Incenter - Point of Concurrency of Angle Bisectors of a Triangle (with proof)

In this post, we will talk about the Incenter, another one of those centers of a triangle. It also has the word ‘center’ in it.

Incenter, often denoted by letter I, is the center of the INCIRCLE of a triangle. Incircle is the name given to a circle of maximum possible radius that completely sits inside of the triangle. Out of all possible circles contained in a triangle, the biggest of all is the incircle, marked with a green outline in the picture below.

One property of an Incircle is that it “touches” the three edges of a triangle. Or the three edges are three tangents to the incircle, as shown in the picture above.

In general, tangent at any point of a circle is perpendicular to the radius at that point. So the three edges of a triangle must be perpendicular to the three radii of the incircle at points of tangencies, as depicted in the image below.

Because of this, the incenter is also said to be the point which is same ‘perpendicular distance’ away from three sides of a triangle. Most texts don’t use the word ‘perpendicular’, and simply assert that the incenter is the point same distance away from the three sides of a triangle.

One more thing we should know about the incenter is that it will always lie inside of the triangle, irrespective of what type of triangle it is. This is unlike the Circumcenter whose location can be either inside or outside or on the edge of a triangle, depending on what type of triangle it is.

There is also another way to describe the incenter of a triangle. It is the point where three angle bisectors of a triangle intersect. An angle bisector is a line that divides an interior angle of a triangle into two equal smaller angles. Every triangle has three interior angles, and hence every triangle has three angle bisectors. In any triangle, all three angle bisectors always intersect at a single point - the Incenter.

Ok. Now my next task will be to provide some sort of proof for everything that I have said so far. I will show you a typical math proof for the concurrency of three angle bisectors of a triangle, i.e. a proof that the three angle bisectors in any triangle always intersect at a single point. You can scroll to the end of this post to find that proof.

But for the existence of the circle that touches all three sides of a triangle, i intend to provide a deeper explanation to help us better understand why this circle has to exist for any given triangle. And that’s what I am going to do first. In the process we will see how and why angle bisectors come into the picture.

So let’s get started!

Consider a line L1 as shown below. Let’s start by answering a very simple question. Line L1 will be tangent to how many circles in the plane?

Clearly countless circles centered anywhere on the plane can be drawn touching the line. As a matter of fact, there are infinite circles that touch any one point on the line.

Take another line L2 not parallel to L1. The two lines meet each other at a point which is represented by letter P in the figure below. Now here comes my second question. How many circles can we draw touching both L1 and L2?

Imagine there is a circle with center O and the lines L1 and L2 are tangents to it at points P1 and P2 respectively.

Clearly OP1 = OP2, as they are radii of the same circle. Moreover, OP1 and OP2 are perpendicular to L1 and L2 because the radius is perpendicular to the tangent.

Any other circle to which L1 and L2 are tangents must also have its center equidistant from the lines. This fact must be true for any circle touching both the lines. We can use this neat little piece of observation to define a criteria for any circle to touch both the lines.

Two lines are tangent to a circle if and only if two perpendicular segments dropped from the center of that circle onto the two lines are equal in length; or in simple words, if the center of the circle is equidistant from the lines.

Now comes the angle bisector into the frame. As it turns out, any point on the angle bisector of the angle between the two lines satisfies this criteria. Perpendiculars dropped from any point on the angle bisector onto the two lines are equal in length.

This is a simple thing to prove. In the figure above, just ignore the circle for a moment. O is the point on the angle bisector which is bisecting the angle P between the lines L1 and L2. Let OP1 and OP2 be two perpendiculars dropped from O onto the lines. Clearly, triangles OPP1 and OPP2 are congruent by ASA or AAS test of congruence.

So, OP1 = OP2

I.e., perpendiculars from any point on the angle bisector onto the two lines are equal.

Our initial question was, how many circles touching both L1 and L2 can be drawn?

Clearly there are infinitely many possible circles, centered anywhere on the angle bisector, that can be drawn touching both L1 and L2. But the crucial point here is that the center must be on the angle bisector of the lines as that’s only when perpendiculars dropped from the center of the circle onto the lines will be equal.

Note : Before we move ahead, I just want to clarify that it is not one but a total of four angle bisectors for any pair of intersecting lines because there are total of four angles between any two intersecting lines. But I have shown only one of the four angle bisectors to save space, and also because that’s all I needed to prove my point.

