# $ABCD$ is a square and $AEB$ is an equilateral triangle. Find $\angle DEC$.

I found this question in an exam question paper.
But I was stumped by it. It seemed that all the answers were correct.
I tried:

Let $\angle ADE=x$ and $\angle EDC=y$.
$\angle DAB=90^{\circ}=\angle DAE+\angle EAB=30^{\circ}+60^{\circ}$.
$\text{ref}\angle AEB=300^{\circ}$.
And we know that $x+y=90^{\circ}$
But now I am stuck. How to proceed?

• Explain the downvote, downvoter. Sep 14, 2015 at 4:20
• See the edit, retract the downvote please. Sep 14, 2015 at 4:21
• – Blue
Sep 14, 2015 at 4:22
• Please dont mark duplicate. It is indeed different. Sep 14, 2015 at 4:23
• We have $x = \frac{180^\circ-30^\circ}{2} = 75^\circ$. So, $y = 15^\circ$. Then, $\angle DEC = 150^\circ$.
– GAVD
Sep 14, 2015 at 4:30

$\triangle AEB$ is equilateral, so $|AE|=|AB|=|AD|$. Hence $\triangle DAE$ is isosceles, so $\angle ADE = \angle AED$, and so $\angle ADE = \frac{1}{2}(\pi-\frac{\pi}{6}) = 5\pi/12$.
The same procedure works on the other side of the square, and hence $\angle DEC = 2\pi-\pi/3-2 \times 5\pi/12 = 5\pi/6 = 150˚$
• Ohh! I forgot that $BE=BC$!! Sep 14, 2015 at 4:32
$DAE$ & $CEB$ are triangles having $BC=BE$ & $AD=AE$ now you can find $x$ ($=75^\circ$) and the others are easy.