The incenter of a triangle is the point of intersection of the triangle's three angle bisectors.
Incenter, concurrency of the three angle bisectors
The three angle bisectors have to meet in a single point because...
If a circle is inscribed in an angle, the angle bisector passes through the center of the circle. The two sides of the angle each form a "point of tangency" where they intersect the circle (not shown on the diagram). At each point of tangency, there is a radius that meets the side of the angle at a right angle. You can there are two congruent triangles formed by the vertex of the angle, a point of tangency, and the intersection of the two radii, which, of course has to be at the center of the circle.
There are several interesting relationships in a triangle between the inscribed circle, the angle bisectors, and the three "exscribed" circles.
As a review of the barycentric coordinates of point P in triangle ABC, I'll remind you they are the three weights you need to give points A, B, and C so that P is the centroid (weighted average) of the three vertices. So, for example, A=(1,0,0), B=(0,1,0), and C=(0,0,1).
Another way to view barycentric coordinates is as "proportional altitudes". Let me explain. A point, P, can be identified by its distance from each of the three sides as a proportion of the altitude to that side. So, if hA is the altitude of point A from its opposite side, BC, and hB and hC are the other altitudes, then barycentric coordinates of a point Q = (x,y,z) indicate that the distances of Q from the three sides are x�hA, y�hB, z�hC, respectively.
Now, the incenter is equidistant from each of the sides, distant from each side by the length of the radius of the incircle. So, viewing the barycentric coordinates as proportional altitudes, and letting the incenter I = (x, y, z), we see that in order to put point I equidistant from the three sides, we need x=1/hA, y=1/hB, and z=1/hC, so the barycentric coordinates of I are
(1/hA, 1/hB, 1/hC)
The scale of each of the points is arbitrary, because these barycentric coordinates are not normalized. So, for example, the barycentric coordinates (1,2,3) and (2,4,6) represent the same point. Now, observe that the altitudes hA, hB, hC are inversely proportional to their bases, because the product of base�altitude is constant, the area of the triangle. So the barycentric coordinates of the incenter can be more simply represented,
(a, b, c)
You can calculate the vector coordinates of the incenter from its barycentric coordinates by multiplying each vertex (as a vector) by the corresponding barycentric coordinate, adding the results, and then dividing by the sum of the barycentric coordinates.
If the vertices of the triangle are given on a two-dimensional plane as A (a,b), B (c,d), and C (e,f) then the incenter (h,k) is given by
h = (a sqrt((c-e)2+(d-f)2) + c sqrt((e-a)2+(f-b)2) + e sqrt((a-c)2+(b-d)2) ) /
(sqrt((c-e)2+(d-f)2) + sqrt((e-a)2+(f-b)2) + sqrt((a-c)2+(b-d)2) )
k = (b sqrt((c-e)2+(d-f)2) + d sqrt((e-a)2+(f-b)2) + f sqrt((a-c)2+(b-d)2) ) /
(sqrt((c-e)2+(d-f)2) + sqrt((e-a)2+(f-b)2) + sqrt((a-c)2+(b-d)2) )
Do you see how we derived the two equations, above, from the barycentric coordinates of the incenter? If not, consider this: The calculation, above, is easier than it looks to carry out, because of the common subexpressions for the lengths of the sides. If we let the three side lengths be
A = sqrt((c-e)2+(d-f)2),
B = sqrt((e-a)2+(f-b)2), and
C = sqrt((a-c)2+(b-d)2),
then the values of h and k can be more simply represented as
h = (aA + cB + eC ) / (A+B+C), and
k = (bA + dB + fC ) / (A+B+C)
Consider triangle ABC, pictured on the left side of this page. Now, if we inscribe a circle in any of its angles, say, angle ACB, then the center, O, of the circle bisects the angle. This is true because of symmetry: The quadrilateral formed by C, O, and the two points where the circle is tangent to the angle is symmetrical about line CO, which means angle ACO is equal to angle BCO. So the center of any circle inscribed in an angle defines the angle bisector of that angle.
The reverse is also true. That is, if O is any point in the interior of
an angle such that ray CO bisects angle ACB, then the circle with its center at
O that is tangent to AC is also tangent to BC.
take this idea a step further. If we draw two angle bisectors, one
bisecting angle C, and the other bisecting angle B, then they
will intersect at some point inside the triangle, which we will label point "I".
