Hello. In this mini-lesson, I’ll talk about a special point in a triangle – called the incenter. Let’s jump right into it.

Take any triangle, say ΔABC. Draw the three angle bisectors, AD, BE, and CF. Then:

AD , BE , and CF always intersect at a point. In other words, they are concurrent.

, , and always intersect at a point. In other words, they are concurrent. The point of concurrency of the three angle bisectors is known as the triangle’s incenter.

Let’s observe the same in the applet below. Press the play button to start.

Drag the vertices to see how the incenter (I) changes with their positions. Once you’re done, think about the following:

does the incenter always lie inside the triangle? (it’s in the name)

can the incenter lie on the (sides or vertices of the) triangle?

Go, play around with the vertices a bit more to see if you can find the answers. Also, why do the angle bisectors have to be concurrent anyways? (This one is a bit tricky!)

A few more questions for you. In terms of the side lengths (a, b, c) and angles (A, B, C),

what is the length of each angle bisector?

how far does the incenter lie from each vertex?

how far does the incenter lie from each side?

To find these answers, you’ll need to use the Sine Rule along with the Angle Bisector Theorem.

Turns out that the incenter is equidistant from each side. Play around with the vertices in the applet below to see this in action first.

So, what’s going on here?

The triangles IBP and IBR are congruent (due to some reason, which you need to find out). This would mean that IP = IR.

And similarly (a powerful word in math proofs), IP = IQ, making IP = IQ = IR.

We call each of these three equal lengths the inradius of the triangle, which is generally denoted by r.

I think you know where this is going – incenter, inradius, in______? Well, no points for guessing.

Taking the center as I and the radius as r, we’ll get a nice little circle which touches each side of the triangle internally. This circle is known as the incircle of the triangle.

Here’s the culmination of this lesson. Press the Play button to start the show.

Hope you enjoyed reading this. Drop me a message here in case you need some direction in proving IP = IQ = IR, or discussing the answers of any of the previous questions.

Goodbye.