Question

如图所示,△ABC中,AB=AC,AD⊥BC于点D,DE⊥AB于点E,F为DE中点,AF,CE交于点H,求证：AH⊥CE

Answer 1

∵AB=AC AD⊥BC

∴BD=CD=1/2BC

∵CE⊥AB

∴∠B+∠BDE=∠BDE+ADE=90

∴∠B=∠ADE

而∠BED=∠ADB=90

∠B=∠B

∴△BDE∽△BAD

∴AD∶BD=DE:BE

∵DE=2DF,BD=1/2BC

∴AD:1/2BC=2DF:BE

∴2AD:BC=2DF:BE

∴AD:BC=DF:BE

而∠B=∠ADE

∴△ADF∽△CBE

∴∠BCE=∠DAF

设AD与CE交O

∠AOH=∠COD

∴∠AHO=∠CDA=90

即AF⊥CE