The centres of a triangle

A triangle has three corners, connected by three edges. There are various positions related to it that can be thought of as centre in one sense or another – even though two of them may actually lie outside it. In each case, there is a line one can construct in relation to a vertex, an edge or the pair of a vertex and the opposite edge; this gives us three lines; and the interesting fact is that they meet in a single point, rather than each missing the point where the other two meet (which would lead to them forming a triangle), as is more usual for three lines in a plane.

Centre of a circle touching each edge

What's the largest circle we can fit inside a triangle ? Imagine expanding a circle inside the triangle until it bumps up against an edge, without crossing it; the edge shall then be touching the circle. After that, we can expand it away from the edge, with the edge remaining tangent to the circle. As we do so, it'll sooner or later bump up against another edge, that'll now also be tangent to it. The two edges the circle touches necessarily meet in one of the vertices of the triangle. The radii of the circle to the points where it touches the two edges shall meet those tangents to the circle in right angles; the line from the circle's centre to the vertex, at which the two edges meet, is then the hypotenuse of two right angle triangles, one on either side of it, that each have a radius of the circle as one of the other perpendicular sides of the triangle. The two right-angle triangles are thus mirror images of one another in their shared hypotenuse, so their angles in the corner of the triangle are equal and the circle's centre lies on the bisector of the original triangle's angle at that corner. We can then move the circle away from that corner, along this angle bisector, expanding it as we go, to keep it touching the same two sides, until it finally expands to bump into the third side. At that point, it must surely be on the angle-bisectors of all three corners of the triangle, and no further expansion is possible within the triangle. But that can only work if all three angle bisectors meet in one point.

So let's bisect the angles at the three vertices and look at where two of these (grey) angle-bisectors meet, dropping a (green) perpendicular from this point to each of the sides of the triangle. Between each of the bisected angles and the meeting point we have a segment of the angle's bisector as the hypotenuse of two right-angle triangles, one on each side, with an edge into the vertex as one side of each triangle and a perpendicular as the remaining side. Since the two right-angle triangles have the same angle at the vertex, as their shared hypotenuse bisects the original triangle's angle there, and each has right angles facing the bisector, their third angles, at the meeting point, must in fact be equal. As they share their hypotenuse, they are thus necessarily each other's mirror images in it, making the two perpendiculars' lengths equal. By considering the matching pair of right-angle triangles for the other vertex whose bisector met, we find one of these two perpendiculars equal to the third, hence all three are equal and the place where two angle bisectors meet is necessarily the same distance from (the nearest point to it on) each of the three sides.

If we now draw a straight line from the third vertex to the given meeting-point of the first two's bisectors, it forms two right-angle triangles with the perpendiculars and the edges into this third vertex; those triangles share their hypotenuse and have a same-length side opposite the third vertex; that (e.g. via Pythagoras) makes their third sides equal, so the two triangles are again mirror images of one another and we can infer that our line from the third vertex to where the other two bisectors met is in fact the bisector of the angle at the third vertex. So the three angle bisectors all meet in a single point; and it's the same distance from each of the sides of the triangle. This point is known as the incentre of the triangle; it's the centre of the largest circle that lies wholly inside the triangle (with the perpendiculars we dropped as its radii). This circle might even be called the incircle; but the name doesn't ring any bells in my memory.

The Law of Cotangents

a b c r s−a s−a s−b s−b s−c s−c α α β β γ γ Notice that the points where the incentre touches the edges split each edge in two parts; each part forms, with a radius of the incircle and the angle bisector of the vertex at its end away from the incircle, a right-angle triangle that's the mirror image for the matching triangle on that vertex's part of the other edge into it. Thus the two edge-parts that end in each vertex are equal.

