# Find the sides of a parallelogram

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!

## 1 Answer

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.