Conjecture: An interesting concurrency concerning incircles

Question

Conjecture: Choose any point $P$ in the interior of the incircle of triangle $ABC$, and from the points of tangency, $D$, $E$, and $F$, draw lines through $P$ intersecting the incircle a second time at points $K$, $L$, and $M$, respectively. Then the straight lines $AK$, $BL$, and $CM$ are concurrent (in the picture at $Q$).

I would be particularly interested in a synthetic proof for this conjecture. It seems that the condition that $P$ lies in the interior of the incircle is not necessary. But a proof for this case would be sufficient for me.

I had already another post with nearly the same title, where I forgot to pose the question, the missing proof. So in this version it is clearer for the reader and maybe the chances better to get a solution.

Answer 1

Here's an outline of a solution in barycentric coordinates, leaving out all of the algebraic manipulation (performed by Mathematica):

Let $P$ (which need not lie in the incircle's interior) have barycentric coordinates $u:v:w$. That is, suppose we can write $$P = \frac{u A + v B + w C}{u+v+w}$$

Let $D'$, $E'$, $F'$, be the "other" points of intersection of $\overleftrightarrow{PD}$, $\overleftrightarrow{PE}$, $\overleftrightarrow{PF}$ with the incircle. Defining $$\alpha := \frac12\angle A \qquad \beta := \frac12\angle B \qquad \gamma :=\frac12\angle C$$ the barycentric coordinates of these points are $$ \begin{align} D'&= 2 \tan\alpha \left(\frac{\cos^2\beta}{v}+\frac{\cos^2\gamma}{w} \right)^2: \frac{\cos^2\beta \sin2\beta}{v^2}: \frac{\cos^2\gamma \sin2\gamma}{w^2} \\[4pt] E'&= \frac{\cos^2\alpha \sin2\alpha}{u^2} : 2 \tan\beta \left(\frac{\cos^2\gamma}{w}+\frac{\cos^2\alpha}{u} \right)^2: \frac{\cos^2\gamma \sin2\gamma}{w^2} \\[4pt] F'&= \frac{\cos^2\alpha \sin2\alpha}{u^2} : \frac{\cos^2\beta \sin2\beta}{v^2} : 2 \tan\gamma \left(\frac{\cos^2\alpha}{u}+\frac{\cos^2\beta}{v} \right)^2 \end{align}$$ Lines $\overleftrightarrow{AD'}$ and $\overleftrightarrow{BE'}$ meet at a point $P'$ with these barycentric coordinates

$$P' = \frac{\cos^2\alpha \sin2\alpha}{u^2} : \frac{\cos^2\beta \sin2\beta}{v^2} : \frac{\cos^2\gamma \sin2\gamma}{w^2}$$

The symmetry of these coordinates with respect to the elements of the triangle guarantees that $P'$ also lies on $\overleftrightarrow{CF'}$. It is indeed a point of concurrency. $\square$

Answer 2

This is a well-known lemma, called Steinbart's Lemma if I recall correctly. Fairly straightforward by trig Ceva. Let $d$ be the diameter of the incircle. Note that $$\sin \angle ECM = \sin \angle MEC \cdot \frac{EM}{MC} = \frac{EM}{d} \cdot \frac{EM}{MC} = \frac{EM^2}{d\cdot MC}.$$ Similarly, $$\sin \angle MCD = \frac{MD^2}{d\cdot MC},$$ hence $$\frac{\sin\angle ECM}{\sin \angle MCD} = \frac{EM^2}{MD^2}.$$ Multiplying this and two analogous equalities we obtain $$\frac{\sin\angle ECM}{\sin \angle MCD}\cdot \frac{\sin\angle DBL}{\sin \angle LBF}\cdot \frac{\sin\angle FAK}{\sin \angle KAE} = \frac{EM^2}{MD^2}\cdot \frac{DL^2}{LF^2}\cdot \frac{FK^2}{KE^2}$$ which equals $1$ because $DK$, $EL$, $FM$ are concurrent. Therefore, by trig Ceva, $AK$, $BL$, $CM$ are concurrent.