Question

如图7,在三角形ABC中,角B=46°三角形的外角角DAC和角ACF的平分线交于点E,求角AEC的度数

Answer 1

∠AEC=180-(∠CAE+∠ACE)

=180-1/2(∠CAD+∠ACF)

又∠CAD=∠B+∠ACB

∠ACF=∠B+∠BAC

所以 ∠AEC=180-1/2(∠CAD+∠ACF)

=180-1/2(∠B+∠ACB+∠B+∠BAC)

=180-1/2(180+∠B)

=90-1/2∠B

=67