# In a triangle ABC with medians $AE$, $BF$, $CD$, we have $\overrightarrow{AE}+\overrightarrow{BF}+\overrightarrow{CD}=0$ with Synthetic Geometry?

I am trying to set up Mathematica Synthetic Geometry stuff to prove the following:

Given a triangle ABC with medians $$AE$$, $$BF$$, $$CD$$. Show that we have $$\overrightarrow{AE}+\overrightarrow{BF}+\overrightarrow{CD}=0$$.

I wrote a piece of code I thought would be useful for this but FindGeometricConjectures won't yield anything useful. How can I create a scene where such conclusion can appear? I know that this can be done much more elementary (without Mathematica's Synthetic Geometry) but I want to learn about Synthetic Geometry.

G = RandomInstance[GeometricScene[
{a, b, c, d, e, f},
{
d == Midpoint[{a, b}],
e == Midpoint[{b, c}],
f == Midpoint[{c, a}],
Triangle[{a, b, c}],
Line[{a, e}],
Line[{b, f}],
Line[{c, d}]
}
]];
FindGeometricConjectures[G]


Or prove directly.(This proof also work for 3-dimension or n-dimension)

Clear[n, a, b, c, d, e, f];
n = 2;
d = Mean[{a, b}];
e = Mean[{b, c}];
f = Mean[{c, a}];
Reduce[(d - c) + (e - a) + (f - b) ==
0, {a, b, c, d, e, f} ∈ Vectors[n]]


True.

G = RandomInstance[
GeometricScene[{a, b, c, d, e, f}, {d == Midpoint[{a, b}],
e == Midpoint[{b, c}], f == Midpoint[{c, a}],
Triangle[{a, b, c}], Line[{a, e}], Line[{b, f}], Line[{c, d}]}]];


Not entirely satisfactorily but using the points in G:

Chop@((d - c) + (f - b) + (e - a) /. G["Points"] ) == {0, 0}


True

• Interesting that there are other ways of using this, I thought that we could only use this by combining GeometricScene and FindGeometricConjectures. Is there a way to make FindGeometricConjectures yield this conclusion or at least something equivalent to it? Commented Aug 14, 2023 at 21:06
• Perhaps there is, but I don't know how to use vectors with a GeometricScene.
– Syed
Commented Aug 15, 2023 at 1:14