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!