<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 $ABE_1E$  be a parallelogram. 
Thus, $BECE_1$ is isosceles trapezoid ($BE=E_1C=13$) and since $EE_1\perp BC$,
we obtain: 
$$BE_1^2+EC^2=BE^2+E_1C^2$$ or
$$BE_1^2+17^2=13^2+13^2,$$ which gives $$BE_1=7.$$ 
Id est, $$AC=AE+EC=7+17=24$$ and
$$S_{ABCD}=\frac{1}{2}\cdot24^2=288.$$
</s>