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 $\angle IBC$

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,
$\angle IBC=\frac{1}{2}\angle ABC=\frac{1}{2}\times {80}^{\circ }={40}^{\circ }$