Question

如图,△ABC中,∠C=70°,AD是∠CAB的平分线,BD是△ABC的外角平分线,AD与BD交于D,求∠D的度数.

Answer 1

∵AD是∠CAB的平分线

∴∠BAD=1/2∠BAC

∵∠DBC=1/2(180°-∠ABC)=90°-1/2∠ABC

∴∠D=180°-(∠BAD+∠ABC+∠DBC)

=180°-(1/2∠BAC+∠ABC+90°-1/2∠ABC)

=90°-1/2(∠BAC+∠ABC)

=90°-1/2(180°-∠C)

=1/2∠C

=35°