Question

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

Answer 1

In $\Delta ABC$, produce BC to D and the bisectors of $\angle ABCand\angle ACD$ meet at point T.
To prove: $\angle BTC=\frac{1}{2}\angle BAC$

Proof:
In $\Delta ABC,\angle C$ 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]
$\Rightarrow \frac{1}{2}\angle ACD=\frac{1}{2}\angle CAB+\frac{1}{2}\angle ABC$
$\Rightarrow \angle TCD=\frac{1}{2}\angle CAB+\frac{1}{2}\angle ABC$
$[\because CTisabisectorof\angle ACD\Rightarrow \frac{1}{2}\angle ACD=\angle TCD]$
$In\Delta BTC,\angle TCD=\angle BTC+\angle CBT$
[exterior angle of a triangle is equal to the sum of two opposite interior angles]
$\Rightarrow \angle TCD=\angle BTC+\frac{1}{2}\angle ABC$
$[\because BTbisectsof\angle ABC\Rightarrow \angle CBT=\frac{1}{2}\angle ABC]$
From equations (i) and (ii),
$\frac{1}{2}\angle CAB+\frac{1}{2}\angle ABC=\angle BTC+\frac{1}{2}\angle ABC$
$\Rightarrow \angle BTC=\frac{1}{2}\angle CAB$
$or\angle BTC=\frac{1}{2}\angle BAC$