Prove that $X,G$ and $B$ are collinear

Question

Given a triangle $\triangle A_1B_1C_1$ with points $D \in B_1C_1, E \in A_1C_1$ and $F \in A_1B_1$ being the vertices of its tangential triangle and points $A,B$ and $C$ being the second meetings of $A_1D, B_1E$ and $C_1F$ with the incircle of $\triangle A_1B_1C_1$

Prove that points $B, G = AD \cap FE$ and $X = AF \cap CE$ are collinear

I thought about Pascal theorem but point $B$ is a bit weird and Brianchon makes $X$ weird so I got no clue. I wonder if there is a more projective solution

Answer 1

Proof by projective theorems

Let $P = AD ∩ BF$, $Q = BC ∩ EF$, $R = AE ∩ DB$.
By Brianchon’s theorem on $A_1FB_1DC_1E$, lines $A_1D, B_1E, C_1F$ concur at some point $H$.
By Pascal’s theorem on $BCFFEE$, points $Q, C_1, B_1$ are collinear.
By Pascal’s theorem on $ADDBEE$, points $H, C_1, R$ are collinear.
By Pascal’s theorem on $ADDBFE$, points $P, Q, R$ are collinear.
Therefore, triangles $CFB$ and $EAG$ are perspective about line $PQR$, and by Desargues’s theorem, they’re perspective about point $X$. ∎

Proof by projective transformation

Apply the projective transformation that preserves the circle and sends $DEF$ to an equilateral triangle. The rest of the diagram can be recovered uniquely, and the conclusion is now obvious from a triangular grid. ∎

Answer 2

Invert about $E$, and let the image of each point $P$ be denoted by $P'$. We have $$(EF;AD)=(EB;DF)=(ED;CF)=-1,$$ so $$(\infty F';A'D')=(\infty B';D'F')=(\infty D';C'F')=-1,$$ so $F'$ is the midpoint of $A'D'$, $B'$ is the midpoint of $D'F'$, and $D'$ is the midpoint of $C'F'$. This gives the following picture: We now have $G=AD\cap EF\mapsto G'=EF'\cap (EA'D')$ and $X=EC\cap AF\mapsto EC'\cap (EA'F')$. Also, let $T$ be the reflection of $F'$ over $A'$. We need to show that $B',E,G',X'$ are concyclic; we'll do this by showing that $(B'EG'T)$ and $(EX'B'T)$ are concyclic. Let $s$ be the length of segment $B'F'$.

• For the first, we have $$F'B'\cdot F'T=s\cdot 4s=2s\cdot 2s=A'F'\cdot F'D'=\operatorname{Pow}_{(EA'D')}(F')=F'E\cdot F'G.$$
• For the second, we have $$C'B'\cdot C'T=3s\cdot 8s = 4s\cdot 6s = C'F'\cdot C'A'=\operatorname{Pow}_{(EA'F')}=C'E\cdot C'X.$$

Answer 3

Let's prove the generalized form of the proposition:

Given a triangle $A_1B_1C_1$ and an inscribed ellipse with tangent points $D{\in}B_1C_1$, $E{\in}A_1C_1$ and $F{\in}A_1B_1$, and A, B and C being the second meetings of $A_1D$, $B_1E$ and $C_1F$ respectively on this inscribed ellipse, and $G=AD{\cap}EF$, then $AF$, $BG$ and $CE$ are concurrent.

According to 3-tangents degeneration of Brianchon's theorem, the inscribed ellipse is determined by an arbitrary Ceva point H (the inscribed ellipse becomes the incircle if H is the Gergonne point), which can be represented as $H=A_1+B_1+C_1$ by homogeneous coordinates. Take $A_1$, $B_1$ and $C_1$ as basis then we can get homogeneous coordinates for these points: $A_1(1,0,0)$, $B_1(0,1,0)$, $C_1(0,0,1)$, $H(1,1,1)$, $D(0,1,1)$, $E(1,0,1)$, $F(1,1,0)$ and $G(2,1,1)$.

It's not very easy to get the equation of the inscribed ellipse. However, we can get $A$ by $(A,D;A_1,G)=-1$ because $A_1$ and $EF$ are pole and polar. So we get $A(4,1,1)$. Analogously, we get the other two points $B(1,4,1)$ and $C(1,1,4)$.

Now we can calculate the coordinates of three lines $AF$, $BG$ and $DE$:

• $AF = [-1,1,3]$
• $BG = [3,1,-7]$
• $CE = [1,3,-1]$

which are concurrent at X because: $$\det\left[\begin{matrix}-1&1&3\\3&1&-7\\1&3&-1\end{matrix}\right]=0$$