# 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.

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Why do you need to find the length of the diagonal? – Mark Bennet Jul 3 '12 at 14:45
So that i could show that when two diagonals intersect each other the triangles formed are right angle issocles. – MistyD Jul 3 '12 at 14:47
If you think about the information you know, you will see that you can build a similar rhombus of any size. It is the angles which matter, not the lengths, and you have enough isosceles triangles already to work out the angles without having to try to manufacture another one (the right angled triangle will not be isosceles because the two diagonals clearly have different lengths). – Mark Bennet Jul 3 '12 at 14:51
If we let $X=\vec{AD}$ and $Y= \vec{AB}$ then the diagonals are $X+Y$ and $X-Y$, their dot product is $(X+Y)\cdot (X-Y) = \| X \|^2 - \|Y\|^2=0$ so they are perpendicular. – Ragib Zaman Jul 3 '12 at 14:56
If you draw the two diagonals, you can by angle-chasing show that the four triangles so formed are all congruent. That will show that the angle between the diagonals is a right angle. I do not believe you can find the diagonals without some trigonometry. The square of the long diagonal is $6^2(2-2\cos 100^\circ)$, so "trig ratio" is built into the answer. – André Nicolas Jul 3 '12 at 15:13

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.

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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.

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