$P, Q, R$ and $S$ are respectively the mid-points of the sides $AB, BC, CD$ and $DA$ of a quadrilateral $ABCD$ in which $AC = BD$. Prove that $PQRS$ is a rhombus.

(N/A) quadrilateral $ABCD$ in which $AC = BD$ and $P, Q, R$ and $S$ are respectively the mid-points of the sides $AB, BC, CD$ and $DA$ of quadrilateral $ABCD$.
To prove: $PQRS$ is a rhombus.
Proof: In $\Delta ABC, P$ and $Q$ are the mid-points of $AB$ and $BC$ respectively. That is,$PQ$ joins mid-points of sides $AB$ and $BC$.
$\therefore PQ \parallel AC$ $... (1)$
And $PQ = \frac{1}{2} AC$ $... (2)$ [Mid-point theorem]
In $\Delta ADC, R$ and $S$ are the mid-points of $CD$ and $AD$ respectively.
$\therefore SR \parallel AC$ $... (3)$
And $SR = \frac{1}{2} AC$ $... (4)$ [Mid-point theorem]
From $(1)$ and $(3)$,we get $PQ \parallel SR$.
From $(2)$ and $(4)$,we get $PQ = RS$.
Since $PQ \parallel SR$ and $PQ = RS$,$PQRS$ is a parallelogram.
In $\Delta DAB, SP$ joins mid-points of sides $DA$ and $AB$ respectively.
$\therefore SP = \frac{1}{2} BD$ $... (5)$ [Mid-point theorem]
Given $AC = BD$ $... (6)$
From equations $(2), (5)$ and $(6)$,we get $SP = PQ$.
Since adjacent sides of the parallelogram $PQRS$ are equal $(SP = PQ)$,$PQRS$ is a rhombus.
Hence,proved.