Let us prove that

In ΔACD, as E and H are the midpoints of AC and CD respectively.

So by applying the theorem:-

The line segment joining the midpoints of two side of a triangle is parallel to the third side and equal to half of it, we get,

and EH || AD ……… (1)

Also, by using above theorem in ΔABD, as F and G are midpoints of AB and BD respectively,

and FG || AD ……… (2)

From equations (1) and (2) we see that, EH = FG and EH || AD.

Similarly for ΔBCD, we have G and H as midpoints of BD and CD respectively, by applying above theorem, we get,

and GH || BC ……… (3)

And also in ΔABC,

and EF || BC ……… (4)

From equations (3) and (4) we see that, GH = EF and GH || EF.

As ABCD is a square, the length of diagonals are equal

⇒ AD = BC

⇒ HG = GF = FE = EH

⇒ AE = AF, this means ∠AEF = ∠AFE = ∠ABC = ∠ACB = 45°

∴ We can easily prove that angle between the sides of the figure FGHE is equal to 90°

Hence, Quadrilateral FGHE is a square.