Lines from vertices through centroid of a triangle bisect opposite sides.

Question

It is well known that the lines from the vertices through the centroid of a general triangle bisect the sides opposite to each vertex. Is there a simple geometrical proof for this? I've managed to solve it using vector-algebra, but it's not nice to say the least. Below is a short Maple-Code for the proof: I started with the line starting at vertex $A$ (with angle $a$) that bisects $CB$ and similarly the line starting at vertex $B$ (with angle $b$) that bisects $AC$. Those 2 lines intersect at some point that turns out to be the centroid. Now I construct a third line that starts at vertex $C$ and goes through that intersection/centroid and I show that (in the coordinate-system) at $y=0$ this line will have x-coordinate $AB/2$ which is the midpoint of the side AB.

The equations that I use below read $$t\begin{pmatrix} AB-\frac{CB}{2}\cos b \\ \frac{CB}{2} \sin b \end{pmatrix} = \begin{pmatrix} \frac{CA}{2} \cos a \\ \frac{CA}{2} \sin a \end{pmatrix} - s\left[\begin{pmatrix} \frac{CA}{2}\cos a \\ \frac{CA}{2} \sin a \end{pmatrix} - \begin{pmatrix} AB \\ 0 \end{pmatrix}\right]$$ where the LHS is the line from the vertex $A$ ($t=0$) at the origin to the midpoint of side $CB$ ($t=1$) and the RHS is the line from the midpoint of $CA$ ($s=0$) to the vertex $B$ ($s=1$) at $x=AB$.

restart;
eqs := [t*(AB - cos(b)*CB/2) = cos(a)*CA/2 - s*(cos(a)*CA/2 - AB), t*sin(b)*CB/2 = sin(a)*CA/2 - s*sin(a)*CA/2];
sol := solve(eqs, [t, s]);
simplify~(eval(eqs, sol[1]));
coords := [lhs(%[1]), lhs(%[2])];
coords2 := [(1 - t)*cos(a)*CA + t*coords[1], (1 - t)*sin(a)*CA + t*coords[2]];
solve(coords2[2], t);
simplify~(eval(coords2, t = %));
simplify~(eval(%, cos(b) = (AB - cos(a)*CA)/CB));
simplify~(eval(%, sin(b) = sin(a)*CA/CB));

Answer 1

Let ABC be a triangle , with vertices $A(0,0) ,B(x_1,y_1)$ and $C(x_2 , y_2)$.

Therefore , coordinates of centroid G are $G({x_1 + x_2 \over 3 }) , ({{ y_1 + y_2} \over 3})$.

Let the line through A and G cut BC at D(x,y), where D divides BC in the ratio $k:1$.

Therefore $$(x,y)=\left ({k x_2 + x_1 \over k+1 } \right) ,\left( {k y_2 + y_1 \over k+1 }\right)$$

Now slope(AG) = slope(GD) .

Therefore , $${{{ky_2 + y_1 \over k+1} - {y_1 + y_2 \over 3}} \over {{kx_2 + x_1 \over k+1} - {x_1 + x_2 \over 3}}} = {{y_1 + y_2 \over x_1 + x_2 }}$$

Which on solving gives you k=1.

Therefore , AG divides BC in the ratio 1:1 , hence ,D is the mid point of B and C .

Answer 2

The natural setting for this problem is affine space. Lines are preserved under an affine transformation and the midpoint of a line segment is also preserved. Any triangle is an affine transform of an equilateral triangle for which the

lines from the vertices through the centroid of a general triangle bisect the sides opposite to each vertex

is clearly true. Since it is true for the special case of equilateral triangles, and this property is preserved by affine transformations, it is true in general.