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Question
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 the given figure). Show that
- ΔDAP ≅ ΔEBP
- AD = BE
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Solution
We have, P is the mid-point of AB.
∴ AP =BP
∠EPA = ∠DPB ...[Given]
Adding ∠EPD to both sides, we get:
∠EPA + ∠EPD = ∠DPB + ∠EPD
⇒ ∠APD = ∠BPE
i. Now, in △DAP and △EBP, we have
∠PAD = ∠PBE ...[∵ ∠BAD = ∠ABE]
AP = BP ...[Proved above]
∠DPA = ∠EPB ...[Proved above]
∴ △DAP ≌ △EBP ...[By ASA congruency]
ii. Since △DAP ≌ △EBP
⇒ AD = BE ...[By Corresponding Parts of Congruent Triangles]
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