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. 7.22). Show that
(i) Δ DAP ≅Δ EBP
(ii) AD = BE
It is given in the question that:
P is the mid-point of AB
∠BAD = ∠ABE and,
∠EPA = ∠DPB
(i) ∠EPA = ∠DPB (Adding ∠DPE both sides)
∠EPA + ∠DPE = ∠DPB + ∠DPE
∠DPA = ∠EPB
In 
∠DPA = ∠EPB
AP = BP (P is the mid-point of AB)
∠BAD = ∠ABE (Given)
Therefore,
By ASA congruence,
(ii) AD = BE (By c.p.c.t)
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