已知如图,BC为半圆O的直径,AD⊥BC,垂足为D,过点B作弦BF交AD于点E,交半圆O于点F,弦AC与BF交于点H,且
1个回答

证明:(1)∵AE=BE,

∴∠BAD=∠ABE,

∵BC是直径,AD⊥BC,

∴∠ADB=∠BAC=90°,

∴∠ABD+∠BAD=∠ABC+∠C=90°,

∴∠BAD=∠C,

∴∠C=∠ABF,

AB =

AF ;

(2)∵∠C=∠ABF,

Rt△ABH ∽ Rt△ACB,

∴AH:BH=AB:BC,即AH•BC=AB•BH,

∵∠EAH+∠BAD=∠AHB+∠ABH=90°,∠BAD=∠ABE,

∴∠EAH=∠AHB,

∴AE=EH=BE=

1

2 BH,

∴AH•BC=2AB•BE.