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) ΔDAP ≅ ΔEBP (ii) AD = BE

i)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
ii)Since ΔEBP ≅ ΔDAP
AD = BE (using CPCT)