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Incircles or incenters! All triangles have them!
And the proof is rather perplexing...
I will try to explain it to you like you are 5 - don't worry.
Assuming here that you didn't know why incircles even exist. And you are hooked to understand why it works? Below I will try to motivate you...
1. What is an incircle to a triangle?
It is a circle that lives inside the triangle and touches the three sides of the triangle. That means that when the circle touches the side, that line is tangential to the circle at that point.
2. The motivation
Observe the above figure. Ask why should an incircle even exist?
If you take the diagram at face value. There is a circle with center $D$. It touches sides $AC, AB, \& BC$ at $E, F, \& \space G$ respectively. But that is just "the property of a circle" isn't it?
Here is another funny fact...that point D actually bisects $\angle A , \angle B \space \& \space \angle C$
Motivated enough? Are you curious as to why $D$ does exactly that? If not your brain is not cut out to understand geometry today. That is OK. Maybe revisit this post another time.
But if you are hooked...read on.
3. The first hint - angle bisection
In schools they teach about angle bisection earlier to the topics leading to triangle's - centroids, incircles, etc. Here is a small animation showing how to bisect angle.
4. Equidistance theorem
The above construction method somewhat hints at this idea. But to put it in words,..
Theorem: Any point on the line that bisects the angle made by two intersecting lines is always equidistant to the two lines.
There is a formal proof for this. It deals a little with the idea behind similar triangles. But I will just finish this with a simple animation.
5. The converse equidistance theorem
The converse of the above is also valid.
But to appreciate it...you really have to state the theorem in your own words. Here is some animation to that effect.
6. The real reason why incircles should exist
Armed with these two ideas we can argue against our initial train of thought (Section 2).
Here is the statement of the proof (or what we are trying to prove)
Theorem: The angle bisectors of a triangle all meet at a point.
Proof:
1. Draw $\triangle ABC$, and the angle bisectors of $\angle A$ and $\angle B$.
Let them meet at some point $D$.
2. Draw perpendicular lines from $D$ to sides $AB$, $BC$, and $CA$.
Let them meet the sides at $E$, $F$, $G$ respectively, so that
$DE \perp AB$, $DF \perp BC$, $DG \perp CA$.
3. By the equidistance theorem (Section 2), a point on an angle bisector is
equidistant from the two sides of the angle. Applying it to the bisector
of $\angle A$ gives $DE = DG$, and to the bisector of $\angle B$ gives
$DE = DF$. Hence
$$DE = DF = DG.$$
4. Consider $\angle C$. Construct line $DC$. From above, $DG = DF$, so $D$ is
equidistant from sides $CA$ and $CB$. By the converse of the equidistance
theorem (Section 3), $DC$ bisects $\angle C$.
That's it! All three angle bisectors pass through the single point $D$ —
therefore the angle bisectors of $\triangle ABC$ are concurrent.
The point $D$ is called the **incenter**: it is the center of the inscribed circle, whose radius is $r = DE$. $\blacksquare$
For the animations and the mathjax bookmarklet visit: https://gist.github.com/deostroll/05ba5173418abe227afb9b56eee36d32
