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$ABCDEF$ is the complete quadrangle determined by four arbitrary lines, so that $E=AD\cap BC$ and $F=AB\cap CD$.

Reflect $D$ across the line $AC$ to a point $D'$, and reflect $F$ across the same line $AC$ to a point $F'$.

Show that the two lines $BD'$ and $EF'$ intersect the diagonal $AC$ at the same point $H$.

Equivalently, prove that $$ BD' \cap AC = EF' \cap AC. $$

Background and motivation.
This configuration arose from studying a generalized “incenter map” associated to four lines $\ell_1,\ell_2,\ell_3,\ell_4$ in the plane. It is a rational map defined (via reflections across diagonals and perpendicular constructions) for any four lines, and restricts to the usual incenter when the four lines bound a tangential quadrilateral. The question above asks for a direct geometric explanation why this map is symmetric with respect to permutations of the four input lines: In the present setting, the quadrilateral $ABCD$ and the quadrilateral $EBFD$ are determined by the same four lines, which suggests that they should give the same point $H$ on $AC$ (and hence give the same generalized incenter).

$ABCDEF$ is an arbitrary complete quadrangle, so that $E=AD\cap BC$ and $F=AB\cap CD$.

Reflect $D$ across the line $AC$ to a point $D'$, and reflect $F$ across the same line $AC$ to a point $F'$.

Show that the two lines $BD'$ and $EF'$ intersect the diagonal $AC$ at the same point $H$.

Background and motivation.
This configuration arose from studying a generalized “incenter map” associated to four lines $\ell_1,\ell_2,\ell_3,\ell_4$ in the plane. It is a rational map defined (via reflections across diagonals and perpendicular constructions) for any four lines, and restricts to the usual incenter when the four lines bound a tangential quadrilateral. The question above asks for a direct geometric explanation why this map is symmetric with respect to permutations of the four input lines: In the present setting, the quadrilateral $ABCD$ and the quadrilateral $EBFD$ are determined by the same four lines, which suggests that they should give the same point $H$ on $AC$ (and hence give the same generalized incenter).

$ABCDEF$ is the complete quadrangle determined by four arbitrary lines, so that $E=AD\cap BC$ and $F=AB\cap CD$.

Reflect $D$ across the line $AC$ to a point $D'$, and reflect $F$ across the same line $AC$ to a point $F'$.

Show that the two lines $BD'$ and $EF'$ intersect the diagonal $AC$ at the same point $H$.

Equivalently, prove that $$ BD' \cap AC = EF' \cap AC. $$

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

$$\frac{BH}{D'H}=\frac{BE}{EC}\cdot\frac{CF'}{F'D'}$$ $$\implies\frac{BH}{DH}=\frac{BE}{EC}\cdot\frac{CF}{FD}$$

Background and motivation.
This configuration arose from studying a generalized “incenter map” associated to four lines $\ell_1,\ell_2,\ell_3,\ell_4$ in the plane. It is a rational map defined (via reflections across diagonals and perpendicular constructions) for any four lines, and restricts to the usual incenter when the four lines bound a tangential quadrilateral. The question above asks for a direct geometric explanation why this map is symmetric with respect to permutations of the four input lines: In the present setting, the quadrilateral $ABCD$ and the quadrilateral $EBFD$ are determined by the same four lines, which suggests that they should give the same point $H$ on $AC$ (and hence give the same generalized incenter).

$ABCDEF$ is the complete quadrangle determined by four arbitrary lines, so that $E=AD\cap BC$ and $F=AB\cap CD$.

Reflect $D$ across the line $AC$ to a point $D'$, and reflect $F$ across the same line $AC$ to a point $F'$.

Show that the two lines $BD'$ and $EF'$ intersect the diagonal $AC$ at the same point $H$.

Equivalently, prove that $$ BD' \cap AC = EF' \cap AC. $$

Background and motivation.
This configuration arose from studying a generalized “incenter map” associated to four lines $\ell_1,\ell_2,\ell_3,\ell_4$ in the plane. It is a rational map defined (via reflections across diagonals and perpendicular constructions) for any four lines, and restricts to the usual incenter when the four lines bound a tangential quadrilateral. The question above asks for a direct geometric explanation why this map is symmetric with respect to permutations of the four input lines: In the present setting, the quadrilateral $ABCD$ and the quadrilateral $EBFD$ are determined by the same four lines, which suggests that they should give the same point $H$ on $AC$ (and hence give the same generalized incenter).

$ABCDEF$ is the complete quadrangle determined by four arbitrary lines, so that $E=AD\cap BC$ and $F=AB\cap CD$.

Reflect $D$ across the line $AC$ to a point $D'$, and reflect $F$ across the same line $AC$ to a point $F'$.

Show that the two lines $BD'$ and $EF'$ intersect the diagonal $AC$ at the same point $H$.

Equivalently, prove that $$ BD' \cap AC = EF' \cap AC. $$

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

$$\frac{D'H}{BH}=\frac{BE}{EC}\cdot\frac{CF'}{F'D'}$$ $$\implies\frac{DH}{BH}=\frac{BE}{EC}\cdot\frac{CF}{FD}$$

Background and motivation.
This configuration arose from studying a generalized “incenter map” associated to four lines $\ell_1,\ell_2,\ell_3,\ell_4$ in the plane. It is a rational map defined (via reflections across diagonals and perpendicular constructions) for any four lines, and restricts to the usual incenter when the four lines bound a tangential quadrilateral. The question above asks for a direct geometric explanation why this map is symmetric with respect to permutations of the four input lines: In the present setting, the quadrilateral $ABCD$ and the quadrilateral $EBFD$ are determined by the same four lines, which suggests that they should give the same point $H$ on $AC$ (and hence give the same generalized incenter).

$ABCDEF$ is the complete quadrangle determined by four arbitrary lines, so that $E=AD\cap BC$ and $F=AB\cap CD$.

Reflect $D$ across the line $AC$ to a point $D'$, and reflect $F$ across the same line $AC$ to a point $F'$.

Show that the two lines $BD'$ and $EF'$ intersect the diagonal $AC$ at the same point $H$.

Equivalently, prove that $$ BD' \cap AC = EF' \cap AC. $$

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

$$\frac{BH}{D'H}=\frac{BE}{EC}\cdot\frac{CF'}{F'D'}$$ $$\implies\frac{BH}{DH}=\frac{BE}{EC}\cdot\frac{CF}{FD}$$

Background and motivation.
This configuration arose from studying a generalized “incenter map” associated to four lines $\ell_1,\ell_2,\ell_3,\ell_4$ in the plane. It is a rational map defined (via reflections across diagonals and perpendicular constructions) for any four lines, and restricts to the usual incenter when the four lines bound a tangential quadrilateral. The question above asks for a direct geometric explanation why this map is symmetric with respect to permutations of the four input lines: In the present setting, the quadrilateral $ABCD$ and the quadrilateral $EBFD$ are determined by the same four lines, which suggests that they should give the same point $H$ on $AC$ (and hence give the same generalized incenter).