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

题目

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

答案

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