# Trigonometry/Circles and Triangles/Other Centres of a Triangle

The circumcenter and incenter can both be regarded as in some senses the center of a triangle, though the circumcenter can lie outside the triangle. There are other points that can be regarded as the center.

**The Orthocenter**

If lines are drawn passing through each vertex of a triangle, perpendicular to the side opposite that vertex (these lines are called *altitudes*), they intersect at a point called the **orthocenter** of the triangle. The orthocenter lies inside an acute triangle, at the right angle of a right triangle, and outside an obtuse triangle.

The existence of the Orthocenter (in other words, the fact that the three altitudes intersect at one point) is a consequence of Ceva's Theorem, which will be discussed later.

If R is the circumradius of ABC, the distance between the orthocenter and circumcenter is R^{2}(1-8cos(A)cos(B)cos(C)).

**Medians and the Centroid**

If lines are drawn passing through each vertex of a triangle and the midpoint of the opposite side (these lines are called *medians*), they intersect at a point called the **centroid** of the triangle. Again, the fact that the centroid exists is a consequence of Ceva's Theorem. The centroid lies ^{2}⁄_{3} of the way from the vertex to the opposite side. It is the center of mass of three equal masses, one at each vertex, and also of a uniform piece of material in the shape of the triangle. If the sections of the medians lying between the centroid and the midpoints of the sides are erased, they will cut the triangle into three smaller triangles that are equal in area.

A median divides a triangle into two triangles of equal area.

If a median through a vertex is drawn, then a line from another vertex that bisects this median, it will cut the side opposite the second vertex at a point ^{1}⁄_{3} of the way from the first to the remaining vertex.

**Euler's Theorem**

The circumcenter O, orthocenter H and centroid G lie on a straight line and GH = 2OG. (Euler published this in 1765.)

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