Hi

In a parallelogram ABCD, the diagonal BD is perpendicular to AB. If the diagonals intersect at T, prove that $\displaystyle AD^2 = AT^2 + 3TD^2$.

Ummm I don't seem to get this problem (e.g. how the diagram's supposed to look - diagonal BD is perpendicular to AB?). Could someone please give me a few hints?

Thanxx