如图,CD是Rt△ABC斜边AB上的高,E为BC上任意一点,EF垂直AB于F,求证：AC^2=AD*AF+CD*EF

从E向 CD做垂线,垂足为 G
∠ECG + ∠DCA = 90
∠A + ∠DCA = 90
所以 ∠ECG = ∠A
同时 ∠EGC = ∠CDA = 90
所以 △EGC ∽ △CDA
EG/GC = CD/DA
EFDG 是矩形,所以 EG = FD
FD/GC = CD/DA
FD*DA = GC*CD
(AF-AD)*AD = (CD-GD)*CD
又因为 GD = EF
所以 (AF-AD)*AD = (CD-EF)*CD
AF*AD - AD^2 = CD^2 - EF*CD
CD^2 + AD^2 = AF*AD + CD*EF
AC^2 = AD*AF + CD*EF