Question

在△ABC中,∠C=62°,∠A的角平分线与∠B的外角平分线交于点D,求∠D的度数

Answer 1

∠EBD=∠D+∠2

1/2(180°-∠ABC)=∠D+1/2∠BAC,

∴90°-1/2∠ABC=∠D+1/2∠BAC,

∠D=90°-1/2(∠ABC+∠BAC)

=90°-1/2(180°-∠C)

=1/2∠C

=31°.