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 ∠ BEC = 130° and ∠ ECD = 20°. Find BAC.

 

Solution:

We know that the angles in the segment of the circle are equal.

So,

∠ BAC = ∠ CDE

Now, by using the exterior angles property of the triangle

In ΔCDE we get,

∠ CEB = ∠ CDE+∠ DCE

We know that ∠ DCE is equal to 20°

So, ∠ CDE = 110°

∠ BAC and ∠ CDE are equal

∴ ∠ BAC = 110°