Question

如图,△ABC的两外角平分线交于点P,△ABC两内角的平分线交于点Q,△ABC的内角的平分线交于点Q,△ABC的内角BM

Answer 1

∠BQC=180 °-∠QBC-∠QCB

=180°-1/2(∠ABC+∠ACB)

=180°-1/2(180°-∠A)

=90° +1/2 ∠A

∠QBP=∠QCP=90°

=>∠P=360°-90°-90°-∠BQC=90° -1/2 ∠A

∠M=90° - ∠MQC=90° -1/2(∠ABC + ∠ACB)

=90° -1/2(180°-∠A)=1/2∠A