Finding the diagonal of a rhombus 

I am trying to show that the diagonals of a RHOMBUS intersect each other at 90 degrees
    However I need to find the length of the diagonal without using Trig. ratios. Any ideas how i could find that ?



The length of each side in the figure is 6.
 A: We give a traditional "angle-chasing" argument. Draw lines $AC$ and $AD$. Label their intersection point $I$.
Note that by the definition of rhombus, $\triangle ABC$ is isosceles, so $\angle BAC=\angle ACB$.  Since $\triangle ABC$ and $\triangle ADC$ are congruent, $\angle DAC=\angle DCA=\angle BAC=\angle ACB$.
Work now with the other diagonal. The same argument as in the preceding paragraph shows that all the angles it forms with the sides are equal.
Now look at $\triangle AIB$ and $\triangle CIB$. Since $\angle IAB=\angle ICB$, and $\angle IBA=\angle IBC$, their remaining angles must be equal. 
So $\angle AIB=\angle CIB$. But these two angles add up to a "straight angle" ($180^\circ$), so each of them must be $90^\circ$.
The lengths of the diagonals are inextricably tied to trig functions of $80^\circ$ or relatives. These trig functions are not at all "nice," so there is no way to sneak around them.
A: There is no way to find the length of the diagonal without introducing trigonometric ratios. Suppose that angle at D is $x.$ Then the diagonal length is $$6\sqrt{2-2\cos x}=12\sin(x/2).$$ Here, $x=100^{\circ}.$ There is no nice value for $\sin (50^{\circ})$ - the best answer for the diagonal length is $12\sin (50^{\circ})$ and trig ratios are inherently involved.
One way to prove this is by using vectors and dot products. Letting $X=\vec{AD}$ and $Y=\vec{AB}$, we see the diagonals are $X+Y$ and $X-Y.$ Their dot product is then $(X+Y)\cdot(X-Y) = \|X\|^2 - \|Y\|^2=0$ so they are perdendicular.
Another approach is Cartesian geometry: Let $A=(0,0), B=(b,0), D= (p,q), C=(p+b,q)$ with the condition that $b^2=p^2+q^2$ (for the same side lengths). The gradients of the diagonals are $$m_1 = \frac{q}{p+b}, \ \ \ m_2 = \frac{q}{p-b}$$
and their product is $m_1m_2 = \displaystyle \frac{q^2}{p^2-b^2} = -1$ so they are perpendicular.
