Question

如图,O是△ABC的内角平分线的交点,过O作DE⊥AO交A、B,AC于D、E.求证：BD×CE=OD×OE.

按三角形相似来做，过程要完整

Answer 1

∵∠DAO=∠EAO，∠AOD=∠AOE=90°，AO为公共边，∴△AOD≌△AOE
∴∠AED=∠ADE，∴∠BDE=∠CED
∵∠AED+∠ADE+∠BAC=180°，即2∠AED+∠BAC=180°，∴∠AED=(180°-∠BAC)/2
∴∠BDE=180°-∠AED=180°-(180°-∠BAC)/2=90°+∠BAC/2
∵∠BOC=180°-∠OBC-∠OCB=180°-∠ABC/2-∠ACB/2=180°-(∠ABC+∠ACB)/2
=180°-(180°-∠BAC)/2=90°+∠BAC/2，∴∠BOC=∠BDE
在△BOD和△COE之间，∠BDE=∠CED，∠BOC=∠BDE，∴△BOD∽△COE
∴BD:OD=OE:CE，即BD×CE=OD×OE