Q:

Parallelogram diagonals angles proof

Given: Parallelogram $ABCD$, $AC$=$\sqrt{2}AB$
Prove: Angle ABD = Angle DAC

So far, I showed the following:
$\triangle DCA\sim\triangle BAC$ and $\triangle DOC\sim\triangle BOA$
DC = AB; DA = BC
$\angle D$ = $\angle B$, $\angle A$ = $\angle C$
$\angle CDB = \angle DBA$; $\angle ADB = \angle CBD$
How would I go on to prove it?

A:

Use this well know identity Parallelogram law
$$2(AB^2+AD^2)=AC^2+BD^2$$
since
$$AC=\sqrt{2}AB\Longrightarrow BD=\sqrt{2}AD$$
let
$$AB=x,AD=y$$
then we have
$$\cos{\angle ABD}=\dfrac{AB^2+BD^2-AD^2}{2AB\cdot BD}=\dfrac{x^2+2y^2-y^2}{2x\sqrt{2}y}=\dfrac{x^2+y^2}{2\sqrt{2}xy}$$
and
$$\cos{\angle DAC}=\dfrac{AD^2+AC^2-DC^2}{2AD\cdot AC}=\dfrac{y^2+2x^2-x^2}{2y\sqrt{2}x}=\dfrac{x^2+y^2}{2\sqrt{2}xy}$$
so
$$\angle ABD=\angle DAC$$