Since point I is on the bisector of angle C, the circle centered at I and
tangent to AC is also tangent to BC. Also, since point I is on the
bisector of angle B, that same circle, tangent to AC, will also be tangent to
AB. So, you see, this circle is tangent to all three sides of the
From this we see that the intersection of any two angle bisectors is the center if the inscribed circle. It follows that all three internal angle bisectors intersect at one point, which is the center of the inscribed circle, or "incircle". The perpendicular distance from point I to any of the sides is the radius, r, of the incircle. The area, K, of triangle ABC is the sum of the areas of AIB, BIC, and CIA. If we label the lengths of the sides of triangle ABC in traditional form, with side a opposite vertex A, b opposite B, and c opposite C, then the areas of these three small triangles are cr/2, ar/2, and br/2 because the height of each small triangle is the radius of the incircle. The sum of these areas is
K = (a+b+c)(r)/2, or
K = sr. Also,
r = K/s
where s = (a+b+c)/2 is the semiperimeter of triangle ABC.
By extending two of the sides of the triangle, AC, and AB, we can bisect the exterior angles of the triangle as well. Here we have drawn two exterior bisectors which intersect at point Ea. The same logic that we used for the incircle can be used again to show that a circle can be drawn at Ea that is tangent to all three sides, suitably extended, of the triangle. A circle centered at Ea and tangent to AC is also tangent to CB, because Ea is on the bisector of the angle formed by those two lines. Similarly, the same circle, which I remind you is tangent to CB is also tangent to AB because it lies on the external bisector of angle B. Finally, since this circle is tangent to both AC and AB, the internal bisector of angle A also passes through point Ea. This circle is called an "excribed" circle, or excircle.
Just as the three "in-triangles" AIB, BIC, and CIA add up to triangle ABC, three "ex-triangles" AEaB, BEaC, and CEaA add up, in a way, to the same triangle ABC. That is, if you subtract the area of BEaC from the sum of the areas AEaB and CEaA, the result is the area of ABC.
K = (-a+b+c)(Ra)/2, or
K = (s-a)(Ra). Also,
Ra = K/(s-a)
where s is the semiperimeter of triangle ABC, and Ra is the radius of the
excircle with center Ea.
Why stop there? Both external bisectors of angle C are on the same line, as are those of angles B and A. In this diagram, these bisectors were extended to show two other excribed circles. As you can see, a triangle has three internal angle bisectors and three external angle bisectors. These six lines intersect in exactly seven points: the three vertices of the triangle (pairwise intersections), the centers of the four in- and excircles (triple intersections).
The internal angle bisector of a given vertex is perpendicular to the external angle bisector. This makes each of the internal angle bisectors an altitude of triangle EaEbEc formed by the three excenters. The point I then, which is the incenter of triangle ABC, is also the orthocenter of triangle EaEbEc.
The radii of the four circles shown are related by the equation
1/r = 1/Ra+1/Rb+1/Rc because Ra = K/(s-a), etc. so
1/Ra+1/Rb+1/Rc = (3s-a-b-c)/K = s/K = 1/r
The radii of the four circles shown are related to the area of triangle ABC by
the formula K=sqrt(r Ra
Rb Rc), because from
K2 = s(s-a)(s-b)(s-c). So,
sqrt(r Ra Rb Rc) =
sqrt(K/s K/(s-a) K/(s-b) K/(s-c)) =
sqrt(K4 / (s(s-a)(s-b)(s-c)) ) =
sqrt(K4 / K2) = K.
If we let R stand for the radius of the circle that circumscibes triangle ABC (the larger gold circle in the diagram, below), we get one more interesting theorem, Ra+Rb+Rc-r=4R. This is because 4R=abc/K (proof) along with the following rather tedious algebra:
Ra+Rb+Rc-r = K/(s-a) + K/(s-b) + K/(s-c) - K/s
now, multiplying each fraction by 1 = s(s-a)(s-b)(s-c)/K2,
Ra+Rb+Rc-r = s(s-b)(s-c)/K + s(s-a)(s-c)/K + s(s-a)(s-b)/K - (s-a)(s-b)(s-c)/K
= (2s3 - as2 - bs2 - cs2 + abc)/K
= (2s3 - (a+b+c)s2 + abc)/K
The three excircles, Ea, Eb, and Ec are externally tangent to the Nine Point Circle, the smaller gold circle shown here. The Nine Point Circle is so-called because it passes through the three "feet" of the altitudes (or their extensions), the midpoints of the three sides, and the midpoints of the segments connecting each of the vertices to the orthocenter (where the three altitudes meet).
In addition to being externally tangent to the three excircles, the Nine Point Circle is internally tangent to the incircle at a point called the Feuerbach point. This result is known as Feuerbach's theorem.
