Question

In the figure, ∠CPD = ∠BPD and AD is the bisector of ∠BAC. Prove that ΔCAP ≅ ΔBAP and CP = BP.

Answer 1

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.