1. Geometry Applications

Hi

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

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

2. never mind

3. Hello, xwrathbringerx!

I'm still working on a proof, but I'll give you a sketch . . .

In parallelogram $ABCD$, the diagonal $BD$ is perpendicular to $AB.$
If the diagonals intersect at T, prove that: . $AD^2 \:=\:AT^2 + 3TD^2$
Code:
     A                 B
o - - - - - - - o
\ *           * \
\   *       *   \
\     *   *     \
\       o T     \
\     *   *     \
\   *       *   \
\ *           * \
o - - - - - - - o
D                 C