Bisectors of interior $\angle B$ and exterior $\angle ACD$ of a $\triangle ABC$ intersect at the point $T$. Prove that $\angle BTC = \frac{1}{2} \angle BAC.$

(N/A) Given : $\triangle ABC$,produce $BC$ to $D$ and the bisectors of $\angle ABC$ and $\angle ACD$ meet at point $T$.
To prove : $\angle BTC = \frac{1}{2} \angle BAC$
Proof : In $\triangle ABC$,$\angle ACD$ is an exterior angle.
$\therefore \angle ACD = \angle ABC + \angle CAB$
[Exterior angle of a triangle is equal to the sum of two opposite interior angles]
$\Rightarrow \frac{1}{2} \angle ACD = \frac{1}{2} \angle CAB + \frac{1}{2} \angle ABC$ [Dividing both sides by $2$]
$\Rightarrow \angle TCD = \frac{1}{2} \angle CAB + \frac{1}{2} \angle ABC .....(1)$
[Since $CT$ is a bisector of $\angle ACD$,$\angle TCD = \frac{1}{2} \angle ACD$]
In $\triangle BTC$,$\angle TCD$ is an exterior angle.
$\therefore \angle TCD = \angle BTC + \angle CBT$
[Exterior angle of the triangle is equal to the sum of two opposite interior angles]
$\Rightarrow \angle TCD = \angle BTC + \frac{1}{2} \angle ABC .....(2)$
[Since $BT$ is a bisector of $\angle ABC$,$\angle CBT = \frac{1}{2} \angle ABC$]
From equation $(1)$ and $(2)$,we get
$\frac{1}{2} \angle CAB + \frac{1}{2} \angle ABC = \angle BTC + \frac{1}{2} \angle ABC$
$\Rightarrow \frac{1}{2} \angle CAB = \angle BTC$ or $\angle BTC = \frac{1}{2} \angle BAC$
Hence,proved.