Question

In the above figure $E$ is any point on median $AD$ of $△ABC$. $A,B,C$ and $D$ are four points on a circle. $AC$ and $BD$ intersect at a point $E$ such that $\angle BEC={130}^{\circ }$ and $\angle ECD=20$.Find $\angle BAC$

Answer 1

In $\Delta EDC$

$\angle BEC=\angle EDC+\angle DCE$ (by external angle)

${130}^o=\angle EDC+{20}^o$

$\angle EDC={110}^o$

$\angle BAC=\angle BDC$ (by common chord. preperation)

$\therefore \angle BAC={110}^o$