Question

bisector of two interior angles intersect at O . prkve that angleBOC =90+ 1/2angle a​

Answer 1

In △ BOC, we have

∠1+∠2+∠BOC=180° ....(1)

In △ ABC, we have,

∠A+∠B+∠C=180°

∠A+2(∠1)+2(∠2)=180°

∠A/2+∠1+∠2=90°

∠1+∠2=90°-∠A/2

Therefore, in equation 1,

90°-∠A/2+∠BOC=180°

∠BOC=90°+∠A/2