Question

ΔABC内接于⊙O,AD是直径,CE⊥AD于E,CE延长线交AB于点F.

Answer 1

证明:连接CD.AD为直径,则:∠ACD=90°=∠CEA.

∴∠ACF=∠ADC;(均为∠CAE的余角)

又∠B=∠ADC.

∴∠B=∠ACF;又∠BAC=∠CAF.

∴⊿CAF∽⊿BAC,AC/AF=AB/AC,AC²=AF×AB.