$ABCD$ is a square with an internal point $E$. $AE=1, CE=2$ and $DE=3$. Find $\angle AEC$

Question

Given a square $ABCD$ with a point $E$ inside, $AE=1, CE=2$ and $DE=3$.

Find the angle $\angle AEC=?$

I solved this problem in a pretty synthetic way, which I’ll post below. I want to see if there are any other possible approaches or if my answer has faults.

Answer 1

Let $$\angle AEC=x, ~~\angle CED=y,~~ \angle AED=z,~~AC=a\Rightarrow AD=\sqrt2a$$

Cosine law:

$$\begin{align}\triangle ACE\Longrightarrow\cos x&=\frac{5-a^2}{4}\tag{1}\\ \\ \triangle CDE\Longrightarrow\cos y&=\frac{13-a^2}{12}\\ \\ \triangle ADE\Longrightarrow\cos z&=\frac{5-a^2}{3} \end{align}$$ We also have $$\cos(x+z)=\cos(2\pi-y)=\cos (y)$$ hence

$$\cos(x)\cos(z)-\sin(x)\sin(z)=\cos (y)$$

Move terms

$$\cos(x)\cos(z)-\cos (y)=\sin(x)\sin(z)$$

We get $$\frac{5-a^2}{4}\cdot\frac{5-a^2}{3}-\frac{13-a^2}{12}=\sqrt{1-\left(\frac{5-a^2}{4}\right)^2}\cdot \sqrt{1-\left(\frac{5-a^2}{3}\right)^2}$$

Square both size and simplify

$$a^2 (17 - 10 a^2 + a^4)=0$$

We get

$$a^2 =5\pm 2\sqrt2$$

From $\triangle ACE$ we know $a^2>1^2+2^2=5$, hence

$$a^2 =5+2\sqrt2$$

Plug into (1), we get

$$\cos x=-\frac{\sqrt2}{2}$$

Therefore,

$$x=135^\circ$$

Answer 2

This is my approach:

Rotate $\triangle AEC$ counterclockwise by $90^\circ$ about point $C$ and form $\triangle CFD$. Since $CE=CF=2$, and $\angle ECF=90^\circ$, we know that $EF=2\sqrt{2}$, and that $\angle CEF=\angle CFE=45^\circ$. Notice that the sides of $\triangle EFD$ are Pythagorean Triples $(3^2=(2\sqrt{2})^2+1^2)$, that means $\angle EFD=90^\circ$.

Therefore $x=45^\circ+90^\circ=135^\circ$