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Is it true that the fact that medians of any two sides meet at a distance of 2/3 from vertex to median on the opposite side implies that all three medians intersect at a single point?

If yes, how is it so?

Ross Millikan
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Adam
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  • Related : http://math.stackexchange.com/questions/411709/proving-that-the-medians-of-a-triangle-are-concurrent – lab bhattacharjee Nov 25 '13 at 15:36
  • Yes, but if you want to prove this fact can you make use of the fact that...and how? I suppose that is the question. I hope it makes sense. wink – Adam Nov 25 '13 at 15:41

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Draw any one of the three medians. There is a unique point two-thirds of the way from the vertex to the opposite edge along that median. Your statement says that both of the other medians must pass through that point, hence the three are coincident at that point.

EuYu
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  • "There is a unique point two-thirds of the way from the vertex to the opposite edge along that median." I understand this part, but as far as other medians go, arent they supposed to lie 2/3 of the distance from their own vertex? (as opposed to all medians lying 2/3 of the distance from vertex along a single median)?? – Adam Nov 25 '13 at 16:00
  • I understood your claim as "any median intersects another at a point two-thirds of the way from vertex to opposite edge". If that's not what you meant, then what exactly did you mean by "medians of any two sides meet at a distance of 2/3 from vertex to median on the opposite side"? – EuYu Nov 25 '13 at 16:03
  • Well, I meant that if we take the segment from vertex A to the midpoint of the opposite side, there is a point at 2/3 distance from A, and if we take the segment from B to midpoint of opposite side then there is a point 2/3 distance from B. These two points coincide. – Adam Nov 25 '13 at 16:10
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    Right. Two lines intersect at most once. This means that each median intersects another two-thirds of the way from the vertex to edge. Take the median from $A$ and let $X$ be the point two-thirds of the way along the median. Then the median from $B$ meets the median from $A$ at $X$ and the median from $C$ meets the median from $A$ at $X$. Hence all three meet at $X$. – EuYu Nov 25 '13 at 16:15