Question

一道数学证明题如图,在△ABC中,AD,AE分别是△ABC的高与角平分线,且∠C＞∠B.试证明：∠DAE=1/2(∠C-

Answer 1

∵AE平分∠BAC

∴∠BAE=∠CAE=1/2∠BAC=1/2(180°-∠B-∠C)=90°-1/2∠B-1/2∠C

∵AD⊥BC

∴△ACD是直角三角形

那么∠CAD=90°-∠C

∴∠DAE=∠CAE-∠CAD

=90°-1/2∠B-1/2∠C-(90°-∠C)

=-1/2∠B+1/2∠C

=1/2(∠C-∠B)