Question

如图:RT△ABC中，AB=AC,∠BAC=90°,D为AC中点，AF⊥BD于E交BC于F，求证:∠ADB=∠CDF

跪求···········

Answer 1

证明:AB=AC,角BAC=90度,则∠ACB=45°.
过点C作CA的垂线,交AF的延长线于M.则∠MCF=∠DCF=45°.
∠ABD+∠BAE=90°;∠CAM+∠BAE=90°.则∠ABD=∠CAM;
又AB=AC;∠BAD=∠ACM=90°.则⊿BAD≌⊿ACM(ASA),得∠ADB=∠M;AD=CM.
又AD=DC,则CM=DC.
又CF=CF;∠DCF=∠MCF.故⊿DCF≌⊿MCF(SAS),得∠CDF=∠M.
所以,∠ADB=∠CDF.(等量代换)