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:

(i) $\angle DBC$ (ii) $\angle IBC,$ (iii) $\angle BIC.$

Answer 1

Given $\angle BAC={66}^{\circ }$ and $\angle ABC={80}^{\circ }$

(i) $\angle DBC=\angle DAC$ (angles in the same segment)

Since I is the incenter. The incentre of the triangle is the point of intersection of the internal bisector of angles.

$\angle DAC=\frac{1}{2}\angle BAC=\frac{1}{2}\times {66}^{\circ }={33}^{\circ }$

Therefore $\angle DBC={33}^{\circ }$

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

(iii) In $△ABC,\angle ACB=180-\angle ABC-\angle BAC=180-80-66={34}^{\circ }$

Also $\angle ICB=\frac{1}{2}\angle BCA=\frac{1}{2}\times {34}^{\circ }={17}^{\circ }$

ie. $\angle ICB=\frac{1}{2}\times {34}^{\circ }={17}^{\circ }$

$\angle BIC={180}^{\circ }-\angle IBC-\angle ICB={180}^{\circ }-{40}^{\circ }-{17}^{\circ }={123}^{\circ }$