Question

AB is a line and p is its mid point D. and E are points on the same side of AB such that Angle BAD = Angle ABE and Angle EPA = Angle DPB show that (1) Traingle DAP = traingle EBP. (2) AD=BE​

Answer 1

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Given,

P is mid-point of AB.

∠BAD = ∠ABE and ∠EPA = ∠DPB

(i) ∠EPA = ∠DPB (Adding ∠DPE both sides)

∠EPA + ∠DPE = ∠DPB + ∠DPE

⇒ ∠DPA = ∠EPB

In ΔDAP ≅ ΔEBP,

∠DPA = ∠EPB

AP = BP (P is mid-point of AB)

∠BAD = ∠ABE (Given)

Therefore, ΔDAP ≅ ΔEBP by ASA congruence condition.

(ii) AD = BE by CPCT.