Put the y-axis along the line AB and put the x-axis along the line BC. => A(0 ; 1) ; B(0 ; 0) ; C(1 ; 0) ; D(1 ; 1)

Let a,b,c,d be the lines through A,B,C,D with the equations y = hx + i ; y = kx + l ; y = mx + n respectively (not necessary to find the equation of d)

a passes through \(\left(\dfrac{2}{3};0\right)\)and (0 ; 1)\(\Rightarrow\left\{{}\begin{matrix}i=1\\-\dfrac{i}{h}=\dfrac{2}{3}\end{matrix}\right.\)

=> The equation of a is y = \(-\dfrac{3}{2}x+1\).

c // a,then their slope are the same : m = h = \(-\dfrac{3}{2}\)

c passed through (1 ; 0)\(\Rightarrow0=-\dfrac{3}{2}+n\)

=> The equation of c is y = \(-\dfrac{3}{2}x+\dfrac{3}{2}\).

b passes through (0 ; 0) ; \(\left(1;\dfrac{2}{3}\right)\)and has equation y = \(\dfrac{2}{3}x\).

The product of the slopes of a and b is :\(hk=-\dfrac{3}{2}.\dfrac{2}{3}\)= -1

\(\Rightarrow a\perp b\) but a // c \(\Rightarrow b\perp c\) but b // d\(\Rightarrow c\perp d\)

=> The shaded region is a rectangle. In fact, because of the symmetry in the problem, each edge of the rectangle will have the same length so that the rectangle is a square

Let E,F be the intersections of a and b ; b and c respectively