Question

如图②,若CE是△ABC的外角∠ACF的角平分线,交BA延长线于点E,且α-β=30°,求∠DCE的度数.

Answer 1

∠ACF=α+β

所以∠ACE=(α+β)/2

∠DCA=90°-α

所以∠DCE=∠DCA+∠ACE

=90°-α+(α+β)/2

=90°+(β-α)/2

=90°-30°/2

=75°