Q. 74.3( 4 Votes )

# Let us prove that

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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