Question

在△ABC中,∠C＝90°,DE垂直平分斜边AB,分别交AB、BC于D、E,若∠CAE＝∠B+30°,求∠AEC的度数

Answer 1

∠AEC=45°.

∠AEC=90°- ∠CAE

=90°- ∠B - 30°

=60°-∠B

=∠B+30°

60°-∠B =∠B+30° 解得∠B=15°

因此∠AEC=45°.