Question

In the given figure, I is the incentre of $\Delta ABC$. $BI$ when produced meets the circumcircle of $\Delta ABC$ at $D$. Given $\angle BAC={55}^o,\angle ACB={65}^o$. Calculate (i) $\angle DCA$ (ii) $\angle DAC$ (iii) $\angle DCI$ (iv) $\angle AIC$

Answer 1

Join $IA,IC$ and $CD$.

(i) $IB$ is bisector of $\angle ABC$

$\angle ABD=\frac{1}{2}\angle ABC=\frac{1}{2}({180}^{\circ }-{65}^{\circ }-{55}^{\circ })={30}^{\circ }$

$\angle DCA=\angle ABD={30}^{\circ }$ -----Angles in the same segment

(ii) $\angle DAC=\angle CBD={30}^{\circ }$ -----Angles in the same segment

(iii) $\angle ACI=\frac{\angle ACB}{2}=\frac{{65}^{\circ }}{2}={32.5}^{\circ }$ -----$CI$ is the angular bisector of $\angle ACB$

$\therefore \angle DCI=\angle DCA+\angle ACI={30}^{\circ }+{32.5}^{\circ }={62.5}^{\circ }$

(iv) $\angle IAC=\frac{\angle BAC}{2}=\frac{{55}^{\circ }}{2}={27.5}^{\circ }$-----$AI$ is the angular bisector of $\angle BAC$

$\angle AIC+\angle IAC+\angle ICA={180}^{\circ }$

$\angle AIC={180}^{\circ }-\angle IAC-\angle ICA={180}^{\circ }-{27.5}^{\circ }-{32.5}^{\circ }={120}^{\circ }$