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 ∠ BAD = ∠ ABE and ∠ EPA = ∠ DPB. Show that (i) ∆ DAP ≅ ∆EBP (ii) AD = BE​

Answer 1

Step-by-step explanation:

Refer the attachment ❤️

Answer 2

$\large\underline{\sf{Solution-}}$

Given that, P is the midpoint of line segment AD.

$\implies\sf \: AP = PB$

Further given that,

$\sf \: \angle EPA = \angle DPB$

can be further rewritten as

$\sf \: \angle EPA + \angle EPD= \angle DPB + \angle EPD$

$\implies\sf \: \angle APD = \angle BPE$

Now, Consider

$\sf \: In \: \triangle \:DAP \: and \: \triangle \: EBP$

$\boxed{\begin{aligned}&\sf \: AP = PB \: \: (given)\\ \\& \sf \:\angle PAD = \angle PBE \: \: (given)\\ \\& \sf \: \angle APD = \angle BPE \: (proved)\end{aligned}}\implies\sf \: \triangle DAP \: \cong \: \triangle EBP$

$[ \sf \: By \: ASA \: Congruency \: rule \: ]$

$\implies\sf \: AD = BE \: \:(By \: CPCT)$

Hence,

$\implies\sf \:\triangle \: DAP\:\cong\: \triangle\: EBP \: \: and \: \: AD = BE$