Question

等腰Rt三角形ABC中,AC=AB,角BAC=90°,点D,E是AC上两点,且AD=CE,AF垂直BD于点F,连接DF,

Answer 1

证明：作CM⊥AC交AF延长线于M.

∵∠BAD=∠AGB=90°

∴∠ABD+∠BAF=∠BAF+∠CAM=90°

∴∠ABD=∠CAM

∵AB=CA ∠BAD=∠ACM=90°

∴△BAD≌△ACM

∴∠1=∠M AD=CM

∵AD=CE

∴CE=CM

∵AB=AC ∠BAC=90°

∴∠ABC=∠ACB=45°

∴∠MCF=∠ACM-∠ACB=∠ECF=45°

∵CF=CF

∴△MCF≌△ECF

∴∠M=∠2

∴∠1=∠2