Question

2在三角形ABC中,AB=AC,角BAC=90°,点DE在BC上,连接AD AE,角DAE=45°,CE=3,BD=4,

Answer 1

旋转AEB到AE‘B使AC与AB重合连接E’D

∠AED=45°=∠BAE+∠CAD=∠DAE’

AE‘=AE,AD=AD => △AED=△AE’D

=> ED=DE'

又∠B=∠C=45° =>∠DCE‘=90°

所以 ED^2=ED' ^2=DC^2+CE'^2=DC^2+BC^2

以上回答你满意么?