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enter image description here

In $ABCD$ rectangle $AB=6$, $AD=8$, $AE=ED$, $BF=FC$, $EP=PQ=QF$. Find the area of $PXQY$.

Source: Bangladesh Math Olympiad 2014 Junior Category

I cant prove that which type of quadrilateral it is. How can I get the diagonal or the sides of $PQXY$? Please help me with a hint.

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$PF=FC$ and $PF\perp FC,$ which says $$\measuredangle QPY=\measuredangle FPC=45^{\circ},$$ which gives that $PXQY$ is a square and $$S_{PXQY}=\frac{XY\cdot PQ}{2}=\frac{2^2}{2}=2.$$

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with Q being in the middle of PF, and the symmetry of the graph making XY cut the middle of PQ, you can actually visualise triangle PBC as the following diagram:

enter image description here

Can you continue?

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  • $\begingroup$ how can you prove that $PY$ and $XQ$ have are same? $\endgroup$ – Shromi Feb 19 at 6:41
  • $\begingroup$ do you know how to prove PYQX is parallelogram? $\endgroup$ – qsmy Feb 19 at 7:00
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As $\triangle XAB\sim \triangle XQP$ and $AB= 3PQ$, the distance from $X$ to $AB$ is equal to 3 times the distance from $X$ to $PQ$.

Therefore, the distance from $X$ to $PQ$ is $\displaystyle 8\times\frac{1}{2}\times\frac{1}{4}=1$.

$PQ=6\div3=2$.

The area of $PXQY$ is $\displaystyle 2\times\frac{1}{2}\times2\times 1=2$.

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