Answer :


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.


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