如图,A、B、C三点在圆O上,△ABC的高AD、CE交于点H,CE的延长线交圆O于F.求证：AF=AH.

证明：
∵AD⊥BC,CE⊥AB
∴∠ABC+∠BCE＝90,∠DHC+∠BCE＝90
∴∠ABC＝∠DHC
∵∠AHF＝∠DHC
∴∠ABC＝∠AHF
∵∠ABC、∠AFC所对应圆弧都为劣弧AC
∴∠ABC＝∠AFC
∴∠AFC＝∠AHF
∴AF＝AH