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The Centroid of a triangle is the point formed by the intersection of its three medians.  A median is the line from a vertex to the midpoint of the opposite side.

 Centroid, concurrency of the three medians

### Concurrence of the medians

The three medians have to intersect in a single point because...

Let A' be the midpoint of BC, B' be the midpoint of AC, and C' be the midpoint of AB.  Ceva's Theorem tells us that AC'/C'B � BA'/A'C � CB'/B'A = 1 iff cevians AA', BB', and CC' concur on a point.  Here, AC'=C'B, BA'=A'C, and CB'=B'A, so each of the three ratios is 1, and so their product is 1.  Thus, by Ceva's Theorem, we have proved that the three medians concur in a point.

### Barycentric coordinates of the centroid

(1, 1, 1)

You can think of the barycentric coordinates of the "weights" of the vertices, so the weighted average of the vertices is the centroid.  You can calculate the centroid from the barycentric coordinates by multiplying each vertex (vector) by the corresponding barycentric coordinate, adding the results, and then dividing by the sum of the barycentric coordinates.

Each median divides the triangle into two equal areas.  (Proof: equal bases, identical altitudes −> equal area.)

All three medians divide the triangle into six segments of equal area. (Proof: pairwise, the triangles have equal bases, identical altitudes −> equal area; since each median divides the big triangle into two equal areas, pairs of sets of three small triangles are equal; algebra completes the result.)

The triangle will balance at the centroid, and it will balance along any line that passes through the centroid.

The centroid divides each median into two pieces that have a 2:1 ratio.  That is, each median is cut 1/3 of the way along its length by the other two medians.

Proof: take any three adjoining small triangles bounded by a median, M, and then observe that the portion of M on one side of the median is the base of a "double triangle" composed of a pair of small triangles, and the portion of M on the other side of the median is the base of one of the small triangles, and that the altitudes of the double and single triangles are the same.  Therefore, the portions of M are in the same proportion as the areas of the double and single triangles, i.e. 2:1)

Generalization of the 2:1 ratio property: van Aubel's second theorem: if triangle A'B'C' is the Cevian triangle of a point P, then

AP/PA' = AB'/B'C + AC'/C'B

### Related pages in this website:

Other triangle centers: Circumcenter, Incenter, Orthocenter, Centroid.  The Orthocenter and Circumcenter of a triangle are isogonal conjugates, and the Incenter is its own isogonal conjugate.

Summary of geometrical theorems summarizes the proofs of concurrency of the lines that determine these centers, as well as many other proofs in geometry.

Pappus' Centroid Theorems - the first one gives the surface area of a solid of rotation; the second gives the volume.  Each one uses the distance traveled by the centroid multiplied by the appropriate property of the plane figure to simplify the problem.

Barycentric Coordinates, which provide a way of calculating these triangle centers (see each of the triangle center pages for the barycentric coordinates of that center).

Ceva's Theorem -- AC'/C'B � BA'/A'C � CB'/B'A = 1 iff cevians AA', BB', and CC' concur on a point.

van Aubel's second theorem:  if triangle A'B'C' is the Cevian triangle of a point P, then AP/PA' = AB'/B'C + AC'/C'B, a generalization of the fact that the centroid of a triangle cuts each Cevian segment through it in the ratio 2:1.

The webmaster and author of this Math Help site is Graeme McRae.