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
It is given that ∠EPA=∠DPB Now, ∠EPA+∠DPE=∠DPB+∠DPE Therefore, ∠DPA=∠EPB In ΔEBP and ΔDAP, ∠EBP=∠DAP (given) BP=AP (P is midpoint of AB) ∠EPB=∠DPA [proved above] By ASA criterion of congruence, ΔEBP≅ΔDAP
