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 Fig.) . Show that (i) $\Delta DAP\cong \Delta EBP$ (ii) $AD = BE$

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Solution

## Given$P$ is the midpoint of line segment $AB$.$\angle BAD=\angle ABE$ and $\angle EPA=\angle DPB$ To Prove(i) $∆DAP\cong ∆EBP$(ii) $AD=BE$Proof$\angle BAD=\angle ABE$ and $\angle EPA=\angle DPB$On adding $\angle DPE$on both the sides of two equal angle $\angle EPA=\angle DPB$, we get$\angle EPA+\angle DPE=\angle DPB+\angle DPE$$\angle DPA=\angle EPB$From $∆DAP$ and $∆EBP$$\angle DPA=\angle EPB$ {from above }…… (i)$AP=BP$ { Given }……………… (ii)$\angle BAD=\angle ABE$ { given }…………..(iii)From above three equation both the triangle satisfies $“ASA”$ congruence criterionSo, $\Delta DAP\cong \Delta EBP$(ii) $AD$ and$BE$are equal as they are corresponding parts of congruent triangles$\left(CPCT\right)$.So that, $AD=BE$Hence Proved.

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