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 $|ED|=\sqrt{12}$
 A: 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 $\angle AEB = 135 \deg$. 
A: Take F outside of square such that $\bigtriangleup AEB \equiv \bigtriangleup CFB$.
Then $EF = 2\sqrt2$, that means $\angle CFE = 90^{\circ}$.
$\angle AEB = \angle CFA = 90+45^{\circ}$
image
A: 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, $\angle AEB = 135^\circ$.
A: 
Let $E^\prime$ be the reflection of $E$ through $AB$, $E^{\prime\prime}$ the reflection of $E$ through $BD$. It follows that

*

*$AE^{\prime\prime}=CE=3$.

*$BE^\prime=BE^{\prime\prime}=BD=2$.

*$AE^\prime=AE=1$.

*$\angle CBE^{\prime\prime}=\angle E^\prime BA=\angle EBA$.

It follows further that $\angle E^\prime BE^{\prime\prime}=90^\circ$. Thus we have
$$E^\prime E^{\prime\prime}=BE^\prime\sqrt{2}=2\sqrt{2}, \angle BE^\prime E^{\prime\prime}=45^\circ\tag{5}.$$
So in $\triangle AE^\prime E^{\prime\prime}$ we have
$${AE^\prime}^2+{E^\prime E^{\prime\prime}}^2={AE^{\prime\prime}}^2,$$
or equivalently $\angle AE^\prime E^{\prime\prime}=90^\circ$. Together with (5), we have $\angle BEA=\angle BE^\prime A=135^\circ$.

Btw, I think that this is way too difficult for a multiple choice question.
