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Let E be a point outside of the square $ABCD$ such $\triangle ABE$ is an equilateral triangle. What is the measure of $\angle CED$, in degrees?

I need help with this problem. I made a diagram beforehand to help me with this problem. From doing this, I find that $m(\angle CAE)$ is the same as $m(\angle DBE)$. I think these angles are $150^\circ$ because the angles of a square are each $90^\circ$, plus when we have point $E$ makes an equilateral triangle (which has three $60^\circ$ angles). So $60^\circ + 90^\circ = 150^\circ$. This is all I have so far.

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Correction: I labeled points of square differently. But our solutions mirror one another.

The statement of the problem is below the image.

enter image description here

Note that both $\angle EBC$ and $\angle EAD$ are $150^\circ$ ($90^\circ$ (measure of an angle of the square + $60^\circ$ (measure of each of three angles in an equilateral triangle)). Note also that $AE=AD=EB=BC$, since the lengths of all sides of square and triangle are equal (for triangle $ABE$ to be equilateral, both $EA$ and $EB$ must be the same length as $AB$.)

Since $AE=AD$, triangle $ADE$ is isosceles (and likewise, triangle $EBC$ is isosceles), and so the angles opposite the equal sides are equal. Since we have that $\angle EBC$ and $\angle EAD$ are $150^\circ$, $m(\angle ADE) + m(\angle AED) = m(\angle CEB) + m(\angle BCE) = 180 - 50 = 30^\circ$. So, in particular, $m(\angle AED) = m(\angle CEB) = 15^\circ$. So we have that $m(\angle CED) = m(\angle AEB) - 2m(\angle AED) = 60 - 2(15) = 30^\circ$.

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J. M.: Thank you, I still have not found out how to insert a picture. Actually (not that it matters) it is not the labelling that I described, mine went counterclockwise, with $A$ at top right. – André Nicolas May 17 '11 at 20:57
@user6312: Yes, I do remember you mentioning counter-clockwise...oh, well, we do the best we can when the question is ambiguous leaves it to us to figure out: like "point E outside of square"...of course, since it is placed so as to form an equilateral triangle with A and B, there wasn't much choice! Inserting pix: If you click on "answer", there's a "menu" at the toptop (font, hyperlink, etc.) If you hover over the icons, you'll find one for "image" on it and you can then upload an image file (jpeg., etc), from your computer. – amWhy May 17 '11 at 21:21
It doesn't matter, though; your two solutions are just mirror images of each other. – Michael Lugo May 17 '11 at 21:50

Draw the figure. For definiteness, I would, like most mathematicians do, arrange the labels of the square counterclockwise around the square. So put $A$ at top right, $B$ at top left, $C$ at bottom left, and $D$ at bottom right. Then $E$ is somewhere above the line segment $AB$. (I am doing this to make sure that we are using the same diagram. Yours was I think labelled differently; if so, please relabel.)

It is clear that $\angle EBC$ is $150^\circ$ ($90+60$). So the two angles $BEC$ and $BCE$ add up to $180-150$, which is $30$. But $BE=BC$, so the two small angles are equal. It follows that each is $15^\circ$.

Note: I did not see your diagram, but it looks as if you were probably very close to a solution. The only thing that you likely missed is the fact that what I called $\triangle EBC$ is isosceles.

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@Briana791: Please dont forget to accept an answer. – user9413 May 17 '11 at 20:23

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