在△ABC中,AB=AC,AD⊥BC,CF‖AB,BF交AD于P ,AC于E.求证:BP平方=PE乘PF

连接PC,∵AB=AC,AD⊥BC,∴AD是BC的垂直平分线,PB=PC；还有∠PBC=∠PCB；∵AB=AC,∴∠ABC=∠ACB；还有∠ABP=∠ACP；∵CF∥AB,∴∠ABP=∠PFC=∠ACP.在△PFC与△PCE中,∠CPE是公用角,∠PFC=∠PCE,∴△PFC∽△PCE,PF/PC=...