This can be easily done.
Consider a triangle ABC with centroid O and medians AD; BE and CF.
Now;
Let;
\(ar(AOE) = ar(COE) = a\);
\(ar(COD) = ar(BOD) = b\);
\(ar(BOF) = ar(AOF) = c\);

But; since \(ar(ADB) = ar(BEA) = \frac{ar(ABC)}{2} \)
\( \implies ar(AOE) = ar(BOD)\)
OR \(a = b\)

Similarly, we can prove that in fact;
\(a = b = c = \frac{ar(ABC)}{6} \)

Now, using the fact that triangles AOF and AOC share the same height but the area of the latter is twice the former; we can easily deduce that CO:FO = 2:1.

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TopNewestThis can be easily done.

Consider a triangle ABC with centroid O and medians AD; BE and CF.

Now;

Let; \(ar(AOE) = ar(COE) = a\);

\(ar(COD) = ar(BOD) = b\);

\(ar(BOF) = ar(AOF) = c\);

But; since \(ar(ADB) = ar(BEA) = \frac{ar(ABC)}{2} \)

\( \implies ar(AOE) = ar(BOD)\)

OR \(a = b\)

Similarly, we can prove that in fact;

\(a = b = c = \frac{ar(ABC)}{6} \)

Now, using the fact that triangles AOF and AOC share the same height but the area of the latter is twice the former; we can easily deduce that CO:FO = 2:1.

Similarly we can do for the other 2 medians.

Sorry for being unable to upload picture.

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Eureka , Nice approach . But can be done even more simply

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Keep it pure keep it geometric .

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this seems to be your signature dialogue hahahaha

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No

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