Question

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

Answer 1

$\angle BEC={130}^o,\angle ECD={20}^o,\angle BAC=?$
Angle formed by arc $AD$ are $\angle ABD,\angle ACD$ and both are equal.
$\therefore \angle ABD=\angle ACB={20}^o$ {an arc in a circle subtend equal angles anywhere on the circumference}
$\angle ABD={20}^o$
$\angle AEB+\angle BEC={180}^o$ (linear pair)
$\angle AEB+{130}^o={180}^o$
$\angle AEB={180}^o-{130}^o$
$\therefore \angle AEB={50}^o$
Now, in $△BAE$,
$\angle BAE+\angle ABE+\angle AEB={180}^o$
$\angle BAE+{20}^o+{50}^o={180}^o$
$\angle BAE+{70}^o={180}^o$
$\angle BAE={180}^o-{70}^o$
$\therefore \angle BAE={110}^o$
But $\angle BAE$ and $\angle BAC$ are same
$\therefore \angle BAC={110}^o$