △EBP$ and $\left(ii\right)AD=BE$

Answer 1

Given:$P$ is the mid-point of $AB$,

So,$AP=BP$ ......$\left(1\right)$

$\angle BAD=\angle ABE$ ......$\left(2\right)$

$\angle EPA=\angle DPB$ ......$\left(3\right)$

To prove:$\left(i\right)△DAP\cong △EBP$

$\left(ii\right)AD=BE$

Proof:Since $\angle EPA=\angle DPB$

We add $\angle DPE$ both sides, we get

$\angle EPA+\angle DPE=\angle DPB+\angle DPE$

$\angle DPA=\angle EPB$ ......$\left(4\right)$

In $△DAP$ and $△EBP$,

$\angle DAP=\angle EBP$ from $\left(2\right)$

$AP=BP$ from $\left(1\right)$

$\angle DPA=\angle EPB$ from $\left(4\right)$

$\therefore △DAP\cong △EBP$ by ASA congruence rule

$\therefore AD=BE$ by C.P.C.T