What is $\angle AEB$? Let $E$ be a point inside the square $ABCD$. and $|EC|=3,|EA|=1,|EB|=2$
What is the angle $\widehat {AEB}$?
I can only find $|ED|=\sqrt{12}$

There has to be an easier way, but this works...
Use the law of cosines to get:
$s^2 = 5 - 4\cos(AEB)$
$s^2 = 13 - 12\cos(BEC)$
$2s^2 = 10 - 6\cos(AEB+BEC)$
Eliminate $s^2$, and use $\cos(AEB + BEC) = \cos(AEB)\cos(BEC) - \sin(AEB)\sin(BEC)$. Exchange the sines for cosines, eliminate $\angle BEC$, and a whole bunch of algebra later we get:
$(4\cos(AEB) - 5)(2\cos^2(AEB) - 1) = 0$
Reject the first root for being too large, and we get $\cos(AEB) = \pm \frac{1}{\sqrt{2}}$.
Reject $45^\circ$ because that implies that $s = \sqrt{5 - \frac{4}{\sqrt{2}}} \approx 1.473 < \frac{3}{\sqrt{2}}$, so $E$ would not be in the square. We can similarly reject $225^\circ$ and $315^\circ$.
Therefore, $\angle AEB = 135^\circ$.
