One needs to get it right.
What I have tried: can do it (see my comment below), would like a simpler solution. Will post a complete solution after a while.
Any constructive feedback would be appreciated!
$\bf{Added:}$ My attempt, using trigonometry.
The projections of the top and bottom vertices on the horizontal diagonal divide it in ratios $\frac{\tan 10^{\circ}}{\tan 20^{\circ}}$, and $\frac{\tan 30^{\circ}}{\tan 50^{\circ}}$. So let's show that these ratios are equal, equivalently
$$\tan 10^{\circ} = \tan 20^{\circ} \tan 30^{\circ} \tan 40^{\circ}$$
The equation $\tan x = \tan 2x \tan 3x \tan 4x$ appears in some high school math journal in the 60's.
Write $\tan x = \frac{e^{i x} - e^{-i x}}{i(e^{i x} + e^{ix})} = \frac{z^2-1}{i(z^2+1)}$, so, with $f(z) = \frac{z^2-1}{z^2+1}$ we need to solve
$$f(z)+ f(z^2) f(z^3) f(z^4)=0$$
The LHS factors as
$$\frac{2 (z-1)(z+1) (z^{12}-z^6+1)}{(z^2+1) (z^4-z^2+1) (z^8+1)}$$ and now notice that $z=e^{\frac{2\pi i}{36}}$ is a root of the cyclotomic polynomial $\Phi_{36}(z) = z^{12}-z^6+1$.
Conclusion: the ratios are equal, the top and bottom vertices project onto the same point of the diagonal, and therefore, the diagonals are perpendicular.
A geometric proof would be nice to have.
$\bf{Added:}$
@timon92: brought this post to my attention. @Andrew Hwang had an intepretation almost identical to this question; should undelete his answer.
Disclosure: discovered this cute question by accident/luck, while playing with Geogebra some 15 years ago. Now the question itself has to do with adventitious quadrilaterals, specifically those with orthogonal diagonals, a classic subject. Yes, there are some other ones.
Consider two equilateral triangles ( in red) with a common vertex, one rotation of the other by $40^{\circ}$. There are formed two isosceles triangles with base angles $40^{\circ}$ (equal sides in blue). Now draw the perpendicular bisector of one of the equilateral triangle ( in yellow). We get another isosceles triangle with base angle $40^{\circ}$. Now conclude that we have an isosceles triangle with base angle $70^{\circ}$. The purple quadrilateral has the required shape.