Question

AB is aline segment And P is its mid point. D and E are point on the same side of AB such that < BAD = < ABE and <EPA = < DPB. Show that. 1.∆ DAP is congruent to ∆EPB. 2.AD =BE.

Answer 1

To Prove : 1.∆ DAP is congruent to ∆EPB. 2.AD =BE.

∠ EPA = ∠ DPB (Given)

⇒ ∠ EPA + ∠DPE = ∠ DPB + ∠DPE
Therefore∠APD = ∠BPE

Consider Triangles DAP and EBP
AP = BP (Given P is midpoint of AB)
∠BAD = ∠ABE (Given)
∠APD = ∠BPE (Proved)

Hence Triangle DAP is congruent to Triangle EBP (By ASA congruence rule)
⇒ AD = BE (CPCT)

Hence Proved

Answer 2

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