Question

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

Answer 1

$\angle TBC=\frac{1}{2}\angle B$ ..(i) (Bisector of $\angle B$)

$\angle TCD=\frac{1}{2}\angle ACD$ ..(ii)

In triangle$ABC,\angle A+\angle B=\angle ACD$ ..(iii)(exterior angle theorem)

In triangle$TBC,\angle T+\angle TBC=\angle TCD$

$\angle T+\frac{1}{2}\angle B=\frac{1}{2}\angle ACD$ (from (i) and (ii))

$\angle T=\frac{1}{2}(\angle ACD-\angle B)$

$\angle T=\frac{1}{2}\angle A$ (from (iii))

So $\angle BTC=\frac{1}{2}\angle BAC$

Hence proved.