# Incenter lies on the Newton-Gauss line

Let $ABCD$ be a tangential quadrilateral with incenter $I$. Prove that $I$ lies on the line joining the midpoints of $\overline{AC},\overline{BD}, \overline{EF}$, where $E=AB\cap CD, F=AD\cap BC$.

In this topic, I proved that the midpoints of the three diagonals are collinear, but I'm struggling to show that the incenter lies on this line too.

As a note, taking a projective transformation implies that the result holds for any conic inscribed in the quadrilateral.

• I didn't get who is $I$. There is a circumcircle for $ABCD$. Do you mean that there is also a incircle for $ABCD$ and $I$ is the center? Dec 21, 2016 at 16:38
• A tangential quadrilateral is one with an incircle: I have named the center of this circle $I$. Dec 21, 2016 at 20:09
• So you also have to prove that there is an incircle. Have you done it? Dec 21, 2016 at 20:18
• No. Given a quadrilateral with an incircle, prove the result. Dec 21, 2016 at 20:19

Let $$M$$, $$N$$ and $$P$$ be the midpoints of $$AC$$, $$BD$$ and $$EF$$. The problem is trivial when $$ABCD$$ is a parallelogram (it becomes a rhombus), so suppose that it is not one. Consider the locus of points $$X$$ such that $$[XAB] + [XCD] = [XDA] + [XBC].$$ where $$[ABC]$$ denotes the signed area of $$\triangle ABC$$. Since this is a linear equation in the coordinates of $$X$$, it is either satisfied trivially for all $$X$$ in the plane, be contradictory, or characterizes a line.

If it is satisfied for all points in the plane, then it holds for $$X = A$$, which gives $$[ACD] = [ABC]$$. This implies that $$B$$ and $$D$$ lie on different sides of $$\overline{AC}$$, and their distances to $$\overline{AC}$$ are the same. Therefore, it follows that $$\overline{AC}$$ bisects segment $$BD$$. Similarly, $$\overline{BD}$$ bisects segment $$AC$$, so $$ABCD$$ is a parallelogram, which is a contradiction to the assumption.

Meanwhile, it is easy to see that $$M$$ and $$N$$ satisfy the equation, so the locus must be a line through them, say $$\ell$$. To see that $$P$$ also lies on this line, note that $$[PAB] = \frac{1}{2}[FAB]$$ and $$[PCD] = \frac{1}{2}[FCD]$$. Therefore, $$[PAB] + [PCD] = \frac{1}{2}[ABCD] = [PDA] + [PBC]$$ and hence $$M$$, $$N$$, $$P$$ all lie on $$\ell$$. It remains to show that $$I$$ also lies on this line, which is equivalent to showing that $$I$$ also satisfies the equation. Notice that by Pitot's theorem, we must have $$AB + CD = AD + BC$$. Therefore, if we let $$r$$ be the radius of the incircle, we have $$[IAB] + [ICD] = \frac{r}{2}(AB + CD) = \frac{r}{2}(AD + BC) = [IDA] + [IBC]$$ as desired. $$\blacksquare$$

Without loss of generality you can choose your coordinate system in such a way that the incircle becomes the unit circle, and none of the tangents touches it at $(-1,0)$. In that case you can describe all four points of contact using the tangent half-angle substitution as homogeneous coordinate vectors of the form $[1-\alpha^2:2\alpha:1+\alpha^2]$. The tangent in this point would be described by the equation $(1-\alpha^2)x+(2\alpha)y=1+\alpha^2$. Do the same for $\beta,\gamma,\delta$ and you have four lines tangent to the unit circle and intersecting in six points, your $A$ through $F$. I'll use $\alpha$ for the line $AB$, $\beta$ for $BC$, $\gamma$ for $CD$ and $\delta$ for $DA$. Doing the math the homogeneous coordinates of these points turn out to be

\begin{align*} A &= \begin{pmatrix}1-\alpha\delta\\\alpha+\delta\\1+\alpha\delta\end{pmatrix} & B &= \begin{pmatrix}1-\alpha\beta\\\alpha+\beta\\1+\alpha\beta\end{pmatrix} & E &= \begin{pmatrix}1-\alpha\gamma\\\alpha+\gamma\\1+\alpha\gamma\end{pmatrix} \\ C &= \begin{pmatrix}1-\beta\gamma\\\beta+\gamma\\1+\beta\gamma\end{pmatrix} & D &= \begin{pmatrix}1-\gamma\delta\\\gamma+\delta\\1+\gamma\delta\end{pmatrix} & F &= \begin{pmatrix}1-\beta\delta\\\beta+\delta\\1+\beta\delta\end{pmatrix} \end{align*}

To compute the midpoint of two points in homogeneous coordinates, you multiply each point by the last coordinate of the other, then add them. So the midpoint between $A$ and $C$ would be

$$(1+\beta\gamma)A+(1+\alpha\delta)C=\begin{pmatrix} 2-2\alpha\beta\gamma\delta\\ \alpha+\beta+\gamma+\delta+\alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta\\ 2+2\alpha\beta\gamma\delta+2\alpha\delta+2\beta\gamma \end{pmatrix}$$

Likewise for the others. You can already see how the first two coordinates are completely symmetric under permutations of the four parameters. Only the third is not. So all your points lie on the line with homogeneous coordinates

$$\begin{pmatrix} \alpha+\beta+\gamma+\delta+\alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta\\ 2\alpha\beta\gamma\delta-2\\ 0 \end{pmatrix}$$

as the scalar product between this line and each point is zero. You could also read the equation of the line as

$$(\alpha+\beta+\gamma+\delta+\alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta)x + (2\alpha\beta\gamma\delta-2)y = 0$$

From the zero in the last coordinate in the vector form resp. the zero constant term in the equation form you can read that this line will pass through the origin, which is the center of the unit circle, your incircle. It has homogeneous coordinates $I=[0:0:1]$.

Note that the above computation makes use of midpoints, which are not invariant under projective transformations, only under affine transformations. So I've got serious doubts that the result would apply equally well to arbitrary conics. It should hold for ellipses, though, as any non-degenerate ellipse can be mapped to the unit circle using an affine transformation only.