已知:如图,AB为⊙O的直径,AC为弦,CD⊥AB于D.若AE=AC,BE交⊙O于点F,连接CF、DE. 求证:(1)AE2=AD•AB; (2)∠ACF=∠AED.

问题描述:

已知:如图,AB为⊙O的直径,AC为弦,CD⊥AB于D.若AE=AC,BE交⊙O于点F,连接CF、DE.
求证:(1)AE2=AD•AB;
(2)∠ACF=∠AED.

证明:(1)连接BC,
∵AB为⊙O的直径,
∴∠ACB=90°.
∵CD⊥AB,
∴△ACD∽△ABC.

AC
AD
AB
AC

∵AC=AE,
∴AE2=AD•AB.
(2)∵AE2=AD•AB,∠EAD=∠BAE,
∴△ADE∽△AEB.
∴∠AED=∠B.
∵∠ACF=∠B,
∴∠ACF=∠AED.