AB is a line segment and P is the mid-point. D and E are points o… - Vidyadip

Question

$AB$ is a line segment and $P$ is the 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:
$i. \Delta DAP \cong \Delta EBP$
$ii. AD = BE ($See figure$)$

✓

Answer

Given that $\angle E P A=\angle D P B$
Adding $\angle EPD$ on both sides, we get
$\angle EPA+\angle EPD=\angle DPB+\angle EPD$
$\Rightarrow \angle A P D=\angle B P E......\text { (i) }$
Also given, $\angle B A D=\angle A B E$
$\Rightarrow \angle P A D=\angle P B E......\text{ (ii) }$
Now in $\Delta APD$ and $\Delta BPE$,
$\angle P A D=\angle P B E. [$from $(ii)]$
$AP = PB [ P$ is the mid$-$point of $AB ]$
$\angle A P D=\angle B P E [$From $(i)]$
Hence, by $\text{ASA}$ congruency criteria;
$\Delta D A P \cong \Delta E B P$
$\Rightarrow \text{AD=BE}[$ By $\text{C.P.C.T.}]$ Proved

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