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.


So, EHGF is a parallelogram and as IH lie on EH and HJ lie on HG.


IHJO is also a parallelogram.


As we know that, the angle subtended by the diagonals in a triangle is equal to 90°, and also in a parallelogram, the opposite angles are equal.


EHG = IOJ = 90°


Hence, quadrilateral HGFE is a rectangle


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