Question

如图已知在△abc中∠b与∠c的平分线交于点p,当∠a=112°求∠bpc的度数；当A=α时求∠bpc的度数

Answer 1

∵∠A+∠ABC+∠ACB=180°

∴∠ABC+∠ACB=180°-∠A

∵BP平分∠ABC,PC平分∠ACB,∠1+∠2+∠3+∠4=180°-∠A

∴∠2+∠4=1/2(180°-∠A)

∠A=112°时∠bpc=180°-（∠2+∠4）=90+1/2∠A=146°

A=α时,∠bpc=180°-（∠2+∠4）=90°+1/2α