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.
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.
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.
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.