Label the edges by their lengths a, b, c; let A be the length of the edge-parts into the vertex opposite a, B the length of the edge-parts into the vertex opposite b and C the length of the edge-parts into the vertex opposite c. Then a = B+C, b = C+A and c = A+B; the triangle's perimeter is then a+b+c = 2.(A+B+C); half of this, known as the semi-perimeter, is thus s = A+B+C, from which we can recover A = s−a, B = s−b and C = s−c; so we can now forget we had names for the part-lengths and just deal with the edge-lengths a, b, c and the semi-perimeter s = (a+b+c)/2.

Let r be the radius of the incircle. The area of the pair of right-angle triangles opposite side a is just r.(s−a) and similar for the other pairs gives, thanks to a+b+c = 2.s, a total area of r.(3.s −2.s) = r.s, so the area of a triangle is its semi-perimeter times the radius of its incircle.

Now, dividing the radius r of the incircle by each of the edge-part lengths, we get the tangent of half the triangle's angle at the vertex associated with the given edge-parts. Let the angle opposite the edge of length a be 2.α, that opposite b be 2.β and that opposite c be 2.γ. Thus

This law is more usually stated in terms of cotangents, with r.coTan(α) = s−a, &c; the cotangent of an angle is just one divided by the tangent. I'll use that form of it below, but I find it clearer in the present form.

Now the sines, cosines and thus tangents of sums of angles can be expressed in terms of those of the angles summed; and α +β +γ = turn/4, half the sum of the interior angles of a triangle, so

0 = Cos(turn/4)
= Cos(α +β +γ)
= Cos(α).Cos(β).Cos(γ) −Cos(α).Sin(β).Sin(γ) −Sin(α).Sin(β).Cos(γ) −Sin(α).Cos(β).Sin(γ),
= Cos(α).Sin(β).Sin(γ) +Sin(α).Sin(β).Cos(γ) +Sin(α).Cos(β).Sin(γ)

which we can divide through by Sin(α).Sin(β).Sin(γ), which is necessarily non-zero, to obtain

= coTan(α) +coTan(β) +coTan(γ)
= (s−a)/r +(s−b)/r +(s−c)/r
= (3.s−2.s)/r = s/r

but, equally,

whence, scaling by r.r.r.s,

from which we can infer the area, r.s, as the square root of the right hand side. I'll give an alternative derivation of the same conclusion, below, that doesn't need trigonometry.

Balance point

What's average of the positions of our triangle's corners ? The average of any two of them is the mid-point of the edge that connects them; so the average of all three must lie on the line from there to the other, twice as far from that other as from the edge's mid-point (since this point now represents two of the original points). But that can only work if the lines from each vertex to the middle of the opposite side all meet in one place, two thirds of the way along each from the vertex end of it.

Connect each vertex by a straight (grey) line to the mid-point of the opposite edge; these lines are known as the medians of the triangle. Consider where two of the medians meet: construct, through this point, a (green) line parallel to each edge of the triangle.

Opposite each vertex, the lines parallel to those that meet at the vertex form a small triangle with the middle part of the opposite edge; as the edges of this triangle are one from the original and two parallel to the original's others, it is similar to the original. Between each line through the meeting point and the vertex opposite the side it's parallel to, we get an intermediate-sized triangle, again similar to the original, due to sharing two edges and having the third parallel. Each intermediate-sized triangle contains two of the small triangles, along its edge through the meeting point; each small triangle is the intersection of two of the intermediate-sized ones; let's now work out the relative sizes of these triangles.

For either of the medians we intersected, scaling down the full triangle towards the vertex at one end pulls the opposite edge (whose mid-point is the other end of the median) to a line parallel to it, with its ends on the other two edges and its centre-point on the median; so, when the scaling brings this edge to pass through the meeting-point, we get an intermediate-sized triangle, with the meeting point as the centre of its edge that passes through that; the two halves of this intermediate edge are edges of small triangles, inside the intermediate-sized one, telling us that those two small triangles are the same size as each other (since they're similar and have a corresponding edge equal) and half the size of the intermediate-sized triangle. Doing this for the other intersected line, we discover its vertex's intermediate triangle is twice the two small triangles it contains, one of which we already know to be equal to the remaining small triangle; so the three small triangles are all the same size and two of the intermediate triangles are twice as big as them.

But now the third intermediate triangle's edge through this meeting point is made of two corresponding edges of equal small triangles, so is bisected by its edge through the meeting point, which makes the third intermediate triangle the same size as the other two. It also implies that scaling the triangle down towards its third vertex, by the factor that makes the edge opposite it pass through our meeting of medians, puts this meeting point at the centre of the scaled opposite edge; but scaling towards this vertex moves the far end of its median towards it along that median, necessarily to the mid-point of the scaled image of the opposite edge, i.e. our meeting of the first two medians. So the meeting point of the first two medians is in fact on this third median. This meeting point is called the centroid of the triangle. (In particular, balancing point masses at the corners implies that the centroid is the average of the position vectors of the three vertices.)

Balancing on it

Now, I described the centroid in its section heading as a balance point; so now to elaborate on that. Briefly: if your triangle is a sheet of some stiff material with uniform mass per unit area and you place its centroid on a pivot-point, the triangle should balance (in theory; of course, your physical reality shall always be a little off centre, so you might need a rounded point to balance it on in practice); likewise, if you put three equal masses on the corners of such a triangular sheet, it'll still balance on a pivot under the centroid. In contrast, a wire-frame of the triangle, with uniform mass per unit length along the edges, has a different balance point. For the corner or edges cases, of course, you'll need some massless rigid sheet to fill the space between, so as to mechanically be able to support the mass at any interior balance point; but the fact that corners and sheet agree means you can combine a uniform sheet with masses at each corner to get a result that'll still balance on the centroid.

For the plain uniform rigid sheet to tip, part of it must go up while the opposite part goes down and some axis stays still. Since the three lines in the construction of the centroid meet, they're not parallel; so they can't all be parallel to the axis that's staying still; indeed, at least two of them aren't. For the triangle to tilt, then, the net torque about at least two of the medians must be non-zero. So let's consider any one of the medians: it joins a vertex to the centre of the opposite edge. When considering torque about a line, we can move any of the triangle's mass parallel to the line without changing the torque; this doesn't change how far off the axis any mass is, or on which side. So shear the triangle, parallel to the median, until the bisected edge is perpendicular to our pivot line; this doesn't change areas or the torque. Now, in the sheared triangle, the pivot line still connects the vertex to the opposite edge's mid-point, meeting it in a right angle and thus cutting the sheared triangle into two right-angle triangles, with equal halves of the bisected edge perpendicular to their shared edge, the pivot-line; this makes the two triangles mirror images of each other. Now, reflecting in the pivot-line will reverse the torque; but it won't change the figure, so can't change the torque; so the torque is equal to its own reverse, making it zero. So the triangle's net torque about each median is zero and there can't be any net torque at all; the triangle must balance. This argument applies to both corner weights and the rigid sheet of uniform mass density.


The case of a wire-frame triangle is trickier: shearing does change the lengths of edges, so would change the balance-point in this case. Now, to work out whether it'll balance on a pivot, we must work out the net torque between the pivot's support and the weight of the triangle. That weight acts on each part of the weight, so we have to accumulate (by integrating or summing) the torque's contributions from the weight in various places. By using the pivot point as origin, we can eliminate the pivot's supporting force (its position is the zero, so the cross-product with force is then zero, too). The contribution from each element of the triangle's mass will then be the cross product of the mass's position, relative to the pivot, and the weight of the mass.

Since the weight is just the mass times the gravitational field strength, which I'll assume uniform, we can accumulate the mass of each element times its position, taking the cross product only at the end when we have accumulated its contributions from the whole triangle. We can, further, assume that we start out by placing the triangle in a horizontal plane, so that the gravitational field is perpendicular to all displacements within it; this just makes taking the cross product with it rotate our accumulated position times mass and scale it by a constant; that won't affect whether the result is zero, so we'll only get zero torque if the accumulated mass times position is zero.

