Question

$A,B,C$ and $D$ are four points on a circle. $AC$ and $BD$ intersect at a point $E$ such that $\angle BEC={130}^o$ and $\angle ECD={20}^o$, then $\angle BAC$ is __________.

• A. ${110}^o$
• B. ${100}^o$
• C. ${90}^o$
• D. ${120}^o$

Answer 1

The correct option is A ${110}^{\circ }$

$\angle BEC+\angle CED={180}^{\circ }$

$\Rightarrow \angle CED={50}^{\circ }$

Also, in $△CED$, sum of all interior angles is ${180}^{\circ }$

$\therefore \angle CDE={180}^{\circ }-{50}^{\circ }-{20}^{\circ }={110}^{\circ }$

For segment $BADCB$, $\angle BAC$ and $\angle BDC$ are in same segment so must be equal.

$\therefore \angle BAC=\angle CDB$

$\angle BAC={110}^{\circ }$