Comment by thaumasiotes

> Thus over all positions of the vertex on this square-edge, the area either is always the same (if the opposite triangle edge is parallel to this square-edge), or the area is maximised by putting this vertex in a square corner.

I'm sure this is correct, but the proof doesn't seem ironclad. As far as I see, you're relying on the idea that moving two vertices jointly, as one operation, can't produce any benefits relative to moving the two vertices in sequence one after the other. And while I have no reason to doubt that that's correct, it isn't obvious to me.

(Wording it another way - how do you prove that [the position of vertex B on the edge of the square] that maximizes the area of the triangle holding the position of vertex C fixed is the same as [the position of vertex B on the edge of the square] that maximizes the area of the triangle holding only the edge to which vertex C is attached fixed?)