The centre of mass of three equal masses placed each at one vertex of the triangle is called the centroid. It is also the centre of mass of a uniform lamina in the shape of the triangle. If the position vectors of the vertices A, B, C are a, b, c, then the centroid is at ^{1}⁄_{3}(a+b+c). All three medial lines (lines connecting a vertex to the midpoint of the opposite side) meet at that point.

Proof:

Centre of mass of the three equal masses is the same as the centre of mass of the system formed by the mass at A and a double mass at the midpoint M_{A} of BC, which has position vector ½(b+c).

That centre of mass must be on the line AM_{A}, and two-thirds of the way down from A to M_{A}, and so at ^{1}⁄_{3}(a+b+c).

That is also the centre of mass of a uniform lamina in the shape of the triangle.

The centre of mass must be on the medial line AM_{A}, because along any other line passing through A, every slice of the triangle parallel to the base BC like the one shown in green would have greater mass on one side of that line than the other.

Since the expression ^{1}⁄_{3}(a+b+c) is symmetric, and one medial line goes through it, all three medial lines must go through it. So it must be the centre of mass of the lamina.

The centre of mass of a triangular *frame* is a different point, the Spieker centre.

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