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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] 
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2 Answers 2

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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.

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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

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  • $\begingroup$ 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? $\endgroup$
    – Red Banana
    Aug 14, 2023 at 21:06
  • $\begingroup$ Perhaps there is, but I don't know how to use vectors with a GeometricScene. $\endgroup$
    – Syed
    Aug 15, 2023 at 1:14

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