Medians, Altitudes & Centres
Four segments, four points of concurrency, and what each one does.
The explanation
Every triangle has four families of special segments, and each family meets at a single point.
- Median: vertex to the midpoint of the opposite side. The three meet at the centroid, the balance point.
- Altitude: vertex perpendicular to the opposite side. The three meet at the orthocentre.
- Perpendicular bisector of each side. These meet at the circumcentre, which is equidistant from the three vertices — the centre of the circle through them.
- Angle bisector. These meet at the incentre, equidistant from the three sides — the centre of the circle that fits inside.
The centroid has a useful ratio: it sits two thirds of the way from each vertex to the opposite midpoint, splitting every median 2:1.
Which centre a problem wants is decided by one question: equidistant from the vertices means circumcentre, equidistant from the sides means incentre.
Each family of segments is concurrent, and the four points of concurrency have distinct characterisations.
The centroid divides each median in a 2:1 ratio measured from the vertex, and is the triangle's centre of mass. In coordinates it is the average of the vertices, ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3).
The circumcentre is equidistant from the vertices because every point on a side's perpendicular bisector is equidistant from that side's endpoints. It lies inside an acute triangle, on the hypotenuse's midpoint of a right triangle, and outside an obtuse one.
The incentre is equidistant from the sides, by the corresponding property of angle bisectors, and always lies inside the triangle.
The orthocentre also moves outside for obtuse triangles, and for a right triangle it coincides with the right-angle vertex.
The centroid, circumcentre and orthocentre are collinear on the Euler line for every non-equilateral triangle, with the centroid dividing the segment in a 2:1 ratio. In an equilateral triangle all four centres coincide.
Worked example
In △ABC, medians meet at G. If AG = 8 on the median from A to midpoint M, find AM and GM.
- The centroid splits each median 2:1 from the vertex.
- AG is the longer piece, worth 2 parts, so 1 part = 4.
- GM = 4.
- AM = AG + GM = 8 + 4.
Answer: AM = 12 and GM = 4
Common mistakes
- Applying the 2:1 ratio from the midpoint rather than from the vertex.
- Using the incentre when the problem asks for a point equidistant from the vertices.