$AB$ 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:
$(i)$ $\Delta DAP \cong \Delta EBP$
$(ii)$ $AD = BE$
$(i)$ $\Delta DAP \cong \Delta EBP$
$(ii)$ $AD = BE$
(N/A) Given: $P$ is the mid-point of $AB$.
$\therefore AP = BP$
$\angle EPA = \angle DPB$ [Given]
Adding $\angle EPD$ to both sides,we get:
$\angle EPA + \angle EPD = \angle DPB + \angle EPD$
$\Rightarrow \angle APD = \angle BPE$
$(i)$ In $\Delta DAP$ and $\Delta EBP$:
$AP = BP$ [Proved above]
$\angle PAD = \angle PBE$ [Given as $\angle BAD = \angle ABE$]
$\angle APD = \angle BPE$ [Proved above]
By $ASA$ congruence criterion,we have:
$\Delta DAP \cong \Delta EBP$
$(ii)$ Since $\Delta DAP \cong \Delta EBP$,their corresponding parts are equal $(CPCT)$:
$\Rightarrow AD = BE$
$\therefore AP = BP$
$\angle EPA = \angle DPB$ [Given]
Adding $\angle EPD$ to both sides,we get:
$\angle EPA + \angle EPD = \angle DPB + \angle EPD$
$\Rightarrow \angle APD = \angle BPE$
$(i)$ In $\Delta DAP$ and $\Delta EBP$:
$AP = BP$ [Proved above]
$\angle PAD = \angle PBE$ [Given as $\angle BAD = \angle ABE$]
$\angle APD = \angle BPE$ [Proved above]
By $ASA$ congruence criterion,we have:
$\Delta DAP \cong \Delta EBP$
$(ii)$ Since $\Delta DAP \cong \Delta EBP$,their corresponding parts are equal $(CPCT)$:
$\Rightarrow AD = BE$
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