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) $AD=BE$

Answer 1

Given:

In figure,

$P$ is mid point of $AB$.

$\angle EPA=\angle DPB$

To prove:

(i) $\Delta DAP\cong \Delta EBP$

(ii) $AD=BE$

Proof:

Lets take, $\angle EPA=\angle DPB$

$\Rightarrow \angle APE+\angle EPD=\angle BPD+\angle DPE$

$\Rightarrow \angle APD=\angle BPE$ ....(1)

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

$\angle DAP=\angle EBP$ [Since, Given]

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

$\angle APD=\angle BPE$ [Since, Proved]

By $A.S.A$. Congruence rule,

$\Delta DAP\cong \Delta EBP$ ----Proof (i)

Since, if two triangles are congruent then their corresponding parts are equal.

$\Rightarrow AD=BE$ -----Proof (ii)

Hence, proved.