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Question (Step 4. in the above):
Let $ABCD$ be a tangential quadrilateral with incircle center $I$.
How do you prove $H_1$ is exactly the foot of the perpendicular from the incenter $I$ to $AC$?

Question.
When $ABCD$ is a tangential quadrilateral, how do you prove $I$ is exactly the incenter?

This is a rational map $(\mathbb R^2)^4 \dashrightarrow \mathbb R^2$.

This is a rational map $\text{Gr}(1,2)^4 \dashrightarrow \mathbb R^2$ and it is symmetric under permutations of the four input lines.

Why this is surprising:

In vector notation, let $u = A \cdot B$ and $v = A \times B$ (the 2D cross product $x_A y_B - y_A x_B$). The formulas can be concisely written as: $$I = \frac{2(A \cdot B)}{A \times B} \left( \frac{x_B x_D}{x_B + x_D} - \frac{x_A x_C}{x_A + x_C}, \frac{x_A y_C}{x_A + x_C} - \frac{x_B y_D}{x_B + x_D} \right)$$

Why this is surprising: