Question

BC为圆O的直径,AD垂直于BC,过点B作弦BF交AD于E,交半圆O于点F,弦AC于BF交于点H,AE=BE

Answer 1

连接CF、AB、BD、DO

BC是直径.AD垂直ABC,则=∠BAD=∠BDA

又∠BDA=∠BCA

BE=EA

则∠ABE=∠BAE

所以∠ABE=∠BAE=∠BAD=∠BDA=∠BDA=∠BCA

∠EHA=∠BCH+∠HBC=∠BDA+∠HBC=∠BAD+∠HBC

=∠ABE+∠HBC=∠ABO

∠BOA=2∠BCA=2∠BDA=2∠BAD=∠BAD+∠ABE=∠AEH

△ABO∽△AHE

AH/AB=AE/AO

AH/AE=AB/AO

AH/BE=AB/AO

AH/BE=2AB/BC=AB/AO

AH*BC=2AB*BE