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$

Answer 1

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 criterion

So, $\Delta DAP\cong \Delta EBP$

(ii) $AD$ and$BE$are equal as they are corresponding parts of congruent triangles$\left(CPCT\right)$.

So that, $AD=BE$

Hence Proved.