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 $|ED|=\sqrt{12}$

Here is a less algebraicly intensive solution and only requires knowledge of the sine law and the cosine law, and solving quadratics of the form $x^2 = a$(super easy) all of which is material known by the average grade 10 math student in Canada.
From the sine law we know that
\begin{align}
\frac{\sin(ABE)}{1} = \frac{\sin(AEB)}{s} && \frac{\sin(CBE)}{3} =\frac{\sin(90 -ABE)}{3} = \frac{\cos(ABE)}{3} = \frac{\sin(BEC)}{s}
\end{align}
After re arranging and squaring both parts we get
\begin{align} 
\sin^2(ABE) = \frac{\sin^2(AEB)}{s^2} && \cos^2(ABE) = \frac{9\sin^2(BEC)}{s^2}
\end{align}
Recall that $\cos^2(x) = 1 - \sin^2(x)$. 
\begin{align}
\frac{9\sin^2(BEC)}{s^2} &= 1 - \frac{\sin^2(AEB)}{s^2}\\
9(1 - \cos^2(BEC)) &= s^2 - 1 + \cos^2(AEB)
\end{align}
Using the cosine law we know that $5 - 4\cos(AEB) = s^2$ and $13 - 12\cos(BEC) = s^2$. Setting the equations equal and isolating for $\cos(BEC)$ yields $\cos(BEC) = \frac{\cos(AEB)}{3} + \frac{2}{3}$. We can know solve our equation
\begin{align}
9(1 - (\frac{\cos(AEB)}{3} + \frac{2}{3})^2) &= 5 - 4\cos(AEB) - 1 + \cos^2(AEB)\\
9(1 - \frac{\cos^2(AEB)}{9} - \frac{4\cos(AEB)}{9} - \frac{4}{9}) &= 4 - 4\cos(AEB) + \cos^2(AEB)\\
5 - \cos^2(AEB) - 4\cos(AEB) &=  4 - 4\cos(AEB) + \cos^2(AEB)\\
\end{align}
And since the $4\cos(AEB)$ terms cancel out we get
$2\cos^2(AEB) - 1 = 0$ yielding 
\begin{equation}
\cos(AEB) = \frac{\pm 1}{\sqrt{2}} \iff \angle AEB = 45, 135
\end{equation}
And since $AEB$ can't be 45 $\angle AEB = 135 \deg$. 
