Question

In the given figure, $ABCD$ is a cyclic quadrilateral, $\angle ABC=70^{\circ },\overrightarrow{FG}$ bisects $\angle CFA,\overrightarrow{EG}$ bisects $\angle DEB$, $\angle DCE=60^{\circ }$ and $\angle EGF=90^{\circ }$. Find $\angle HEC$.

• A. $20^{\circ }$
• B. $40^{\circ }$
• C. $25^{\circ }$
• D. $45^{\circ }$

Answer 1

Answer: (C)

$\angle DAB=\angle DCE=60^{\circ }$ and $\angle EDC=\angle CBA=70^{\circ }$.
(Exterior angle of a cyclic quadrilateral.)
In $△ABE,\angle DEC=180-(60^{\circ }+70^{\circ })=50^{\circ }$.
Since, $\overline{FG}$ bisects $\angle F$
$\therefore \angle HEC=\frac{\angle DEC}{2}=\frac{50^{\circ }}{2}=25^{\circ }\text{. }$