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 (see Fig). Show that
(i) ∆ DAP ≅ ∆ EBP
(ii) AD = BE
It is given that EPA = DPB
∠EPA +∠ DPE = ∠DPB + ∠DPE
∠DPA = ∠EPB
In ∆DAP and ∆EBP,
∠DAP = ∠EBP (Given)
AP = BP (P is mid-point of AB)
∠DPA = ∠EPB (From above)
∠∆DAP ∠∆EBP (ASA congruence rule)
∴ AD = BE (By CPCT)