In ∆ABC, the bisectors of <B and <C intersect each other at a point 0. Prove the <BOC=90°+1/2+<2.
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Step-by-step explanation:
ANSWER
Given :
A △ ABC such that the bisectors of ∠ ABC and ∠ ACB meet at a point O.
To prove :
∠BOC=90
o
+
2
1
∠A
Proof :
In △ BOC, we have
∠1+∠2+∠BOC=180
o
....(1)
In △ ABC, we have,
∠A+∠B+∠C=180
o
∠A+2(∠1)+2(∠2)=180
o
2
∠A
+∠1+∠2=90
o
∠1+∠2=90
o
−
2
∠A
Therefore, in equation 1,
90
o
−
2
∠A
+∠BOC=180
o
∠BOC=90
o
+
2
∠A
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