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(D) In $\Delta CDE$,the exterior angle $\angle BEC$ is equal to the sum of the interior opposite angles.$\angle BEC = \angle EDC + \angle ECD$Given $\

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$110$

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(D) In $\Delta CDE$,the exterior angle $\angle BEC$ is equal to the sum of the interior opposite angles.$\angle BEC = \angle EDC + \angle ECD$Given $\angle BEC = 130^{\circ}$ and $\angle ECD = 20^{\circ}$.$130^{\circ} = \angle EDC + 20^{\circ}$$\angle EDC = 130^{\circ} - 20^{\circ} = 110^{\circ}$Since $\angle EDC$ and $\angle BDC$ represent the same angle,$\angle BDC = 110^{\circ}$.Angles in the same segment of a circle are equal.Therefore,$\angle BAC = \angle BDC$.$\angle BAC = 110^{\circ}$.

In the figure,$A, B, C$ and $D$ are four points on a circle. $AC$ and $BD$ intersect at a point $E$ such that $\angle BEC = 130^{\circ}$ and $\angle ECD = 20^{\circ}$. Find $\angle BAC$. (in $^{\circ}$)

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