Q7 of 24 Page 157

Let us prove that, the quadrilateral formed by joining midpoints of a rhombus is a rectangle.


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


More from this chapter

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5

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.

 6

Let us prove that, the quadrilateral formed by joining midpoints of consecutive sides of a square is a square.

 8

In the triangle ABC, the midpoints of AB and AC are D and E respectively; the midpoints of CD and BD are P and Q respectively. Let us prove that, BE and PQ bisect each other.

 9

In the triangle ABC, AD is perpendicular on the bisector of ∠ABC. From the point D, a straight line DE parallel to the side BC is drawn which intersects the side AC at the point E. Let us prove that AE = EC.