已知在△ABC中,∠BAC=90° ad⊥bc于D,p为ad中点,bp延长线交ac于e,ef垂直bc于F,求证：EF?=

过P点作PQ⊥AB交AB于点Q,在△CEF和△CAD中,由相似三角形原理得：EF/AD=CE/AC , 变形得:EF/CE=AD/AC...①.在△ABD和△ABC中,由相似三角形原理得：AD/AC=BD/AB.②,由①和②得：EF/CE=BD/AB.③同理在△BPD和△BEF中有：PD/EF=PQ/AE,变形得：AE/EF=PQ/PD.④在△APQ和△ABD中可得：PQ/BD=AP/AB,由已知条件p为ad中点可得,PD=AP,所以PQ/BD=PD/AB,变形得：PQ/PD=BD/AB.⑤.由④和⑤得：AE/EF=BD/AB.⑥,由③和⑥可得：AE/EF=EF/CE,变形得：EF=AE×CE.