已知三角形ABC中,∠A=90°,AD是BC边上的高,BE是角平分线,且交AD于P.求证:AE=AP

在BC边上截取BG = AB
连结PG,EG
因为BP是∠ABC的角平分线
∴∠ABP = ∠GBP
∵AB = BG,BP = BP
∴△ABP ≌△GBP
∴AP = PG,
∠BAP = ∠BGP
∵∠BAC = 90°,AD⊥BC
∴∠BAD = ∠C
∴∠PGD = ∠C
∴PG//AC
∵AB = BG,∠ABE = ∠GBE,BE = BE
∴△ABE≌△GBE
∴∠ABE = ∠BGE = 90°
∴EG⊥BC
∵AD⊥BC
∴AD//EG
∴四边形APGE是平行四边形
∴PG = AE
∵AP = PG
∴AE = AP