Question

已知:在△ABC中,∠ACB=90°,外角∠PAC的平分线与BC的延长线相交于D,且EA⊥AD,点E在BC上，EF⊥AB于点F求证

求证EC=EF

Answer 1

没写求证的内容如何证明，把求证补上再研究

∵AD是∠PAC的平分线
∴∠PAD=∠CAD
又∵EA⊥AD
∴EAC+CAD=90
∵AC⊥BC
∴∠EAC+∠AEC=90
∴∠CAD=∠AEC
∵∠EAF+∠PAD=90
∠EAF+∠AEF=90
∴∠PAD=∠AEF
∴∠AEF＝∠AEC
∵AE=AE
∴△AEF≌△AEC
∴EC=EF