Let us draw the figure as below –
Here, we have
BC is extended to D.
Let BT be the bisector of ∠B of the triangle.
Also, let us assume the bisector of ∠ACD to be CT.
It is given that BO and CT intersect at point T.
We have to prove that ∠BTC = 1/2∠BAC
We know, if a side of a triangle is produced, the exterior angle so formed is equal to the sum of the two interior opposite angles. Here, ∠ACD is an exterior angle and the two interior opposite angles are ∠ABC and ∠CAB.
⇒ ∠ACD = ∠ABC + ∠CAB
Dividing both sides of the equation by 2,
1/2∠ACD = 1/2∠ABC + 1/2∠CAB
⇒ ∠TCD = 1/2∠ABC + 1/2∠CAB (∠TCD = 1/2∠ACD, since CT is the bisector of ∠ACD) - - - - (i)
Again, in ΔBTC, ∠TCD is an exterior angle of the triangle and the two opposite interior angles are ∠BTC and ∠CBT.
⇒∠TCD = ∠BTC + ∠CBT
⇒ ∠TCD = ∠BTC + 1/2∠ABC (∠CBT = 1/2∠ABC, since BT is the bisector of ∠ ABC) - - - - (ii)
From equations (i) and (ii), we can say that
1/2∠ABC + 1/2∠CAB = ∠BTC + 1/2∠ABC
⇒ 1/2∠CAB = ∠BTC
⇒ ∠BTC = 1/2 ∠CAB
⇒ ∠BTC = 1/2 ∠BAC
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