RT△ABC中,∠ACB=90°,AD平分∠BAC交BC于D,DF⊥AB于F,CE⊥AB于E;求证：AF²=AE×AB

Question

RT△ABC中,∠ACB=90°,AD平分∠BAC交BC于D,DF⊥AB于F,CE⊥AB于E;求证：AF²=AE×AB

Answer 1

证明：
∵∠ACB=90°,CE⊥AB,∠CAE=∠BAC
∴△ACE∽△ABC
∴AE/AC=AC/AB
即AC²=AE×AB
∵AD平分∠BAC,DF⊥AB,∠ACB=90°
∴AC=AF
∴AF²=AE×AB