Locus of points of the incenter of right isosceles triangles

Question

Good evening to everyone. I was trying to solve an exercise on planar geometry concerning a locus of points. More precisely, the exercise was stating : '' Suppose a line $ \varepsilon $ and a point O outside this line are fixed. We draw an isosceles right triangle $ OAB $ such that A is on the line $ \varepsilon $ and the angle $ OAB $ is right , so $ OA=AB $ . As the point A moves along the line $ \varepsilon $ find the locus of points of the incenters of all those right isosceles triangles. ( The orientation of the letters used to name the vertices must be the same each time )'' . This exercise should be solved only using plane geometry techniques, not Cartesian coordinates for example. Well , I have almost sovled this exercise, namely I found that all these incenters belong to another straight line and I proved it by using the fact that if $ I $ and $ I ' $ are the incenters of the triangles $ OAB $ and $ OA'B' $ , then the triangles $ OAI $ and $ OA'I' $ are similar (because $ \angle{OAI}=\angle{OA'I'}=45^0 $ and $ \angle{AOI}=\angle{A'OI'}=22,5^0 $ ). So, as A moves along the line, we have a collection of similar triangles, which have fixed the vertex O and the vertices A , A', A'' , .... are all collinear,which means that also the other vertices I , I' , I'', ... must be collinear. The final argument was something I remembered from the past, but when I tried to prove it I couldn't ! Is there any idea about how to prove that 3 similar triangles which have the one vertex fixed the other 3 homologous vertices collinear then they necessarily have the third vertices collinear ? Thank you for your help.

Answer 1

The similar-triangles-collinear-vertices argument you seek is a well-known one, but I can't find a resource for it online. So here is my sketch of a proof, which is probably not the most elegant but works.

Note that all below point labels are separate from the main part of your problem. Also, when I refer to similar triangles, assume they are also all oriented similarly.

Consider triangles $\triangle XOY$ and $\triangle YOZ$ that are similar; ie. define the angles \begin{align*} \alpha&:=\angle OXY=\angle OYZ\\ \beta &:= \angle OYX = \angle OZY\\ \gamma&:= \angle XOY = \angle YOZ \end{align*} The idea is to construct a new triangle $\triangle AOB$, firstly by selecting a point $A$ on the $XY$ line, then creating line $OA$, then from this line subtending an angle $\gamma$ at $O$ (oriented correctly) to create a line that meets $YZ$ at point $B$.

If we can show that either $\angle OAB=\alpha$ or $\angle OBA=\beta$ (oriented in the correct way), then we are done, because we will have shown that $\triangle AOB$ is also a similar triangle. From here we can argue that if we choose any auxiliary point $A'$ that lies on $XY$ and construct a similar triangle $\triangle A'OB'$ using $OA'$ as the analogous side to $OX$, then $B'$ will necessarily lie on $YZ$.

There are three cases to consider:

Case 1: $A$ lies between $X$ and $Y$

Consider the quadrilateral $\square OAYB$. We know $$\angle AYB = \angle OYX+\angle OYZ = \alpha+\beta$$ Its opposite angle is $$\angle AOB=\gamma$$ But we know $\alpha+\beta+\gamma=180^\circ$ as they are the interior angles of the similar triangles. So, $\square OAYB$ is cyclic. By the bowtie theorem, we have the final step, after which we are done: $$\angle OAB=\angle OYB(=\angle OYZ)=\alpha$$

Cases 2 and 3: $A$ lies on $XY$ but not between $X$ and $Y$

Again, under these cases we can prove that the respective quadrilaterals $\square OYAB$ or $\square OABY$ are cyclic, and from here use analogous angle-chasing arguments to show $\triangle AOB$ is similar.