Let us happily borrow Inceptio's image:
Note that $\,\Delta ALZ\cong\Delta LFK\cong\Delta FCD\cong \Delta CAB\,$ as these all are right-angled triangles with legs of the same sizes $\,1\,,\,0.5\,$ .
The sum of these four triangle's areas is $\,0.25\cdot 4=1\;\;(**)\,$ , but note that the areas of the little triangles $\,\Delta ABX\,,\,\Delta LZY\,,\,\Delta FKL\,,\,\Delta CED\,$ are summed twice in the above calculation.
Now, we prove all these four little triangles are congruent to each other and right-angled. From the first congruence above we get:
$$\alpha:=\angle LAZ=\angle FLK=\angle DFC=\angle ACB$$
and, for example, in the triangle $\,\Delta ALZ\,$ , we get $\,\beta:=\angle AZL=90^\circ-\alpha\,$ , so in the triangle $\,\Delta LZY\,$ we have the pair of angles $\,\alpha\,,\,\beta\,\,,\,\,\,\alpha+\beta=90^\circ\,$ , and from here that this little triangle is right- angled. The same holds for the other three little triangles.
We also got the additional fact that $\,XEJY\,$ is a rectangle as its four angles are right, and in a little while more it will follow that it is actually a square.
From (**) above and the lines after it, it follows that the area of $\,XEJY\,$ equals the sum of the four little triangles.
We also get that the four trapezoids in the figure are congruent (as two of their sides are parts of the congruent little triangles and a third side equals the length of their hypotenuses, which is $\,0.5\,$ .
Now it is clear that $\,XE=EJ=JY=YX\,$ (since, for example, $\,XE=BC-BX-EC=$ (hypotenuse of big right angled triangle minus short lef of tine right-angled triangle-hypotenuse of tiny right angled triangle) and, as promised above, it follows $\,XEJY\,$ is in fact a square.
I now believe you can complete the exercise.