Centres of Triangle

A square has one obvious centre — the point where the diagonals cross. A circle has one obvious centre — the point equidistant from every point on the boundary. But a triangle? Where is the "centre" of a triangle?

The honest answer is: a triangle does not have one centre. It has several, and each one answers a different question. Where should you place a single support so the triangle balances perfectly? That gives the centroid. Where is the centre of the largest circle that fits inside the triangle? That gives the incentre. Where is the centre of the circle that passes through all three vertices? That gives the circumcentre. Where do the three altitudes meet? That gives the orthocentre.

Each of these points is defined by a different geometric property, and each has a coordinate formula. Here, you will derive every one of them from scratch.

The centroid

Cut a triangle out of stiff cardboard and try to balance it on the tip of a pencil. The point where it balances — where the triangle's weight is evenly distributed in every direction — is the centroid.

Geometrically, the centroid is the point where the three medians of the triangle intersect. A median is the line segment from a vertex to the midpoint of the opposite side.

Why the three medians meet at a single point

This is not obvious. Three lines drawn arbitrarily inside a triangle have no reason to pass through the same point. But the three medians always do — a fact called concurrency.

Here is the proof using coordinates. Let the vertices be A(x_1, y_1), B(x_2, y_2), C(x_3, y_3).

The midpoint of side BC is M_A = \left(\frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2}\right).

The median from A to M_A is a segment from (x_1, y_1) to \left(\frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2}\right).

The point that divides this median in the ratio 2 : 1 from vertex A (i.e., two-thirds of the way from A to M_A) is, by the section formula:

Why: the section formula gives the point dividing a segment in a ratio m:n. Here m = 2 (weight on the far end M_A) and n = 1 (weight on the near end A).

Simplifying the x-coordinate:

Why: the 2 in the numerator cancels with the 2 in the denominator of the midpoint, leaving x_2 + x_3. Adding x_1 gives the sum of all three.

Similarly for y. So the point two-thirds of the way along the median from A is:

Now do the same computation starting from vertex B. The midpoint of AC is M_B = \left(\frac{x_1 + x_3}{2}, \frac{y_1 + y_3}{2}\right). The point two-thirds of the way from B along this median is:

The same expression. The point two-thirds of the way along the median from B is the same point G.

By the same argument starting from C, the median from C also passes through G. All three medians meet at the same point.

The formula is beautifully simple: the centroid is the arithmetic mean of the three vertices. It does not depend on the side lengths or the angles — only on the coordinates. This makes it the easiest of all the triangle centres to compute.

The 2:1 property is worth remembering: the centroid sits one-third of the way from the midpoint of a side, or equivalently, two-thirds of the way from the vertex.

The incentre

The incentre is the centre of the inscribed circle (or incircle) — the largest circle that fits inside the triangle, touching all three sides.

Geometrically, the incentre is the point where the three angle bisectors of the triangle meet. An angle bisector divides an angle into two equal halves.

Why the incentre is equidistant from all three sides

The incentre must be the centre of a circle tangent to all three sides. A circle tangent to a line touches it at exactly one point, and the centre of such a circle is perpendicular distance r from that line. So the incentre is equidistant from all three sides — that distance is the inradius r.

Any point on the bisector of angle A is equidistant from the two sides that form angle A (sides AB and AC). Any point on the bisector of angle B is equidistant from sides BA and BC. The intersection of these two bisectors is therefore equidistant from all three sides — and the bisector of angle C automatically passes through the same point. That is why the three bisectors are concurrent.

Derivation of the formula

The incentre has a coordinate formula, but the weights are not equal — they depend on the side lengths.

Let A(x_1, y_1), B(x_2, y_2), C(x_3, y_3) be the vertices, and let the side lengths be:

(The standard convention: side a is opposite vertex A, side b is opposite vertex B, side c is opposite vertex C.)

The angle bisector from vertex A hits side BC at a point D that divides BC in the ratio AB : AC = c : b (by the angle bisector theorem).

Why: the angle bisector theorem says that the bisector of an angle of a triangle divides the opposite side in the ratio of the adjacent sides. This is a standard Euclidean result — a consequence of the sine rule applied to the two sub-triangles.

So D divides BC in the ratio c : b. By the section formula:

Why: in the section formula, the weight on C is c (the side AB, opposite to C in the ratio) and the weight on B is b (the side CA, opposite to B in the ratio). The weights cross to the opposite endpoints.

The incentre I lies on the segment from A to D. The ratio in which I divides AD can be shown (via the property that I is equidistant from all three sides) to place I at:

Here is the direct verification. The incentre divides AD in the ratio a : (b + c - a)... but there is a cleaner way to see why the formula is correct. The incentre is the weighted average of the vertices, where the weight of each vertex equals the length of the opposite side. This can be proven by noting that the bisector from A meets BC at D with BD:DC = c:b, and then I divides AD in the ratio BD + DC : AD'... The simplest proof uses the fact that the position vector of I satisfies:

This follows from the angle bisector property and can be verified by checking that I is equidistant from all three sides.

Compare this with the centroid. The centroid gives equal weight \frac{1}{3} to each vertex. The incentre gives weight proportional to the opposite side length. A vertex opposite a long side pulls the incentre toward it more strongly.

This makes geometric sense. A vertex with a long opposite side has a wide angle, and the bisector of a wide angle sweeps deep into the triangle — pulling the incentre toward that vertex.

The circumcentre

The circumcentre is the centre of the circumscribed circle (or circumcircle) — the unique circle passing through all three vertices of the triangle.

Geometrically, the circumcentre is the point equidistant from all three vertices. It is where the three perpendicular bisectors of the sides meet.

Why the perpendicular bisectors are concurrent

The perpendicular bisector of a segment is the set of all points equidistant from the two endpoints. The perpendicular bisector of AB consists of all points equidistant from A and B. The perpendicular bisector of BC consists of all points equidistant from B and C. Their intersection is equidistant from A, B, and C — and the perpendicular bisector of AC automatically passes through this point. That proves concurrency.

Computing the circumcentre

The circumcentre O_c = (x, y) satisfies:

Squaring the first two:

Expanding both sides:

The x^2 and y^2 terms cancel from both sides:

Why: squaring both distance equations and subtracting eliminates the squared unknowns, leaving a linear equation in x and y. This is the key trick — a distance equation is quadratic, but the difference of two distance equations is linear.

Rearranging:

Why: collect all x- and y-terms on the left, and all constants on the right. The right side is a difference of squares, which you can factor if needed.

This is one linear equation in x and y. Setting O_cA = O_cC gives a second linear equation:

Solving these two simultaneous linear equations gives (x, y) — the circumcentre. There is no single clean closed-form formula like the centroid or incentre; you solve the system for each specific triangle.

An important property: the circumcentre lies inside the triangle if all angles are acute, on the hypotenuse if the triangle is right-angled, and outside the triangle if one angle is obtuse.

The orthocentre

The orthocentre is the point where the three altitudes of the triangle meet. An altitude is the perpendicular from a vertex to the opposite side (or the line containing the opposite side).

Why the altitudes are concurrent

This is not as immediately obvious as for the medians or perpendicular bisectors. A standard proof uses coordinates.

Let A(x_1, y_1), B(x_2, y_2), C(x_3, y_3). The altitude from A is perpendicular to BC. The direction vector of BC is (x_3 - x_2,\; y_3 - y_2). A line through A perpendicular to this direction satisfies:

Why: two vectors are perpendicular when their dot product is zero. The direction from any point (x,y) on the altitude to A must have zero dot product with the direction of BC.

Similarly, the altitude from B perpendicular to AC is:

Solving equations (1) and (2) simultaneously gives the orthocentre H = (x, y). You can then verify that the altitude from C (perpendicular to AB) also passes through this point.

To verify concurrency, check that the altitude from C perpendicular to AB also passes through the intersection of (1) and (2). That altitude has the equation:

Adding equations (1) and (2), expanding, and collecting terms gives an expression that is equivalent to equation (3) — so any point satisfying (1) and (2) automatically satisfies (3). The three altitudes are concurrent.

Like the circumcentre, the orthocentre does not have a single compact coordinate formula — you solve a system of linear equations. Unlike the circumcentre, the orthocentre lies inside the triangle only when all angles are acute. For a right triangle, the orthocentre is at the vertex of the right angle. For an obtuse triangle, it lies outside.

Worked examples

Common confusions

• "The centroid and incentre are the same point." They coincide only in an equilateral triangle. In any other triangle, they are distinct. The centroid depends only on the vertex positions (equal weights); the incentre depends on the side lengths (unequal weights).

• "The circumcentre is always inside the triangle." Only for acute triangles. For a right triangle, the circumcentre is at the midpoint of the hypotenuse. For an obtuse triangle, it lies outside the triangle, on the far side of the longest side.

• "The orthocentre is always inside the triangle." Again, only for acute triangles. For a right triangle, it is at the right-angle vertex. For an obtuse triangle, it is outside.

• "In the incentre formula, the weight of a vertex is proportional to that vertex's angle." No — the weight is the length of the opposite side, not the angle at that vertex. These are related (a larger opposite side often corresponds to a larger angle), but they are not the same.

• "Any three of these centres determine the triangle." Two suffice in some cases. The centroid, circumcentre, and orthocentre are related by the Euler line — a remarkable fact explored below.

Going deeper

Ex-centres

The three ex-centres are the centres of the excircles of the triangle. An excircle is a circle that is tangent to one side of the triangle and to the extensions of the other two sides. Every triangle has exactly three excircles, one opposite each vertex.

The ex-centre opposite vertex A is denoted I_A. It is the intersection of the external bisector of angle A with the internal bisectors of angles B and C.

The formula is a variation of the incentre formula, with one sign change:

Why: the external bisector at A reverses the direction relative to the internal bisector. This flips the sign of the weight on vertex A, changing +a to -a in both numerator and denominator.

The other two ex-centres are obtained by the same pattern — flip the sign of the weight corresponding to the vertex opposite the excircle:

The denominators -a + b + c, a - b + c, and a + b - c are all positive by the triangle inequality (each is the sum of two sides minus the third).

Each excircle's radius (the exradius) is related to the triangle's area \Delta and semi-perimeter s = \frac{a+b+c}{2}:

The inradius similarly satisfies r = \frac{\Delta}{s}. These four radii are connected by several elegant identities. The cleanest is:

To see why: \frac{1}{r_A} + \frac{1}{r_B} + \frac{1}{r_C} = \frac{(s-a) + (s-b) + (s-c)}{\Delta} = \frac{3s - 2s}{\Delta} = \frac{s}{\Delta} = \frac{1}{r}.

Another beautiful relation involves the circumradius R:

The Euler line

Here is a remarkable fact. For any triangle, the centroid G, circumcentre O_c, and orthocentre H are collinear — they lie on a single straight line called the Euler line.

Moreover, the centroid divides the segment from the circumcentre to the orthocentre in the ratio 1 : 2:

or equivalently, G lies one-third of the way from O_c to H. This relationship is independent of the triangle — it holds for every non-equilateral triangle. (In an equilateral triangle, all three points coincide, so the "line" is a single point.)

The Euler line is one of the most surprising results in triangle geometry. There is no obvious reason why the centroid (intersection of medians), the circumcentre (intersection of perpendicular bisectors), and the orthocentre (intersection of altitudes) should be related by such a clean linear relationship. The proof uses the vector identity:

where O is the circumcentre. Once you have this (which follows from the circumcentre being equidistant from all three vertices and some vector algebra), the centroid's position \vec{OG} = \frac{\vec{OA} + \vec{OB} + \vec{OC}}{3} = \frac{\vec{OH}}{3} falls out immediately.

The nine-point circle

The Euler line passes through one more famous point: the centre of the nine-point circle, which passes through:

1. The three midpoints of the sides
2. The three feet of the altitudes
3. The three midpoints of the segments from the vertices to the orthocentre

Nine points, all on one circle — and the centre of that circle sits exactly at the midpoint of the segment from the circumcentre to the orthocentre (i.e., the midpoint of O_c H). The radius of the nine-point circle is exactly half the circumradius.

This is one of the most beautiful theorems in all of triangle geometry, and it connects the centroid, circumcentre, orthocentre, and nine-point centre in a single elegant line.

Where this leads next

The centres of a triangle are a meeting point of coordinate geometry, Euclidean geometry, and algebra. The topics that follow build on what you have learned here.

• Coordinate Geometry — Basics — the distance and section formulas used throughout this article.
• Area and Collinearity — compute the area of any triangle from its vertices; the collinearity test is needed to verify the Euler line property.
• Straight Line Forms — equations of lines (medians, altitudes, perpendicular bisectors) in various forms.
• Locus — the circumcircle is a locus (the set of all points equidistant from the circumcentre), and so are many other curves defined by distance conditions.
• Geometry with Complex Numbers — another coordinate system where each point is a complex number, making rotation and reflection much cleaner.