已知：如图,△ ABC中,（AB＞AC）,AD为高,P为AD上任一点.求证：PB-PC＞AB-AC.

PB²=BD²+PD², PC²=CD²+PD², 两式相减,PB²-PC²=BD²-CD²又AB²=BD²+AD²,AC²=CD²+AD², 两式相减,AB²-AC²=BD²-CD²于是PB²-PC²=AB²-AC²,两边因式分解,(PB+PC)(PB-PC)=(AB+AC)(AB-AC),移项整理,(PB-PC)/(AB-AC)=(AB+AC)/(PB+PC)因为等号右边(AB+AC) 明显大于(PB+PC),即(AB+AC)/(PB+PC)＞1所以等号左边(PB-PC)/(AB-AC)＞1,即PB-PC＞AB-AC