Irish Mathematical Olympiad

The right triangles $\triangle ABE$ and $\triangle BCF$ have legs of the same length, hence they are congruent. This implies that $\angle GBE + \angle BEG = 90^\circ$ and so

$\angle EGB = 90^\circ$. This shows that $\triangle BGE$ is similar to $\triangle BCF$. Using symmetry we see now that $GHKL$ is a square. In particular, $GL$ and $HK$ are parallel, hence the triangles $\triangle BGE$ and $\triangle BHC$ are similar.

From these two similar triangles we obtain $\frac{|HC|}{|BC|} = \frac{|GE|}{|BE|} = \frac{|CF|}{|BF|}$. Using that $|BC| = 1$, $|BE| = |CF| = x$, $|BF| = \sqrt{1+x^2}$ (Pythagoras) and symmetry, we obtain now
$$|BG| = |HC| = \frac{x}{\sqrt{1+x^2}} \quad \text{and} \quad |GE| = \frac{x^2}{\sqrt{1+x^2}}.$$
The area of the quadrilateral $FCEG$, which is equal to the difference of the areas of the triangles $\triangle BCF$ and $\triangle BGE$, is therefore equal to
$$\frac{1}{2} \left( x - \frac{x^3}{1+x^2} \right) = \frac{x}{2(1+x^2)}.$$
By symmetry again, $MDFH$, $NAMK$ and $EBNL$ have the same area. As these four quadrilaterals cover without overlap the space between $GHKL$ and $ABCD$, the area of the quadrilateral $GHKL$ is equal to
$$1 - \frac{2x}{1+x^2} = \frac{(1-x)^2}{1+x^2}.$$