Right Isosceles Triangles on Sides of a Quadrilateral: What is this about?
A Mathematical Droodle

Explanation

1. The quadrilateral is orthodiagonal and the diagonals are equal in length.
2. If right isosceles triangles are drawn on the sides of quadrilateral with successive apexes P, Q, R, S - internal to the quadrilateral and neither P and R nor Q and S coincide, then PR is orthogonal to QS.
3. If right isosceles triangles are drawn on the sides of quadrilateral with successive apexes P, Q, R, S - internal to the quadrilateral, then P and R coincide iff Q and S coincide.
4. If right isosceles triangles are drawn on the sides of quadrilateral with successive apexes P, Q, R, S, so that P and R are internal whilst Q and S are external to the quadrilateral, then if P and R coincide then P, Q, R, S are collinear.

Complex numbers come in handy to help solve this problem. Let a, b, c, d be the complex numbers corresponding to the vertices A, B, C, and D and, similarly, p, q, r, s for the points P, Q, R, S. In general, we underscore this relationship as A(a), B(b), etc.

For two points U(u) and V(v), define W(w) via

where μ = (1 + i)/2, is the apex of the right isosceles triangle with base UV such that UVW is oriented counterclockwise. Thus, assuming ABCD presents the positive orientation, we express points (or rather the corresponding complex numbers) P, Q, R, S in terms of the vertices as follows

The two are orthogonal iff i·pr = qs. In view of (4), we see that

So (4) implies (6). Conversely, assume i·pr = qs. Then from (5)

we reduce (7) to

or,

But μ/(1 - μ) = (1 + i)/(1 - i) = (1 + i)²/2 = 2i/2 = i. So (8) implies i·ac = bd. This is of course assuming neither pr = 0 or qs = 0. Now, what if pr = 0 say? Then (6) implies that also qs = 0. Obviously, (6) also yields the reversed implication.

Finally, assume Q(q) and S(s) are defined so that BCQ and DAS have negative orientation causing Q and S outside ABCD. In this case (2) is replaced by

P = R is still equivalent to (4). We observe from (2') that (p + r)/2 = (q + s)/2, and, since p = (p + r)/2 = r iff P = R, it follows that if P = R then it serves as the midpoint of QS.

(It also appears that in the latter case, the point of intersection of the diagonals also lies on QS. How come?)

Remark

When all four triangles have the same orientation it was shown elsewhere that PR and SQ are equal and perpendicular. However, the argument falls through when either has zero length.