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Question
In a ΔABC, it is given that AB = AC and the bisectors of ∠B and ∠C intersect at O. If M is a point on BO produced, prove that ∠MOC = ∠ABC.
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Solution
Given that in , ΔABC,
AB=AC and the bisector of ∠B and ∠C intersect at O and M is
a point on BO produced
We have to prove ∠ MOC=∠ABC
Since,
AB =AC ⇒ΔABC is isosceles ⇒ ∠B=∠C(or) ∠ABC=∠ACB
Now,
BO and CO are bisectors of ∠ABC and ∠ACB respectively
⇒ABO=∠OBC=∠ACO=∠OB=`1/2` ∠ABC=`1/2`∠ACB ............(1)
We have, in ΔOBC
∠OBC +∠OCB +∠BOC =180° .............(2)
And also
∠BOC +∠COM =180° ..................(3)[Straight angle]
Equating (2) and (3)
⇒ ∠OBC+∠OCB+-∠BOC=∠BOC+∠MOC
⇒ ∠OBC+∠OBC=∠MOC [∵ from (1)]
⇒ 2∠OBBC=∠MOC
⇒2`(1/2∠ABC)`=∠MOC [∵from (1)]
⇒ ∠ABC=∠MOC
∴ ∠MOC=∠ABC
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