Додаток: підсумок і більше про середини дуг і центри вписаного і зовнівписаних кіл

A lemma (or a few?) regarding the angle bisectors

Important: It is assumed that the reader has understood circle basics (angle chasing in a cyclic quadrilateral), the circumcircle of a triangle, and how the incenter and excenters are defined (visit 0.3 if you are not sure). There will be a quick recap of what those are in this text. Knowledge of the nine-point circle, the orthocenter and the orthic triangle is optional but helpful.

Revision time! Let's check our knowledge of incenters and excenters.

Okay, what about the excenter?

The two lines are perpendicular, which is not really surprising.

Here's a 'lazy' explanation, a rigorous proof is left as an exercise to the reader (more like fill in the blanks). The angle between blue-red can be separated as blue-black and red-black. The former is half of something, the latter is half of something, and something combined with something is a straight line. Therefore the angle we want is half of a straight line, in other words, the two lines are perpendicular.

So there are a bunch of right angles around here. So what? Well, right angles are amazing, as many of them lead to these wonderful cute little creatures called cyclic quadrilaterals, as seen here. In this text, since geo currently does not support indices (or I do not know how to access them), we have to stick with points $D, E, F$ being the $A$-, $B$-, $C$- excenters respectively. Now complete the diagram (draw one yourself!) with all of the internal angle bisectors, external bisectors, incenter, and the excenters. Also you may wonder why I drew the circumcircle, think about that as well.

A very fancy diagram indeed. Just by the right angles we found just now, all the following are cyclic quadrilaterals: $DBIC, ECIA, FAIB, BCEF, CAFD, ABDE.$ Due to the symmetry between these cyclic quadrilaterals, there are only two cyclic quadrilaterals we are interested in, $DBIC$ and $BCEF$. Since we already know the diameter of these circles, it is natural to ask for the center of these circles. Think about that.

You might have realized, or if not let me tell you now, that both centers lie on the circumcircle $ABC$, that is the points that circumcircle $ABC$ intersect with lines $AD$ and $EF$, which is distinct from $A$, and call them $X$ and $Y$ respectively. These lines are the internal angle bisector and external angle bisector of angle $BAC$. $X$ and $Y$ are the centers of circles $DBIC$ and $BCEF$ respectively, what's more they are also the midpoints of arc $BC$, the former for the arc not containing $A$, the latter for the arc containing $A$. That is a lot of nice symmetry going on there, but we need to prove it! Observing is not enough in mathematics, so we need fine a rigorous proof for that. Have a go at that and continue when you are ready.

Optional: Observant readers would have realized that $I$ is not just the incenter of triangle $ABC$, but also the orthocenter of triangle $DEF$. This makes triangle $ABC$ to be the orthic triangle of triangle $DEF$. What's more is that the circumcircle of triangle $ABC$ is then the nine-point circle of triangle $DEF$. Then our previous observations can be easily proven, since the nine-point circle passes through the midpoint of $EF$, and $EF$ is the diameter of circle $BCEF$, that's one part done. Next the nine-point circle also passes through the midpoint of $DI$, and $DI$ is the diameter of circle $DBIC$, therefore the other part is done as well. But try to find an elementary proof based on angle chasing!

Let us go through this step by step. First draw the diagram again, only with the necessary lines. Consider the circle $DBIC$ first.

This is because we exploit the fact that $AX$, passing through $I$, is the angle bisector of angle $BAC$. What is next? If we want $X$ to be center, we would also like $BX=IX$ or $BX=DX$. These two statements are equivalent to $\angle XBI=\angle XIB$ and $XBD=\angle XDB$ respectively. But by a simple angle chase in the right-angle triangle $IBD$ tells us that $\angle XBD=90^\circ -\angle XBI$ and $\angle XDB=90^\circ -\angle XIB$. Therefore we can prove either $\angle XBI=\angle XIB$ or $XBD=\angle XDB$ since they are equivalent. Let's try the former. Time to apply a bit of wishful thinking!

This is a slightly trickier angle chase: $$\angle XIB=180^\circ -\angle AIB=180^\circ - (180^\circ - \angle IAB -\angle IBA)=\angle IAB +\angle IBA=\frac{\angle BAC+\angle ABC}{2}=\frac{180^\circ -\angle ACB}{2}\\ =\frac{180^\circ -\angle AXB}{2}\text{ (by the cyclic quadrilateral $ABXC$) }=90^\circ - \frac{\angle IXB}{2}$$ as desired.

Now to complete the proof, we just need $XB=XC=XI$, since three points are enough to define a circle. There are two ways to do this, one is to notice that $XB=XI$ and $XB=XC$, it is trivial enough to say $XC=XI$. Another method is to say that based on the proof of $XB=XI$, we can follow a similar argument to show that $XC=XI$, and we did not need $XB=XC$ at all (also $XB=XC$ follows from this as well)! And at the same time, we would have also shown that $XB=XD$ and $XC=XD$. Thus the proof is complete.

Now for the second circle, $BCEF$. Here's the diagram:

So we just want to show that $YB=YC=YE$ which is enough, similar to what we did before. $YB=YC$ is trivial and left to reader to justify. Note there are two cyclics in this diagram (sort of hidden) that we just shown before already. This time we would do the angle chasing part by filling in the blanks. Remember to draw a diagram and add lines and label angles to it!

The missing part should be somewhat similar to a part of the proof we had before. This is basically it for this article, an application of triangle centers and angle chasing! It is highly encouraged for the reader to now understand the nine-point circle as it is linked to this topic, if not done already. Keep this lemma (or if you think these are multiple lemmas) in mind in future geometry problems, they can be very handy!

The angle bisectors of a triangle meet in the incenter, and here is an easy proof of this fact:

$\triangle ABC$ has incenter $I$.Draw the perpendiculars from $I$ to sides $AB, BC,$ and $AC$, and denote them by $C_I, A_I,$ and $B_I,$ respectively. Because of equal tangents, $BC_I \cong BA_I$, and $IC_I \cong IA_I$ (because they are both radii of the incircle). $BI \cong BI$, so using $SSS$ congruence, we have $\triangle C_IIB \cong \triangle A_IIB$, so $\angle ABI \cong \angle IBC$, thus meaning $BI$ is the angle bisector from $B$. Applying the same method to segments $AI$ and $CI$, we conclude that the angle bisectors of a triangle are concurrent, and concur at the incenter of a triangle.