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$ABCD$ is a rectangle where $AB= \sqrt{2}$ and $BC= \sqrt6$ . $P$ and $Q$ are two points on $AC$ such that $AP=CQ$. From $P$ and $Q$, two perpendicular $PR$ and $QS$ are drawn on $BC$ and $AD$ respectively. If $PRQS$ is a rhombus and $PR=CQ$, then find the length of $PQ$.

from pythagoras's theorem,
$AC^{2}=AB^{2}+BC^{2}$
$AC^{2}=\sqrt{2}^{2}+\sqrt{6}^{2}$
$AC^{2}=2+6=8$
$AC=\sqrt{8}=2\sqrt{2}$

now for $\Delta ABC\Leftrightarrow sin\angle ACB=\frac{AB}{AC}=\frac{\sqrt{2}}{2\sqrt{2}}=\frac{1}{2}$
$\therefore \angle ACB=30$

now for $\Delta PRC\Leftrightarrow sin 30=\frac{PR}{PC}=\frac{1}{2}\Leftrightarrow PC=2PR\Leftrightarrow PQ=PR$

$\therefore PQ=PR=CQ=AP$
$\therefore PQ=\frac{AC}{3}=\frac{2\sqrt{2}}{3}.......(ans:)$ ......

Hi Tahmid. Nice job solving the problem. Just an advice.
Many haven't solved the problem yet. So if you can, please "hide" the solution.
That way, other's fun will not be spoiled and they can always check your solution if they cannot solve it.

ok labib vaia ...... i'll try that .....