Now, in accumulating position times mass, each edge has equal amounts either side of its mid-point at each distance from the mid-point, along a straight line. At each distance from the centre, the average of positions (times equal mass) for parts (of equal length) on opposite sides of the centre is just the centre; so the accumulated position times mass of the whole edge is just the mass of the edge times its mid-position. In effect, then, we reduce the wire-frame of a triangle to point masses at the mid-points of its edges: however, these masses are not equal, as in the cases considered above – each corner's mass is proportional to the length of the edge it was on. As a result, the average of mass times position for these masses is slightly off the centroid in the case illustrated (see the small blue dot near the middle). Indeed, in the case of a triangle with one edge very short compared to the other two, the short edge's mass is tiny compared to the masses of the long edges, whose centre-points are close together; so the balance point of the wire-frame is quite close to the line connecting the mid-points of the two long edges, roughly half-way along the line connecting the short edge to its opposite vertex, rather than roughly a third of the way along it, from the short edge, where the centroid lies.

Notice, however, that the triangle whose vertices are the mid-points of the original's edges is necessarily just a half-scale version of the original (each edge of it is an edge of a triangle in a corner that's the result of scaling the original down by a factor of two towards that corner), half-turned; and each median of the original bisects an edge of this inner triangle on its way to its opposite vertex, so the outer triangle's medians are the inner's medians, so they have the same centroid. Of course, we can repeat this construction indefinitely: at each step, we form a triangle whose vertices are the mid-points of the previous triangle's edges; every two steps shrink the triangle towards its centroid by a factor of four. (Indeed, if we interpret a half turn as a scaling by a factor of −1, each step just scales inwards by a factor of −2.)


Each median is split into a longer part from a vertex to the centroid and a shorter part, half as long, from the centroid to the middle of the opposite edge. The medians cut our triangle into six pieces, each a triangle whose edges are the longer part of one median, the shorter part of another and half an edge of the original triangle. We can group the pieces into three pairs, one pair along each edge of the triangle, and half-turn one of each pair about the mid-point of the edge they lie along. This maps the moved part's half of the edge onto the one that stayed still, taking the moved corner of the original triangle onto the other, and maps the moved part's short part of a median onto an extension of that median past the mid-point of the edge it bisected. This combines with the unmoved part's short part of that median to form an edge equal to the long part of that median, on the other side of the centroid. The moved and unmoved parts thus form a triangle, one (unmoved) side of which is the long part of a median, another is a half-turn image of the long part of another median, while the two short parts of the third median combine to form an edge equal to the long part of that median. Thus two parts we combine in this way form a triangle whose three sides are the long parts of the three medians of the original triangle. Furthermore, each is also parallel to the relevant median.

Furthermore, the half-edges of the original triangle – within each of the three triangles thus formed out of the six parts of the original – are medians of these derived triangles. When we construct all three medians of any one of these derived triangles, each cuts the others in the ratio 2:1 and so the long parts of these derived medians are each two thirds of a half-edge of the original triangle, so one third of an edge of the original. So if we repeat this construction to cut up the derived triangle into six pieces, that we rearrange in the same way, we get a triangle similar to the original, scaled down by a third in each length.

This delicious little result is elementary, as its its proof, and relies only on things that were known to the ancient Greeks (and probably earlier geometers), yet appears to have gone undiscovered until Lee Sallows noticed it in 2014.

Centre of a circle through the three corners

We've found the biggest circle inside a triangle: so what's the smallest circle around the triangle ? As before, we can think of a huge circle, way outside, around the triangle and shrink it until it bumps into one of the vertices; then we can shrink it towards that vertex until it bumps into another vertex. At this point, the triangle formed by its centre and the edge between those two vertices, both the same distance from the centre, is isosceles and the bisector of the angle between the circle's radii to those vertices must bisect the edge that connects them. So now we'll keep the centre on that line and move it closer to these two points as we shrink it further, until the circle bumps into the third point. At this point, it must be on the perpendicular bisectors of all three sides – which can only work if these do meet in a single point.

