证明:延长CN交AB于Q.∠ANC=∠ANQ=90°,AN=AN,∠CAN=∠QAN.则⊿QAN≌⊿CAN,得:S⊿QAN=S⊿CAN;QN=CN.又CN与BM都垂直于AP,故S⊿PNB=S⊿CNM.所以,S⊿PAN+S⊿PNB=S⊿CAN+S⊿CNM,即S⊿ABN=S⊿ACM.