Question

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

Answer 1

According to the problem $\angle TBC=\frac{1}{2}\angle B$......(1) and

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

Now from $\Delta$ ABC we have

$\angle A+\angle B=\angle ACD$.......(3).

And from $\Delta$TBC we have

$\angle T+\angle TBC=\angle TCD$

or, $\angle T+\frac{1}{2}\angle B=\frac{1}{2}\angle ACD$[Using (1) and (2)]

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

or, $\angle T=\frac{1}{2}\angle A$.[ Using (3)]

or, $\angle BTC=\frac{1}{2}\angle BAC$.