\overline{AE} is a transversal crossing through parallel lines \overline{HD} and \overline{BF}. Therefore, the angles around point I have to match their corresponding angles around point J. For example, \angle JIL has to be the same as \angle EJK. The same logic applies for transversal \overline{GC} and points L and K.

\overline{HD} is a transversal through parallel lines \overline{AE} and \overline{GC}. Therefore, the angles around point I have to match their corresponding angles around point L. For example, \angle JIL has to be the same as \angle KLD. The same logic applies for transversal \overline{BF} and points J and K.

So consider internal angle \angle JIL. That angle has some value, x. The opposite angle to it, \angle AIH, has to be x. Also, because of the interlocking transversals, \angle KJE and \angle DLK and \angle FKC must be x. Because of opposite angles, \angle IJB and \angle GLI and \angle JKL must be x.

The congruence of the triangles means that \angle FKC must have the same value as \angle GLD and \angle EJB - namely, x. Opposite angles means that \angle ILK and \angle IJK must likewise be x. Thus the inside of the shaded area is 4x. We also see that around points L and J we have 4 x’s.

The total of all the angles in full rotation around some point has to equal 360 degrees. Therefore, we know that 4x = 360, and we know that the interior of our shaded area has angles that equal 4x. x = 90, so the shaded area is a square.