Question

the internal bisectoes of b and c of triangle abc intersect at d then show that d =90+ a by 2
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Answer 1

Answer:

Given :

A △ ABC such that the bisectors of ∠ ABC and ∠ ACB meet at a point O.

To prove :

∠BOC=90  + 1/2<A

Proof :

In △ BOC, we have

∠1+∠2+∠BOC=180  

o

     ....(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 eq 1,

90  -∠A /2+∠BOC=180  

∠BOC=90 + <A/2

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