[Math] Finding the diagonal of a rhombus

Question

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.

Answer 1

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.