Spherical Kite

Since a kite on a sphere can be projected onto a plane while preserving angles through a stereographic projection , and all kites on a Euclidean plane have perpendicular diagonals, so do all kites in spherical geometry. Therefore, the two diagonals are always perpendicular to each other.

Is the quadrilateral a square?

Imagine the sphere viewed from directly above some point $A$ . Imagine another point, $B$ , directly in front of you on the sphere at some arbitrary distance, and another point $C$ directly behind you. Imagine a great circle of the sphere that appears "horizontal" when viewed from your vantage point, and imagine points $D$ and $E$ on this circle equidistant from $A$ . Imagine another great circle containing $B$ and $C$ ; by definition, this one is vertical. At point $A$ , the plane tangent to the sphere is perpendicular to the direction of your sight, so the projection of these features as you see them becomes true at this point. Therefore, at point $A$ , the two great circles you have constructed are perpendicular.

Imagine connecting points $B$ , $C$ , $D$ , and $E$ via great circles such to form a quadrilateral. The quadrilateral will meet the given requirements. Therefore, it is at least possible to construct such a quadrilateral such that the diagonals are perpendicular. Furthermore, this can be done with any distances $d_1$ and $d_2$ where $BD=BE=d_1$ and $CD=CE=d_2$ . For a given $d_1$ and $d_2$ , there is also a range of possible distances $DE$ , which is a measure of the "squishedness" of the quadrilateral.

Moreover, this is the only way to construct such a quadrilateral. To prove this, first let the location of $B$ be fixed. The locations of $C$ , $D$ , and $E$ each have two degrees of freedom on the two-dimensional surface of the sphere, so in total there are 6 degrees of freedom . You can reduce the degrees of freedom to 2 by identifying 4 independent restrictions inherent to the requirements of the quadrilateral:

1. $BD=d_1$
2. $BE=d_1$
3. $CD=d_2$
4. $CE=d_2$

One of the remaining 2 degrees of freedom is the orientation of the quadrilateral, which you can define such that $C$ is directly behind you (simply by orienting your vantage point), reducing the degrees of freedom to 1.

The last degree of freedom is how much you "squish" the quadrilateral, which can be represented as the distance $DE$ . You can see that this freedom is restricted to varying the "squishedness" $DE$ while keeping $D$ and $E$ on the "horizontal" great circle, simply because this motion is possible and there are no additional degrees of freedom to draw from. Therefore, for any $d_1$ and $d_2$ , you can orient yourself such that $\overline{BC}$ appears vertical and $\overline{DE}$ appears horizontal while looking down on the sphere, meaning they (the diagonals) are perpendicular.

The problem is basically asking if the diagonals in a kite are perpendicular to each other. This can be easily shown by a congruency argument.