Question

If $I$ is the incentre of triangle $ABC$ and $AI$ when produced meets the circumcircle of triangle $ABC$ in point $D$. If $\angle BAC={66}^{\circ }$ and $\angle ABC={80}^{\circ }$. Calculate $\angle BIC$

Answer 1

Join $DB$ and $DC$, $IB$ and $IC$.
Given, if $\angle BAC={66}^{\circ }$ and $\angle ABC={80}^{\circ }$, $I$ is the incentre of the $\Delta ABC$.
As $I$ is the incentre of $\Delta ABC$, $IB$ bisects $\angle ABC$.
Therefore,
$\angle IBC=\frac{1}{2}\angle ABC=\frac{1}{2}\times {80}^{\circ }={40}^{\circ }$
In $\Delta ABC$, by angle sum property
$\angle ACB={180}^{\circ }–(\angle ABC+\angle BAC)$
$\angle ACB={180}^{\circ }–({80}^{\circ }+{66}^{\circ })$
$\angle ACB={180}^{\circ }–{156}^{\circ }$
$\angle ACB={34}^{\circ }$
And since, $IC$ bisects $\angle C$
Thus, $\angle ICB=\frac{1}{2}\angle C=\frac{1}{2}\times {34}^{\circ }={17}^{\circ }$
Now, in $\Delta IBC$
$\angle IBC+\angle ICB+\angle BIC={180}^{\circ }$
${40}^{\circ }+{17}^{\circ }+\angle BIC={180}^{\circ }$
${57}^{\circ }+\angle BIC={180}^{\circ }$
$\angle BIC={180}^{\circ }–{57}^{\circ }$
Therefore, $\angle BIC={123}^{\circ }$