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 circumcenter.
You can calculate the circumcenter 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.
Factoids about the centroid
Each median divides the triangle into two equal areas. All three
medians divide the triangle into six segments of equal area. The triangle
will balance at the centroid, and it will balance along any line that passes
through the centroid.
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.
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.
