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 $\angle BEC=130o$ and $\angle ECD=20\odot$ Find $\angle BAC\dot{}$

Solution: Please refer to video for the figure.
Here, $\angle BEC={130}^{\circ }\text{and}\angle ECD={20}^{\circ }$$\angle AEB+\angle BEC={180}^{\circ }$(Sum of angles formed by a straight line is 180^@)
$\angle AEB=180-130={50}^{\circ }$
Also, $\angle ABD=\angle ACD$ as these angles are subtended by a same chord on the circle.
So,$\angle ABD=\angle ABE={20}^{\circ }$
Now, in $\Delta AEB$
$\angle ABE+\angle AEB+\angle EAB={180}^{\circ }$
$EAB+50+20=180$
$EAB={110}^{\circ }$
So, $\angle BAC=EAB={110}^{\circ }$