Concurrence of Medians (1 of 2: Proof via similar triangles)

[leonardofelipesilva]

This proof is very nice, it uses a very simple argument for the ratio of the sides and for the median, that's what's so cool about geometry

I like it more than the coordinate proof

[JohnM...]

Remember, people, perpendicular bisectors meet at the circumcentre ( centre of an excribed circle), altitudes meet ( are concurrent) at the orthocentre, and medians meet at the centroid.

[chordsequencer]

I cant help but to fail to see that concurrency of the medians cannot be "proven" in the sense of the word. To me, its more like an instance of a "magical miracle" that all three of these lines meet at one, unique point.

Question: Is this theorem, being a specific case of the Ceva Theorem itself, named after a famous mathematician? Just seems like it would to make it easier to memorize.

[srovilegend]

I wanted to ask a question. Would that median also be in a ratio 1:2. Coz it trisects right?

[JJclassofLearning]

sir I am not finding the solution of an question?

[manartz]

just use midpoint theorem for the first part then prove ABP similar to EDP ez

[renan]

O cara da frente ao inves de prestar atençao no professor, preferiu copiar a lousa

[leonardofelipesilva]

Actually, it's not clear for me why is ABC similar to DCE

[nononnomonohjghdgdshrsrhsjgd]

so you just criticized the proof, didn't show it by yourself in whole 8 minutes and titled the video: Proof via similar triangles. Nice one!