The diagonals of a quadrilateral ABCD are equal.


The diagonals of a quadrilateral ABCD are equal. Prove that the quadrilateral formed by joining the midpoints of its sides is a rhombus.


Given: In quadrilateral ABCDBD = AC and KLM and N are the mid-points of AD, CD, BC and AB, respectively.

To prove: KLMN is a rhombus.



Since, K and L are the mid-points of sides AD and CD, respectively.

So, $K L \| A C$ and $K L=\frac{1}{2} A C$                ...(1)

Similarly, in ABC,

Since, M and N are the mid-points of sides BC and AB, respectively.

So, $N M \| A C$ and $N M=\frac{1}{2} A C$       ...(2)

From (1) and (2), we get
KL = NM and KL || NM

But this a pair of opposite sides of the quadrilateral KLMN.

So, KLMN is a parallelogram.

Now, iABD,

Since, K and N are the mid-points of sides AD and AB, respectively.
So, $K N \| B D$ and $K N=\frac{1}{2} B D$         ...(3)
But BD = AC            (Given)
$\Rightarrow \frac{1}{2} B D=\frac{1}{2} A C$
$\Rightarrow K N=N M \quad[$ From $(2)$ and $(3)]$
But these are a pair of adjacent sides of the parallelogram KLMN.

Hence, KLMN is a rhombus.

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