Question

在△ABC中,AB=AC,AD是BC上中线,P为AD上一点,过C作BA平行线,交BP延长线于F,BF交AC于E,

Answer 1

因为AB=AC,BD=DC,所以AD垂直平分BC.

连CP,则BP=CP,∠CBP=∠BCP.

又由∠ABC=∠ACB,

∠ABC-∠CBP=∠ACB-∠BCP,即∠ABP=∠ACP.

因为CF‖AB,所以∠F=∠ABP=∠ACP,

又因为公共角∠CPF,ΔCPE∽ΔFPC,

PF/PC=PC/PE,从而PF·PE=PC²=PB²