euclidean-geometrygeometryreference-request
the theorem is this:
Given the $\triangle ABC$; let $O$ be it's incenter, let $D,E$ and $F$ be the points of contact of it's incircle with sides $BC,AC$ and $AB$ respectively and let point $G$ be the foot of the perpendicular line to the inner angle bisector of $C$ in $\triangle ABC$ through $B$. Then $G,F$ and $E$ are colinear.
I don't need proof, just the name of this theorem please.
Here is a possible proof "without inversion". (Note: Generally i am against settings of problems, which discriminate some structural part of mathematics, and ask for a solution "without" some ingredient, which would make the solution straightforward, simple, and easy to remember. But in this case i will accept the challenge. There will be some similitude argument instead... well, same two circles as in the solution using inversion are the main actors.)
Note that even the main wiki page for the inversion uses this problem to illustrate the use and usefulness of the inversion:
https://en.wikipedia.org/wiki/Inversive_geometry#Application
I will slightly change the notation to fit the order in the alphabet. So $M$ is not on $AB$, but on the side opposite to the vertex $A$, and similarly for $N,P$. (Else i would commit errors.) So let us state explicitly...
Let $\Delta ABC$ be a triangle. Let $I$ be the incenter, the intersection of the bisectors of the angles of the triangle. The incircle $(I)$ touches the sides of $\Delta ABC$ in the points $M\in BC$, $N\in CA$, and $P\in AB$.
Let $S,T,U$ be the intersections of the the angle bisectors $AI$, $BI$, $CI$ with the segments respectively perpendicular on them, $NP$, $PM$, $MN$.
Let $O$ be the circumcenter of $\Delta ABC$, the center of the circle $(ABC)$.
Let $o$ be the circumcenter of $\Delta MNP$, the center of the circle $(MNP)$. Of course, $o=I$, but it may be useful to use the notation $o$ when addressing this point for its quality of being the center of $(MNP)$. We use in the notation a small letter as possible, and with the same convention let $h$, $g$, $9$ be respectively for $\Delta MNP$ the orthocenter, the centroid, and the center of the nine point (Euler) circle $(STU)$. Here, $g=MS\cap NT\cap PU$, and $9$ is the mid point of $oh$.
We denote by $A'$ the intersection of the bisector $AI$ with the circle $(O)=(ABC)$. Construct $B',C'$ similarly.
Let $M'$ be the mid point of the segment $Mh$, so $M'\in(STU)$.
The we have:
(1) The Euler line (e) of $\Delta MNP$ is passing through $o=I$, $9$, $g$, $h$, in particular these points are colinear.
(2) $OA'\| oM\|S9M'$, and similarly $OB'\| oN\|T9N'$, $OC'\| oP\|U9P'$.
(3) The triangles $\Delta STU$ and $\Delta A'B'C'$ are perspective / similar / homothetic, the center of the perspectivity being $o=I$, the intersection of the bisectors $ASIA'$, $BTIB'$, $CUIC'$.
(4) The circles $(STU)$ and $(A'B'C')=(ABC)$ respect the same homothety. In particular, the center of homothety $o$ is on the line their centers $9$ and $O$.
(5) The points $o=I$, $9$, $g$, $h$, $O$ are on the Euler line $(e)$ of $\Delta MNP$.
Proof: All points are clear. (2) deserves maybe a slight explanation. The lines $oM=IM$, and $OA'$ are perpendicular on $BC$. ($M$ is the tangency point of $(I)$, and $OA'$ is the side bisector of $BC$.) So $oM\|OA'$. Then in $\Delta MNP$ we have the parallelogram $oMM'S$. (For instance, $IS$, $MM'$ are perpendicular on $NP$, it remains to show they have the same length. Use now that $9$ is the mid point of $oh$, $9$ is also the mid point of the diameter $M'S$, so $ShM'o$ parallelogram, so $SI=So=hM'=M'M$.)
$\square$
It seems that although this problem is solvable (i.e. constructable), there is no simple construction.
The paper Wernickâs List: A Final Update, which is a survey whether the problems in Wernick's list are solvable or not explains:
[I]t is theoretically possible to extract a construction [$\dots$], but it is very difficult to obtain and [$\dots$] is geometrically unappealing
In Example 2, this is explained for Problem 108:
Recall that it is possible to perform additions, multiplications, divisions and root extract by using ruler and compass constructions. This construction might not be elegant, but it is perfectly valid
So in theory, it is possible to construct Problem 82, but this would involve using a ruler and compass for arithmetic operations and square radicals, which seems out of the spirit of synthetic constructions.
They don't rule out the possibility of an elegant construction, but based on what they say I'd advise you to not spend a lot of time looking for one because even automated geometry solvers have come up empty handed so far.
There is a compendium of automatically generated constructions for Wernick's list at http://poincare.matf.bg.ac.rs/~vesnap/animations/compendiums.html. The construction for $O, T_a, I$ is notably non-existent.
Best Answer
This configuration is known as the "Right Angles on Incircle Chord Lemma", also referred to by many as Iran Lemma. The problem (Iran TST 2009/9) which made it popular can be found in Jeffrey Kwan's document (Example $3.1$) along with its solution.