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(B) Let the mid-points of sides $PQ, QR, RS,$ and $SP$ be $A, B, C,$ and $D$ respectively. By the Mid-point Theorem,$AB \parallel PR$ and $AB = \frac{

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diagonals of $PQRS$ are equal.

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(B) Let the mid-points of sides $PQ, QR, RS,$ and $SP$ be $A, B, C,$ and $D$ respectively. By the Mid-point Theorem,$AB \parallel PR$ and $AB = \frac{1}{2} PR$. Similarly,$DC \parallel PR$ and $DC = \frac{1}{2} PR$. Thus,$AB \parallel DC$ and $AB = DC$,making $ABCD$ a parallelogram. For $ABCD$ to be a rhombus,adjacent sides must be equal (i.e.,$AB = AD$). Since $AD = \frac{1}{2} QS$ and $AB = \frac{1}{2} PR$,the condition $AB = AD$ implies $\frac{1}{2} PR = \frac{1}{2} QS$,which means $PR = QS$. Therefore,the quadrilateral formed by joining the mid-points is a rhombus if the diagonals of the original quadrilateral $PQRS$ are equal.

The quadrilateral formed by joining the mid-points of the sides of a quadrilateral $PQRS,$ taken in order,is a rhombus,if

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