Given a triangle, construct the (grey) perpendicular bisectors of its three edges. Each point on the perpendicular bisector of an edge is exactly as far from one end of the edge as from the other; so, where two of these perpendicular bisectors meet, the common vertex in which the two bisected edges meet is the same distance (green lines) from this meeting point as each of the other two vertices of the triangle, so these two others are in fact equal distances from the meeting point, which thus lies on the third edge's perpendicular bisector. Thus, the perpendicular bisectors of the three edges all meet at one point.

This point is known as the circumcentre of the triangle. As the circumcentre is the same distance from each of the vertices, there's a circle – known as the circumcircle – centred on the circumcentre, that passes through all three vertices. Notice, by considering a triangle whose vertices are three points on the same side of some diameter of a circle, that the circumcentre needn't actually lie within the triangle. Even in that case, though, I trust it is clear that the triangle won't fit inside any smaller circle than its circumcircle. In a right-angle triangle, the hypotenuse is a diameter of the circumcircle and its mid-point is the circumcentre.


We've just bisected our edges with lines pependicular to them; previously we bisected each wih a line through the opposite vertex; and we've bisected the angles at those vertices, ignoring the opposite edges; but what happens if we drop lines through each vertex, perpendicular to the opposite edge ? It looks like these all meet in a single point, but why ? What geometric significance does this meeting-point hold ?

When we were constructing the centroid of our triangle, the mid-points of its edges formed a triangle whose edges were necessarily parallel to those of our original triangle but half as long; this triangle is the image of the original triangle under a half-turn about the centroid and scaling by a factor of two towards the centroid, which is thus also the centroid of the shrunk triangle. So if we do the reverse operation – scale away from the centroid by a factor of two and half-turn about the centroid – we must get a triangle whose edge mid-points are our original triangle's vertices, with edges parallel to those of the original triangle.

The perpendiculars we dropped from our vertices to their opposite edges are thus also perpendicular to this enlarged triangle's edges; and pass through the mid-points of its edges – so they are the perpendicular bisectors of the edges of the enlarged triangle. They do thus all meet in one place – the enlarged triangle's circumcentre, which is the image of our original triangle's circumcentre under a half-turn about the centroid and scaling by a factor of two away from that centroid. This circumcentre of the enlarged triangle, obtaind as meeting point of the perpendiculars from each vertex onto the opposite edge, is known as the orthocentre of the original triangle. Like the circumcentre, it can lie outside the triangle when one of its interior angles is bigger than a right angle (obtuse). In a right-angle triangle, it is the right-angle corner.

This gives rise to a nice way to construct a line, from a point on a circle, perpendicular to a given diameter of the circle, using nothing but straight lines (without measuring).

The Euler line

Because the orthocentre is the result of a half-turn and double-scaling of the circumcentre about the centroid, the orthocentre, centroid and circumcentre lie (in that order) along a single straight line, known as the Euler line of the triangle, unless two of them coincide – in which case, of course, all of them coincide. Considering what it means for the circumcentre and orthocentre to coincide, the three perpendicular bisectors of the edges would have to be the three perpendiculars from each vertex to the opposite edge; which implies that each vertex is on the perpendicular bisector of the opposite edge, hence is equidistant from the other two vertices; this being true for all three vertices, the three sides of the triangle are equal, so no two of these centres coincide unless the triangle is equilateral and all of the centres coincide.

When does the incentre lie on the Euler line ? When the triangle is isosceles, the bisector of the angle between equal edges is the perpendicular bisector of the third edge, hence is both a median and the perpendicular, to that edge, through the vertex opposite it; thus all of our centres lie on it. Hence it is the Euler line for an isosceles triangle and the incentre then lies on it.

