Finding the measurement of an angle

Question

I have been stumped on this problem for a couple days now, and I would like some help solving it.

Here is the picture that I drew up:

$ABCD$ is a regular square. Line $FG$ is a perpendicular bisector of line $CB$. You have to find the measurement of $\angle AEC$.

I know that $\angle CEG$ is $30^\circ$ because $\triangle ECB$ is an equilateral triangle since $EC$ and $CB$ are congruent due to both being the radius of the circle. And $EB$ is the reflexive line of $EC$, making them congruent.

What do I do from here? Thank you so much!

Answer 1

$\triangle CGE\cong\triangle BGE$ $\Longrightarrow$ $CE=EB=BC$ $\Longrightarrow$ $\triangle EBA$ is isosceles, $\triangle CBE$ is equilateral.
$\angle EAB=90^{\circ}-\dfrac{\angle EBA}{2}=90^{\circ}-\dfrac{\angle BEG}{2}=90^{\circ}-\dfrac{\angle BEC}{4}=75^{\circ}$
$\therefore$ $\angle AEC=\angle AEB+\angle BEC=135^{\circ}$

Answer 2

nowwe join BE.FG is the perpedicular bisector. SO CE=BE=10

• WE see in $\bigtriangleup$BEC CE=BE=BC=10 so $\angle$BEC=60
• NOW in $\bigtriangleup$AEB BE=AB=10 SO $\angle$AEB=$\angle$EAB
• now FROM $\bigtriangleup$CEB EG=$\frac{\sqrt{3}}{2}$.10 =5$\sqrt{3}$
• FG=10. SO EF= 10$-$5$\sqrt{3}$
• in $\bigtriangleup$AEF
• tan$\angle$EAF=$\frac{EF}{FA}$=$\frac{10-5\sqrt{3}}{5}$=2$-$$\sqrt{3}$
• so $\angle$EAF=15 . SO WE GET $\angle$EAB=75[90-15]
• SO $\angle$AEB=75
• so $\angle$AEC=60+75=135