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Jan 5 at 14:32 comment added Blue @hbghlyj: "For a kite, the formula's result is path-dependent." Indeed. This kind of thing seems a bit problematic for an "incenter map", but it's also fascinating. :) ... Now I'm curious about cases of non-kites that have one diagonal bisecting the other. (The ones that send $H_1$ to infinity.) ... Anyway, neat problem! I'm delighted to have come across the "complex harmonic" representation. I wonder if the right-hand side of $(2')$ evaluating to $-Q$ might amount to a (novel?) characterization of tangential quads. That's in puzzle for another day. :) ... Cheers!
Jan 5 at 12:09 comment added hbghlyj For a kite, the formula's result is path-dependent. A limit of tangential quadrilaterals (AB+CD=AD+BC) converges to the incenter. A limit of ex-tangential quadrilaterals (AB+BC=AD+DC) converges to the excenter. A limit of orthodiagonal quadrilaterals converges to diagonal intersection. For a generic sequence, the limit is unstable and can lie anywhere.
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Jan 3 at 4:23 history answered Blue CC BY-SA 4.0