Answer:
in the triangle shown, AX = (2/3)AG
Explanation:
You want to prove the medians meet at a point 2/3 of the distance from the vertex to the opposite side.
Point Definitions
Consider triangle ABC shown in the attached. Without any loss of generality, we can assign coordinates to the vertices as ...
A(0, 0), B(6a, 0), C(6b, 6c)
Then the segment midpoints are ...
G = (B+C)/2 = (3(a+b), 3c), H = (C +A)/2 = (3b, 3c), J = (A+B)/2 = (3a, 0)
2-Point Line Formula
The lines through a pair of coordinates (x1, y1) and (x2, y2) will have equations ...
y = (y2 -y1)/(x2 -x1)(x -x1) +y1
Median AG has equation ...
y = (3c/(3(a+b))(x -0) +0 = cx/(a+b)
Median BH has equation ...
y = (3c)/(3b -6a)(x -6a) +0
When written in the form px -y = q, we have ...
- c/(a+b)x -y = 0
- c/(b -2a)x -y = 6ac/(b -2a)
Solution
The second attachment shows the solution of these equations is ...
X = (2(a+b), 2c)
This point is 2/3 of the distance from vertex A to the opposite midpoint G:
(2(a+b), 2c) = (2/3)×(3(a+b), 3c)
Hence the intersection of medians is 2/3 of the distance from the vertex to the opposite side.
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Additional comment
We can verify that point X is also 2/3 of the distance from the other vertices to their opposite sides.
The 2/3 point of BH is (2H+B)/3 = (2(3b, 3c) +(6a, 0))/3 = (2(a+b), 2c)
The 2/3 point of CJ is (2J +C)/3 = (2(3a, 0) +(6b, 6c))/3 = (2(a+b), 2c)
That is, the 2/3 point of each median has the same coordinates as for the others. (Perhaps this is the simplest proof of all.)