Get CA=CD, ∠ACP=∠DCP,

Because c this =CP, then there is: Δ δPCA?δPCD,

Reason: SAS.

9. Prove that ∵E is the midpoint of ∴BE=CE BC,

In δ abbe and δDCE:

BE=CE，

∠ 1=∠2

AE=DE，

∴δabe≌δdce(sas)，

∴AB=DC。