Question

已知△ABC中,AD⊥BC于D,DE⊥AC于E,F是DE上一点,且满足DF:FE=CD:BD,求证

AF⊥BE

Answer 1

设AD和EF交于M∵DE⊥AC,DF⊥AB∴△ADE和△ADF是直角三角形∵DE=DFAD=AD∴Rt△ADE≌Rt△ADF∴∠EAD=∠FADAE=AF∵AM=AM∴△AME≌△AMF∴∠AME=∠AMF∵∠AME+∠AMF=180°∴∠AME=90°∴AM⊥AF即AD⊥EF