Let us prove that, the quadrilateral formed by joining midpoints of consecutive sides of a rectangular figure is not a square figure but a rhombus.

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 rectangle, the length of diagonals are equal

⇒ AD = BC

⇒ HG = GF = FE = EH

But, we also see that since ABCD is a rectangle not square BD is not equal to CD,

∴ ∠ DCB is not equal to ∠ DBC.

Hence, the angle between the sides in EFGH cannot be 90° and as a result EFGH is not a square figure but a rhombus.