<s>[INST] Geometry Problem: Find the area of $ABCD$ square 
My Problem is:

The $E-$  point on the $AC$ diagonal is marked on the $ABCD$ square.If $BE= 13$, $CE=17$ find the square area.

My way:
$$a=\frac{k+17}{\sqrt2}$$
$$13^2=k^2+\frac {(k+17)^2}{2}-\frac {2k(k+17)}{\sqrt2}×\frac{\sqrt2}{2} \Rightarrow k^2+289=338 \Rightarrow k^2=49 \Rightarrow k=7$$
$$S_{ABCD}=\frac {(k+17)^2}{2}=\frac {(7+17)^2}{2}=288 $$
It looks ugly solution.
 Is this solution correct?
 [/INST] Let $F$ be the projection of $E$ on $BC$.

$CE=17$ implies $EF=FC=\frac{17}{\sqrt{2}}$. $BE=13$ and the Pythagorean theorem imply
$$ BF = \sqrt{13^2-\frac{17^2}{2}}=\frac{7}{\sqrt{2}} $$
hence
$$ BC=BF+FC = \frac{17+7}{\sqrt{2}} = 12\sqrt{2} $$
and the area of $ABCD$ is clearly $288$. You do not really need the cosine theorem, just the trick behind one of its proofs.
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