I learned about the Euler line from a Numberphile video and one of his interviews with Zvezdelina Stankova.


To obtain the area of a triangle, cut it along the perpendicular, from a vertex with its widest internal angle, to the opposite edge; this decomposes the triangle into a pair of right angle triangles, back-to-back, with a common edge (that isn't the hypotenuse of either). Now cut along the perpendicular bisector of the first cut; on the side of this towards the original vertex, we get a pair of half-scale versions of the two right-angle triangles from the previous cut; half-turn each of these about its corner that wasn't on the original perpendicular, where its perpendicular bisector emerges from the original triangle. This rearranges the original triangle's four pieces into a rectangle that shares one side with the original triangle and whose other sides are half the height of the triangle perpendicular to that edge. So the area of the triangle is half the length of an edge times the height, by which is meant the distance from the opposite vertex to (the nearest point to it on) that edge.

I've only shown this for the edge opposite the widest angle; but it sufficed that the line perpendicular to that edge that passes through the vertex does so within the edge (i.e. not on its extension outside the triangle). Furthermore, we can do the same for any edge. Just shear parallel to that edge (a shear doesn't change area, or distances between lines parallel to the shear's direction) to put the opposite vertex where the perpendicular from it to the chosen edge does indeed lie within the edge. This hasn't changed the vertex's distance, along a line perpendicular to the edge, from the line of which the edge forms a segment, or the length of the edge; so hasn't changed either the area or the result of multiplying half base by height.

Now, still with the perpendicular from a vertex hitting the interior of the opposite side, let it sub-divide that side into two piece of lengths a and c, with the perpendicular being of length b. The perimeter of the triangle then consists of a, c and the hypotenuses of the right triangles with a or c as one side and b as the other; let's name the three whole sides A, B, C with:

and refer to half their sum, the semi-perimeter, as d, with 2.d = A +B +C, so

The triangle's area is b.(a +c)/2 = b.B/2. So now let's look at the product of d with the results of subtracting each side from it, using square = power(2) = (: x.x ←x :):

d.(d −A).(d −B).(d −C)
= (B +A +C).(A +C −B).(C +B −A).(B +A −C)/16
= −(A +C +B).(A +C −B).(C −A +B).(C −A −B)/16
= −(square(A +C) −B.B).(square(C −A) −B.B)/16
= −(A.A +2.A.C +C.C −B.B).(C.C −2.A.C +A.A −B.B)/16
= −(A.A −B.B +C.C +2.A.C).(A.A −B.B +C.C −2.A.C)/16
= (4.A.A.C.C −square(A.A +C.C −B.B))/16

in which A.A = c.c +b.b, C.C = a.a +b.b and B.B = a.a +2.a.c +c.c, so

d.(d −A).(d −B).(d −C)
= (4.(c.c +b.b).(a.a +b.b) −square((c.c +b.b) +(a.a +b.b) −(a.a +2.a.c +c.c)))/16
= (4.(c.c +b.b).(a.a +b.b) −4.square(b.b −a.c))/16
= (a.a.c.c +b.b.c.c +a.a.b.b +2.a.b.b.c −a.a.c.c)/4
= b.b.(a.a +2.a.c +c.c)/4 = square(b.(a +c)/2)

i.e. the square of the area of our triangle. This gives rise to Hero's formula (Hero was a not uncommon name in classical Greece), which gives the area of a triangle as the square root of the product of the semi-circumference and the three results of subtracting one edge's length from it.

Note that a straight line segment is the shortest path between its two end-points. The route along two edges of a triangle – from one end of the third, via the other vertex, to the other end of the third – is manifestly a path from one end of the third to the other hence at least as long as that third edge. Thus the sum of the lengths of any two sides of a triangle is greater than the length of the third side of the triangle. As a result, the half-length d of the perimeter of a triangle is greater than the length of any of its sides, so the factors in the product above are all positive, ensuring we can indeed take its square root.

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