\( A B C D \) is a quadrilateral in which \( P, Q, R \) and \( S \) are mid-points of the sides \( A B, B C, C D \) and \( DA \) (see Fig 8.29\( ) . A C \) is a diagonal. Show that
(i) \( \quad S R \| A C \) and \( S R=\frac{1}{2} A C \)
(ii) \( P Q=S R \)
(iii) PQRS is a parallelogram.

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Given:

\( A B C D \) is a quadrilateral in which \( P, Q, R \) and \( S \) are mid-points of the sides \( A B, B C, C D \) and \( DA \). $AC$ is a diagonal.

To do:

We have to show that:

(i) \( \quad S R \| A C \) and \( S R=\frac{1}{2} A C \) (ii) \( P Q=S R \) (iii) PQRS is a parallelogram

Solution:

We know that,

The Midpoint Theorem states that the segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side.

(i) In $\triangle DAC$, $S$ is the mid-point of $DA$ and $R$ is the mid-point of $DC$. Therefore, by mid-point theorem,

$SR\parallel\ AC$ and $SR=\frac{1}{2}AC$.....(i)

(ii) In $\triangle BAC, P$ is the mid-point of $AB$ and $Q$ is the mid-point of $BC$. Therefore, by mid-point theorem,

$PQ\parallel\ AC$ and $PQ= \frac{1}{2}AC$.

$PQ=SR$  (From (i))

(iii) $PQ\parallel\ AC$ and $SR\parallel\ AC$.

Therefore, $PQ\parallel\ SR$ and $PQ=SR$.

A quadrilateral with opposite sides equal and parallel is a parallelogram.

Therefore, PQRS is a parallelogram.

Hence proved.

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Updated on: 10-Oct-2022

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