The internal bisectors of angle ABC and angle ACB of triangle ABC meet at O. Let's prove that angle BOC = 90°+½BAC
Answers
Answered by
0
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,
∠1+∠2+∠BOC=180
o
In △ABC,
∠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,
90
o
−
2
∠A
+∠BOC=180
o
∠BOC=90
o
+
2
∠A
Answered by
0
Step-by-step explanation:
hfsuIgtgggglklkkkohsooyffhjioo()→^π√
Similar questions