ABCD is a quadrilateral in which $AD = BC$. If P, Q, R, S be the mid-points of AB, AC, CD and BD respectively, show that PQRS is a rhombus.

Given:

ABCD is a quadrilateral in which $AD = BC$. P, Q, R, S be the mid-points of AB, AC, CD and BD respectively.
To do:

We have to show that PQRS is a rhombus.

Solution:
 

In $\triangle ABC$, by mid-point theorem,

$PQ \parallel BC$ and $PQ=\frac{1}{2}BC$......(i)

In $\triangle ABD$, by mid-point theorem,

$PS \parallel AD$ and $PS=\frac{1}{2}AD$......(ii)

In $\triangle CAD$, by mid-point theorem,

$RQ \parallel AD$ and $RQ=\frac{1}{2}AD$......(iii)

In $\triangle CBD$, by mid-point theorem,

$RS \parallel BC$ and $RS=\frac{1}{2}BC$......(iv)

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

$PQ \parallel BC \parallel RS$ and $PQ=RS=\frac{1}{2}BC$....(v)

Therefore,

$PQ \parallel BC$ and $PQ=RS$.

This implies,

PQRS is a parallelogram.

Similarly, from (ii) and (iii), we get,

$PS=RQ=\frac{1}{2}AD$

$PS=RQ=\frac{1}{2}BC$   (Since $AD=BC$)....(vi)

From (v) and (vi), we have,

$PQ=RS=\frac{1}{2}BC=PS=RQ$

$PQ=QR=RS=SR$

Therefore,

$PQRS$ is a parallelogram and all the sides are equal to each other.

This implies PQRS is a rhombus.

Hence proved.

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Updated on: 10-Oct-2022

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