Geometry problem about Nagel line; $IG$ intersects $(AB)$ and $(AC)$ inside

Question

Let $\triangle ABC$ be a scalene triangle, $I$ its incenter and $G$ its centroid. Prove that the line $GI$ intersects both $(AB)$ and $(AC)$ on the line segments (id est between the two points of extremities), if and only if there exists a $t \in \left[0;1\right)$ for which we may write $a = tb + (1 -t)c$, where $a, b, c$ are the sides of the triangle in trigonometric notation (id est $a = BC$, $b = AC$, $c = AB$).

Approach:

Let's observe that $IG$ line is the Nagel line of the triangle. And that the condition $a = tb + (1 - t)c$ means that $a$ is neither the largest, nor the smallest side in the triangle.

By observing these, it's enough to prove two facts:

1. If $a$ is not the medium-sized side in the triangle, the Nagel line will intersect inside the other two lines;
2. If the Nagel line intersects $(AB)$ and $(AC)$ inside, then $a$ is the medium-sized side in triangle.

I thought at trilinear coordinates, by exploiting the fact that the Nagel line equation is well known (see Wolfram MathWorld), and also that the equations of $AB$ and $AC$ are easy to find. Also, thinking at Cartesian (natural) coordinates, as it'ss easier to establish if a point belongs to a line segment, not just the line.

However, because the calculations are terrible, I'm searching for a classical (maybe vectorial) solution to the problem. In this direction, my only idea was to express $\vec{IG}$ in terms of $(\vec{AB}, \vec{AC})$, and then to use a proportionality coefficient $\alpha$ to find an expression of the vector $\vec{MN}$, where $M, N$ are the intersection points with the whole lines $AB$ and $AC$. However, despinde managing to find these expressions, they brought me no benefit to prove the result I'm seeking.

Answer 1

The barycentric coordinates of $G$ with respect to $ABC$ are $(1,1,1)$, and those of $I$ are $(a,b,c)$. If $GI$ intersects $BC$ at a point with barycentric coordinates $(0,y,z)$, we get the equation $$ \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ 0 & y & z \\ \end{vmatrix} = 0. $$

Hence $(b - a)z = (c - a)y$. Line $GI$ intersects line $BC$ outside segment $BC$ if and only if the ratio of $y$ to $z$ is negative, which occurs precisely when $a$ is the middle length.

Analogous statements can be made for lines $AB$ and $AC$.

Answer 2

$\triangle ABC$ is a scalene triangle. Now here is a geometrical argument.

WLOG we assume that $c \gt b$ and then we need to show that $c \gt a \gt b$

For the line $l$ through $G$ and $I$ to cut the segment $AC$ internally, incenter $I$ must be above median $CD$ or in other words, $\displaystyle \frac{AH}{BH} \lt 1$.

But given $CH$ is an angle bisector, $\displaystyle \frac{AH}{BH} = \frac ba \implies b \lt a$.

Now for the line $l$ to cut the segment $AB$ internally, incenter $I$ must be below median $BE$, so $\displaystyle \frac{CK}{AK} = \frac a c \lt 1 \implies a \lt c$.