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Question
In the figure, ∠CPD = ∠BPD and AD is the bisector of ∠BAC. Prove that ΔCAP ≅ ΔBAP and CP = BP.
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Solution
In ΔBAP and ΔCAP
∠BAP = ∠CAP ...(AD is the bisector of ∠BAC)
AP = AP
∠BPD + ∠BPA = ∠CPA + ∠CPA = 180°
∠BPD = ∠CPD
⇒ ∠BPA - ∠CPA
Therefore,
ΔCAP ≅ ΔBAP ...(ASA criteria)
Hence, CP = BP.
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