Q.
In the given figure, $A B C D$ is a cyclic quadrilateral, $\angle A B C=70^{\circ}, \overrightarrow{F G}$ bisects $\angle C F A, \overrightarrow{E G}$ bisects $\angle D E B$, $\angle D C E=60^{\circ}$ and $\angle E G F=90^{\circ}$. Find $\angle H E C$.
Geometry
Solution:
$\angle D A B=\angle D C E=60^{\circ}$ and $\angle E D C=\angle C B A=70^{\circ}$.
(Exterior angle of a cyclic quadrilateral.)
In $\triangle A B E, \angle D E C=180-(60^{\circ}+70^{\circ})=50^{\circ}$.
Since, $\overline{F G}$ bisects $\angle F$
$\therefore \angle H E C=\frac{\angle D E C}{2}=\frac{50^{\circ}}{2}=25^{\circ} \text {. }$
