Question

如图,三角形ABC,角C=90度,角ABC的角平分线与角BAC的外角平分线相交于点E,求证：角E=45度

Answer 1

∠EAC=1/2(∠C+∠ABC)

∠E=180°-∠EAB-∠ABE

=180°-（∠EAC+∠BAC）-∠ABE

=180°-1/2∠C-1/2∠ABC-∠BAC-∠ABE

=180°-45°-(∠BAC+1/2∠ABC+∠ABE)

=135°-（∠BCA+∠ABC）

=135°-90°

=45°