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Note by Rohit Camfar
11 months, 3 weeks ago

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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.

Similarly we can do for the other 2 medians.

Sorry for being unable to upload picture.

- 11 months, 3 weeks ago

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

- 11 months, 3 weeks ago

- 11 months, 3 weeks ago

Keep it pure keep it geometric .

- 11 months, 3 weeks ago

this seems to be your signature dialogue hahahaha

- 11 months, 2 weeks ago

No

- 11 months, 2 weeks ago

TRy trig

- 11 months, 2 weeks ago

It is it's a compliment not anything bad

- 11 months, 2 weeks ago

I know that

- 11 months, 2 weeks ago

- 11 months, 2 weeks ago

No bye

- 11 months, 2 weeks ago

Exams are being held too late

- 11 months, 2 weeks ago

Can;t believe

- 11 months, 2 weeks ago