Q 5. In Fig. 10.39, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that $\angle B E C=130^{\circ}$ and $\angle E C D=20^{\circ}$. Find $\angle B A C$.
$\angle \mathrm{DEC}+\angle \mathrm{BEC}=180^{\circ} \quad \text { (linear pairs) }$
$\Rightarrow \angle \mathrm{DEC}+130^{\circ}=180^{\circ} \quad\left(\angle \mathrm{BEC}=130^{\circ}\right)$
$\Rightarrow \angle \mathrm{DEC}=180^{\circ}-130^{\circ}$
$\Rightarrow \angle \mathrm{DEC}=50^{\circ}$
In $\triangle \mathrm{DEC}$,
$\angle \mathrm{D}+\angle \mathrm{DEC}+\angle \mathrm{DCE}=180^{\circ}$
$\Rightarrow \angle D+50^{\circ}+20^{\circ}=180^{\circ}$
$\Rightarrow \angle D+70^{\circ}=180^{\circ}$
$\Rightarrow \angle D=180^{\circ}-70^{\circ}=110^{\circ}$
$\angle \mathrm{D}=\angle \mathrm{BAC} \quad$ (angles in the same segment are equal)
$\angle B A C=110^{\circ}$