The Medians of a Triangle Are Concurrent, A Vector-Based Proof - As Straightforward as It Gets!

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[@alegian7934]

I was constantly waiting for you to "ingeniously" say that γ must equal δ, but not even that was required!! Beautiful indeed :)

[@MathTheBeautiful]

I'm glad you enjoyed it and thank you for your feedback!

[@KaiseruSoze]

Excellent proof! Not only that, but you just produced a ratio of lengths that produces numbers!

[@Jack_Callcott_AU]

I really enjoyed this. This type of vector proof is elegant, simple, and the way to go, but you forgot to address the 3rd median to show it intersects the same point, to make the proof complete. I know it is fairly obvious.

[@MathTheBeautiful]

Good point!

[@theoremus]

Nice!

[@MathTheBeautiful]

Thanks!

[@Pluralist]

[@ReginaldCarey]

9:11 when did you claim that a and b are linearly independent?

[@MathTheBeautiful]

If they represent the size of a triangle, then they are non-collinear and non-zero and therefore independent

[@ReginaldCarey]

@@MathTheBeautifulyou drew two vectors as sides of a triangle, the problem as stated did not have a restriction on a right triangle. If they are linearly independent then, they must form a right triangle. As of writing this I have not seen the rest of the video.

[@MathTheBeautiful]

@@ReginaldCarey Actually, linearly independent just not non-colinear. See czcams.com/video/7siCweBXxCo/video.html

[@ReginaldCarey]

@@MathTheBeautiful thank you for fixing my understanding of linear independence. I was imposing orthogonality as a condition. Orthogonal vectors are linearly independent but it is not a requirement for independence.