Also, since A, B, and C are the feet of the altitudes of triangle EaEbEc, the circumcircle of ABC is also the nine-point circle of EaEbEc. For a given triangle, the radius of the nine-point circle is half that of the circumcircle, so the radius of the larger gold circle in this diagram is twice that of the smaller gold circle.
The Angle Bisector Theorem of Triangles states that the point where a bisector intersects the opposite side divides that side in the same ratio as that of the other two sides. Referring to the drawing, below, the internal bisector of angle C divides side AB into two segments, x and y. The ratio of the lengths of two segments equals the ratio of the lengths of the two sides that form angle C; in other words, x:y = b:a.
Referring to the diagram, below, the proof of this theorem starts by constructing a line parallel to CC1 that passes through B, meeting line AC at point D. Triangle BCD is isosceles, because angle ACC1 equals angle ADB, and angle C1CB equals angle CBD. Since line CC1 bisects angle ACB, angle ACC1 equals angle C1CB, so angle ADB equals angle CBD. Triangle AC1C is similar to triangle ABD, so AC:AD = x:(x+y), and so AC:CD = x:y. Since BCD is isosceles, BC=CD, so AC:BC = x:y, proving the theorem.
Surprisingly, the external bisector, shown here as CC2, has the same property. That is, the ratios AC2:BC2 and b:a are equal. (. . . . . . proof? ) Note that if triangle ABC is isosceles with AC = BC, then point C1 is halfway between A and B (and so CC1 is an altitude), and point C2 doesn't exist, because if it did, it would have to be infinitely far away from the triangle.
If CC1 were any cevian (not necessarily a bisector), whose length is t, then Stewart's Theorem tells us
xa2 + yb2 = (x+y)(t2+xy),
which has a number of other formulations, such as replacing (x+y) with c, or expanding the right hand side this way, which is the one we will find most useful:
xa2 + yb2 = (x+y)t2 + xy2 + x2y
The following proof of Stewart's Theorem uses the law of cosines on triangles AC1C and BC1C. We will let θ represent angle AC1C. Note that angle BC1C is the supplement of θ, and so its cosine is -cos θ. Then the law of cosines on these two triangles gives us
b2 = x2 + t2 - 2xt cos θ
a2 = y2 + t2 + 2yt cos θ
Solving these two equations for cos θ,
cos θ = (x2 + t2 - b2)/(2xt) = (a2 - y2 - t2)/(2yt), so
(2xt)(a2 - y2 - t2) = (2yt)(x2 + t2 - b2)
x(a2 - y2 - t2) = y(x2 + t2 - b2)
Rearranging this equation gives us the desired result. Since x:y = b:a, ax=by. Solving Stewart's Theorem for t2,
t2 = (xa2 + yb2 - xy2 - x2y)/(x+y)
t2 = (xa2 + yb2 - (x+y)xy)/(x+y)
t2 = (xa2 + yb2)/(x+y) - xy
So far, we haven't taken advantage of the fact that CC1 is not just any cevian, but a bisector. Now, since x:y = b:a, ax=by, and so xa2=aby, and yb2=abx, so
t2 = (abx + aby)/(x+y) - xy
t2 = ab - xy
Pat Ballew points out that each bisector, such as CC1 is divided into two pieces by the incenter, I. The lengths of these two pieces are always in a ratio related to the three sides. Using the figure, above, as our example, the ratio CI:IC1 = (a+b):c.
To see this, look at triangle CC1B. The internal bisector of B cuts CC1 into two pieces at I in the same ratio as that of the lengths other two sides of that triangle, i.e. CI:IC1 = a:y. Similarly, using the bisector of A and triangle CC1A, CI:IC1 = b:x. Since
CI:IC1 = a:y, and
CI:IC1 = b:x, it follows that
CI:IC1 = (a+b):(x+y) = (a+b):c
www.pballew.net/Tribis.html describes the relationship between angle bisectors, the incircle, and the excircles, a very comprehensive summary, by Pat Ballew.
The Nine-point circle, from mathworld, is the circle that passes through the "feet" of the three altitudes of a triangle. It also passes through the midpoints of the three sides, and the three midpoints of the segments connecting the vertices to the orthocenter; hence "nine points".
Stewart's Theorem, from mathworld.
Wikipedia: Nine-point circle
Cut the knot: Feuerbach's Theorem
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).
Orthocenter -- the intersection of the altitudes (or their extensions) of a triangle
Inversion Circle -- a page that describes orthogonal circles and inversion circles.
The webmaster and author of this Math Help site is Graeme McRae.