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A parallelogram $ABCD$ is given. Let $DP$ be perpendicular to the diagonal $AC$ $(P\in AC).$ If $AP=6$ $cm$ and $CP=15$ $cm$ and the difference between the sides of $ABCD$ is $7$ $cm,$ find $BD.$

If we find the sides of $ABCD,$ we will find the other diagonal easily using the fact that $AC^2+BD^2=2(AB^2+AD^2)$. I think that we should try to find other relationship between $AB$ and $AD$ (other than $AB-AD=7$) in order to be able to solve for the sides. Can you give me a hint? Thank you in advance!

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Letting $AB=CD=x+7$, $AD=x$ and $DP=y$ we get $$y^2+6^2=x^2$$ $$y^2+15^2=(x+7)^2$$ by Pythagorean theorem. Subtracting these equations gives you a linear equation in $x$, allowing you to solve $x$ and thus the sides of the parallellogram.

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