Generalized “incenter map” is symmetric with respect to permutations of the four input lines

Let $BD∩AC=P$.

By a property of complete quadrilateral, $$\frac{BP}{DP}=\frac{BE}{EC}\cdot\frac{CF}{FD}$$

Let $BD'∩AC=H_1$.

Let $BD'∩EF'=H_3$.

Since $D',C,F'$ are collinear, apply Menelaus's theorem to $\triangle D'CB$ and $F'H_3E$:

$$\frac{BH_3}{D'H_3}=\frac{BE}{EC}\cdot\frac{CF'}{F'D'}$$ $$\implies\frac{BH_3}{DH_3}=\frac{BE}{EC}\cdot\frac{CF}{FD}$$ $$\implies\frac{BH_3}{DH_3}=\frac{BP}{DP}$$ By angle bisector theorem$$\angle BH_3P=\angle DH_3P$$ Since $H_1$ is the unique point on $BD'$ such that $\angle BH_1P=\angle DH_1P$, we must have $H_1=H_3$.

Let the line $AC$ be the $x$-axis. We can set the coordinates of points $A$ and $C$ as: $$A = (-1, 0), \quad C = (1, 0)$$ The four lines forming the complete quadrangle are determined by their slopes from $A$ and $C$. Let:

• Line $AB$ and $AF$ have slope $m_1$, so its equation is $y = m_1(x + 1)$.
• Line $AD$ and $AE$ have slope $m_2$, so its equation is $y = m_2(x + 1)$.
• Line $BC$ and $EC$ have slope $n_1$, so its equation is $y = n_1(x - 1)$.
• Line $DC$ and $FC$ have slope $n_2$, so its equation is $y = n_2(x - 1)$.

The vertices $B, D, E, F$ are the intersections of these lines:

• $B = AB \cap BC \implies x_B = \frac{m_1 + n_1}{n_1 - m_1}, \quad y_B = \frac{2 m_1 n_1}{n_1 - m_1}$
• $D = AD \cap CD \implies x_D = \frac{m_2 + n_2}{n_2 - m_2}, \quad y_D = \frac{2 m_2 n_2}{n_2 - m_2}$
• $E = AD \cap BC \implies x_E = \frac{m_2 + n_1}{n_1 - m_2}, \quad y_E = \frac{2 m_2 n_1}{n_1 - m_2}$
• $F = AB \cap CD \implies x_F = \frac{m_1 + n_2}{n_2 - m_1}, \quad y_F = \frac{2 m_1 n_2}{n_2 - m_1}$

Let $D'$ be the reflection of $D$ across $AC$ and $F'$ be the reflection of $F$ across $AC$. Their coordinates are simply $(x_D, -y_D)$ and $(x_F, -y_F)$, respectively. The intersection of a line through $(x_1, y_1)$ and $(x_2, -y_2)$ with the $x$-axis ($y=0$) is given by: $$x_H = \frac{x_1 y_2 + x_2 y_1}{y_1 + y_2}$$ Applying this to the line $BD'$ (connecting $B(x_B, y_B)$ and $D'(x_D, -y_D)$), we find the intersection point $H_1$ on $AC$: $$x_{H_1} = \frac{x_B y_D + x_D y_B}{y_B + y_D}$$ Substituting the coordinates: $$x_{H_1} = \frac{m_1 m_2 (n_1 + n_2) + n_1 n_2 (m_1 + m_2)}{n_1 n_2 (m_1 + m_2) - m_1 m_2 (n_1 + n_2)}$$ Notice that this expression is symmetric with respect to the pairs $(m_1, n_1)$ and $(m_2, n_2)$. To find the intersection $H_2$ of $EF'$ with $AC$, we use the coordinates of $E$ and $F$, which effectively swaps $m_1$ and $m_2$ in the calculation for $H_1$ (since $E$ uses $m_2, n_1$ and $F$ uses $m_1, n_2$). Because the expression for $x_{H_1}$ is symmetric in $m_1, m_2$, it follows that: $$x_{H_2} = x_{H_1}$$

Since the lines $BD'$ and $EF'$ intersect the diagonal $AC$ at points with the same $x$-coordinate, they must intersect $AC$ at the same point $H$. $$BD' \cap AC = EF' \cap AC = H$$