Question

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

Answer 1

​Given: In quadrilateral ABCD, BD = AC and K, L, M and N are the mid-points of AD, CD, BC and AB, respectively.

To prove: KLMN is a rhombus.

Proof:

In ∆ADC,

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

So, KL || AC and KL = $\frac{1}{2}$AC ...(1)

Similarly, in ∆ABC,

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

So, NM || AC and NM = $\frac{1}{2}$AC ...(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, in ∆ABD,

Since, K and N are the mid-points of sides AD and AB, respectively.

So, KN || BD and KN = $\frac{1}{2}$BD ...(3)

But BD = AC (Given)

$\Rightarrow \frac{1}{2}$BD = $\frac{1}{2}$AC

$\Rightarrow$KN = NM [From (2) and (3)]

But these are a pair of adjacent sides of the parallelogram KLMN.

Hence, KLMN is a rhombus.