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  "body": "\n## Proof\n\nWe compute the point two thirds of the way along each median.\n\n1. Let $M_A = \\frac{1}{2}(B + C)$, $M_B = \\frac{1}{2}(C + A)$ and $M_C = \\frac{1}{2}(A + B)$ be the [[midpoints|0S706JBZBG]] of the three sides.\n2. We have $A + \\frac{2}{3}(M_A - A) = \\frac{1}{3} A + \\frac{2}{3} \\cdot \\frac{1}{2}(B + C) = \\frac{1}{3}(A + B + C)$.\n3. By the definition of the [[centroid|0X7EEWES5G]], the right side of step 2 is $G$. So $G = A + \\frac{2}{3}(M_A - A)$.\n4. In the same way, $B + \\frac{2}{3}(M_B - B) = \\frac{1}{3} B + \\frac{1}{3}(C + A) = G$ and $C + \\frac{2}{3}(M_C - C) = \\frac{1}{3} C + \\frac{1}{3}(A + B) = G$.\n5. The [[median|0XW8HX8VEV]] from $A$ is the set of the points $A + t (M_A - A)$ with $t \\in [0, 1]$. Since $\\frac{2}{3} \\in [0, 1]$, step 3 shows that $G$ lies on it.\n6. In the same way, step 4 with $t = \\frac{2}{3}$ shows that $G$ lies on the medians from $B$ and from $C$.\n7. So $G$ lies on each of the three medians, and $G = A + \\frac{2}{3}(M_A - A)$, with the same formula at $B$ and at $C$.\n\n## Idea\n\nThe centroid is the average of the three vertices. Go two thirds of the way from a vertex to the midpoint of the opposite side. You reach one third of each of the three vertices, summed. The result is symmetric in the vertices, so all three medians pass through it.\n",
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