Answer :

Let us draw the figure as below –


Here, we have


A ΔABC


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 = 1/2BAC


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,


1/2ACD = 1/2ABC + 1/2CAB


TCD = 1/2ABC + 1/2CAB (TCD = 1/2ACD, 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 + 1/2ABC (CBT = 1/2ABC, since BT is the bisector of ABC) - - - - (ii)


From equations (i) and (ii), we can say that


1/2ABC + 1/2CAB = BTC + 1/2ABC


1/2CAB = BTC


BTC = 1/2 CAB


BTC = 1/2 BAC


Hence, proved.


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