Now consider triangle PQR as shown in the figure below. Let me draw angle bisectors of interior angles P and Q.

The angle bisectors of interior angles P and Q are intersecting at point I. Any pair of angle bisectors will always intersect, be it any triangle. In other words, they can never be parallel because the sides of a triangle cannot be parallel!

Since I is a point on the angle bisector of P, there exists a circle centered at I to which the sides PQ & PR are tangents.

Since I is also a point on the angle bisector of Q, there also exists a circle centered at I to which sides PQ and QR are tangents.

Now are they two different circles?

No, they are the same circle! Because if they are not, they have to be concentric as they have the same center I. They cannot be concentric because the side PQ is tangent to both the circles! A line in geometry cannot be tangent to two concentric circles at the same time! So they have to be one and only one circle, and the important thing to observe here is that it is touching all three sides of a triangle.

And that’s it. This shows that there exists a circle inside the triangle PQR to which the three sides of the triangle are tangents.

We call this circle Incircle as mentioned earlier, and its center the Incenter.

Now the next task is to show the incenter is the point of intersection of all three angle bisectors.

Well, we already see in the figure that it is the point of intersection of at least two angle bisectors. The question is, how can we show that the angle bisector of interior angle R also passes through I?

Again, we just saw there exists a circle centered at I touching all three sides of the triangle. So this circle touches the sides RP and RQ, which means the center, I, of this circle must be on the angle bisector of the interior angle R! Hence proving that the angle bisector of angle R also goes through I, and thus all three angle bisectors are concurrent.

There are also a couple of other formal ways to prove this statement. Let us look at one of them where we are using the combo of Converse of Ceva’s Theorem and Angle Bisector Theorem to prove our result.

Taking the same triangle PQR, let angle bisectors from vertices P, Q and R meet the opposite sides at points A, B and C.

In the terminology of Ceva’s theorem(and its converse), a cevian is a line segment connecting a vertex of a triangle to any point on the side opposite to the vertex. So in the above figure, angle bisectors PA, QB and RC are cevians.

As per the converse of Ceva’s theorem, three cevians from three vertices will meet at a single point if :

To prove : \(\frac{RB}{BP} \cdot \frac{PC}{CQ} \cdot \frac{QA}{AR} = 1\)

So to prove the angle bisectors of triangle PQR are concurrent, we basically need to prove the product of the three fractions on the left hand side is equal to one.

Note : We need to be careful when noting down this equation. The three fractions above are simply the ratios in which the cevians divide (internally) the lengths of the sides of the triangle. But the terms in the fractions should follow a certain cyclic pattern. For instance, if I write \(\frac{CQ}{PC}\) instead of \(\frac{CQ}{PC}\) in the above equation, that would be incorrect as it would no longer be in the required cyclic pattern, even though it is still the ratio of division of side PQ. To know what that pattern is, and more about Ceva’s theorem and its converse in general, you can read my post here.

In the proof, we will use the angle bisector theorem of triangles to show the above relation to be true. This theorem relates the sides of the triangle adjacent to a vertex, with the lengths of the divisions of the side opposite to the vertex created by the angle bisector from the vertex.

Since PA is the angle bisector of P,

\(\frac{PQ}{PR} = \frac{QA}{AR}\)

If i were to translate the above equation in words : the ratio of the two sides of the triangle adjacent to the vertex P, is equal to the ratio of the two divisions of the opposite side made by the angle bisector from P.

There is only one thing that we need to take care of in this equation. If PQ is the side we are taking in the numerator on the left, then on the right, the numerator should also begin with Q, so QA will be in the numerator on the right hand side. So if I’d taken PR in the numerator on the left, the equation will then be

\(\frac{PR}{PQ} = \frac{RA}{QA}\)

which is the same as taking the inverse of both sides of the above equation.

Similarly, for vertices Q and R we can write,

\(\frac{QR}{QP} = \frac{RB}{BP}\)

and, \(\frac{RP}{RC} = \frac{PC}{CQ}\)

Remember our aim was to prove :

\(\frac{RB}{BP} \cdot \frac{PC}{CQ} \cdot \frac{QA}{AR} = 1\)

Consider the LHS

\(\frac{RB}{BP} \cdot \frac{PC}{CQ} \cdot \frac{QA}{AR}\)

And now simply replacing the three fractions

= \(\frac{PR}{PQ} \cdot \frac{QR}{QP} \cdot \frac{RP}{RC}\)

= 1