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. 

Answer 1

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