△ABC中,BA=BCP为AC上一点AD⊥BC于点D,PM⊥BC于M,PN⊥AB于N求证PM+PN=AD

过点P作PQ⊥AD于Q,
∵AD⊥BC于点D,PM⊥BC于M,
∴PMDQ为矩形,
∴PM=QD,PQ//DM,PM//QD,
∴∠APQ=∠C=∠NAP,
又∵PN⊥AB于N
∴∠ANP=∠AQP=90
∵AP=PA
∴△APN全等△PAN
∴PN=AQ
∴PM+PN=QD+AQ=AD