$ABCD$ is a kite having $AB = AD$ and $BC = CD$. Prove that the figure formed by joining the mid-points of the sides, in order, is a rectangle.

Given:

$ABCD$ is a kite having $AB = AD$ and $BC = CD$.

To do:

We have to prove that the figure formed by joining the mid-points of the sides, in order, is a rectangle.

Solution:

Let $P, Q, R$ and $S$ be the mid points of the sides $AB, BC, CD$ and $DA$ respectively.

Join $AC$ and $BD$.

In $\triangle ABD$,

$P$ and $S$ are the mid points of $AB$ and $AD$.

This implies,

$PS \parallel BD$ and $PS = \frac{1}{2}BD$....…(i)

Similarly,

In $\triangle BCD$,

$Q$ and $R$ the mid points of $BC$ and $CD$.

This implies,

$QR \parallel BD$ and $QR = \frac{1}{2}BD$...…(ii)

Similarly,

$PQ \parallel SR$ and $PQ = SR$.....…(iii)

From equations (i) and (ii) and (iii), we get,

$PQRS$ is a parallelogram.

$AC$ and $BD$ intersect each other at right angles.

Therefore,

$PQRS$ is a rectangle.

Hence proved.

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Updated on: 10-Oct-2022

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