Bisectors of angles B and C of a triangle ABC intersect each other at the point O. Prove that ∠BOC = 900+ 1/2∠A.
Answers
Answered by
0
Answer:
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
solution
Answered by
3
Answer:
Step-by-step explanation:
∠OBC=1/2∠ABC
∠OCB=1/2∠ACB
∠OBC+∠OCB+∠BOC=180°(ASP)
∠BOC=180°-(∠OBC+∠OCB) .......(1)
∠A+∠ABC+∠ACB=180°(ASP)
DIVIDE BY 2
∠A/2+∠ABC/2+∠ACB/2=180/2°
1/2∠A=90°-(∠OBC+∠OCB)
ADD 90° ON BOTH SIDES
1/2∠A+90°=90°+90°-(∠OBC+∠OCB)
=180°-(∠OBC+∠OCB)....(2)
RHS OF EQ 1 AND 2 ARE EQUAL
∴∠BOC=1/2∠A+90°
Hence Proved
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