Question

△ABC和△ADE都是等腰直角三角形,∠BAC=∠DAE=90°,点M是BE中点,求证:AM⊥CD

Answer 1

延长BA至F,使AF=AB,连结FE并延长交AC于G,交DC于F.

∵∠BAC=∠DAE=90°,

∴∠BAD=∠CAE=90°-∠DAC,

∵∠DAC=90°-∠BAD,∠EAF=90°-∠CAE,

∴∠DAC=∠EAF.

又∵AD=AE,AC=AB=AF,

∴△ADC≌△AEF.

∴∠DCA=∠F.

又∠HGC=∠AGF,

∴∠GHC=180°-∠C-∠HGC=180°-∠F-∠AGF∠GAF=90°,

即FH⊥DC.

∵M是BE中点,A是BF中点,

∴AM‖FE.

∴AM⊥CD