Question

三角形ABC中,角C>角B,AD垂直BC于D,AE平分角BAC.求证：角DAE=1/2（角C-角D）.

Answer 1

∠EAD=∠EAC-∠CAD

=∠EAC-(90°-∠C)

=∠EAC+∠C-90°

∠EAD=90°-∠AED

=90°-(∠B+∠BAE)

=90°-∠B-∠BAE

2∠EAD=∠EAC+∠C-90°+90°-∠B-∠BAE

=∠C-∠B

=>>∠EAD=1/2（∠C-∠B）