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 $\angle EPA=\angle DPB$. Show that
(i) $△DAP\cong △EBP$
(ii) $AD=DE$

Answer 1

i)

It is given that $\angle EPA=\angle DPB$

Now,

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

Therefore,

$\angle DPA=\angle EPB$

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

$\angle EBP=\angle DAP$ (given)

$BP=AP$ ($P$ is midpoint of $AB$)

$\angle EPB=\angle DPA$ [proved above]

By ASA criterion of congruence,

$\Delta EBP\cong \Delta DAP$

ii)

Since $\Delta EBP\cong \Delta DAP$

$AD=BE$ (using CPCT)