Bisectors of interior `/_`B and exterior `/_`ACD of a `Delta`ABC intersect at the point T. prove that `/_BTC = 1/2 /_`BAC.

Explanation

Given In `Delta` ABC, produce BC to DC and the bisectors of `/_`ABC and `/_`ACD meet at point T.
To prove `/_`BTC = 1/2 `/_`BAC
Proof In `Delta`ABC, `/_`C is an exterior angle.
`:." "/_ACD = /_ABC + /_CAB `[exterior angle of a triangle is equal to the sum of two opposite angles]
`rArr" "1/2/_ACD = 1/2/_CAB + 1/2/_ABC " "`[dividing both sides by 2]
`rArr" "/_TCD= 1/2/_CAB + 1/2/_ABC " "`....(i)
[`:. CT is a bisector of /_ACD rArr1/2 /_ACD = /_`TCD ]
In `DeltaBTC, " " /_TCD = /_BTC + C/_`CBT [exterior angles of a equal to the sum of two opposite interior angles]
`rArr" "/_TCD = /_BTC + 1/2/_ABC " "` ....(ii) `[:.BT bisects of /_ABC rArr /_CBT = 1/2 /_`ABC]
From Eqs. (i) and (ii) ,
1/2`/_CAB + 1/2/_ABC = /_BTC + 1/2/_`ABC
`rArr" " /_BTC = 1/2 /_`CAB
`or" " /_BTC = 1/2/_`ABC