What is the proof that all 3 medians of a triangle will always intersect at a common point? Thank You.
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what are your own thoughts about the problem? – Vinyl_cape_jawa Jan 15 '18 at 08:23
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2The key property here is to prove two medians cut themselves in the ratio $\frac 23,\frac 13$. Since this is verified for any two of them, it means all $3$ lines passes through the same point. – zwim Jan 15 '18 at 09:28
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$$\frac{a/2}{a/2}\cdot\frac{b/2}{b/2}\cdot\frac{c/2}{c/2}=1,$$ hence the medians of a triangle are concurrent by Ceva's theorem. – Jack D'Aurizio Jan 15 '18 at 12:35
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https://artofproblemsolving.com/wiki/index.php?title=Centroid – Intelligenti pauca Jan 15 '18 at 14:37
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Closely related: How to prove using Plane Geometry that Centroid divides in ratio 2 :1 – user1551 Jan 15 '18 at 15:22
2 Answers
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First take an arbitrary triangle with vertices $(x_1,y_1),(x_2,y_2),(x_3,y_3)$
Then find the midpoint of each side. i.e. you will have 3 midpoints.
Then use each midpoint and the opposite vertex to construct the 3 equations for the 3 medians.
Lets call the 3 medians $K$, $L$ and $M$.
The equations get a bit messy due to the nasty slope expressions but follow through carefully and you will get your 3 medians.
Now to prove that they all intersect at the same point, use simultaneous equations to find the point of intersection of each pair of medians.
i.e. Find an expression for the intersection of $K$ and $L$. Then for $L$ and $M$, and finally for $K$ and $M$.
You will find the same answer in each of the 3 cases.
Kantura
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You might show that the point $\frac13(x_1+x_2+x_3,y_1+y_2+y_3)$ lies on all medians.
Michael Hoppe
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