Proof the three intersections are in the same line. Some days ago I discovered a conclusion which is proved true:
Let $AX,AY,AZ$ be three rays (in which $A$ is the initial point) which any two of them form an acute angle, and $AY$ is inside $\angle XAZ$. Let
$$
\begin{align}
&A_1\in AX,\ B_1,B_2\in AZ(AB_2<AB_1),\\
&A_1B_1\cap AY=C_1,\ A_1B_2\cap AY=C_2,\ B_1C_2\cap AX=A_2,\ A_2B_2\cap AY = C_3,\\
&B_1C_2\cap B_2C_1=D_1,\ B_1C_3\cap B_2C_2=D_2,
\end{align}
$$
then $A,D_1,D_2$ are on a same line.

I've been working on it since a month ago, but I can't solve it in a geometric way. Could anyone provide a solution? Thanks!
(It's better that the solution is pure geometric, not analytic.)
 A: Here is a pedestrian way to show the result. (So we avoid Pappus, Desargues, et co. because we can extract an intermediate result which brings more light in the situation.)
Let $P$ be the intersection of the lines $A_1B_1$ and $A_2B_2$. Then the we have the following harmonic proportions, written as cross ratios equal to $-1$:
$$
(A_1,B_1;P,C_1)=(A_2,B_2;P,C_3)=-1\ .
$$
(For the first cross ratio, apply in $\Delta AA_1B_1$ Ceva w.r.t. the point $C_2$, and Menelaus w.r.t. the transversal $PA_2B_2$. For the second cross ratio do the same in $\Delta AA_2B_2$, or better, "move" the first configuration from the line $PA_1C_1B_1$ to the line $PA_2C_3B_2$ using the perspective point $A$.)
Now we look at $(A_1,B_1;P,C_1)=-1$ and move this harmonic proportion from the line $PA_1C_1B_1$ to the line $Ay=AC_3C_2C_1$ using the perspective point $B_2$. In other words, we build the lines $B_2A_1$, $B_2B_1$, $B_2P$, $B_2C_1$ (in this order, the one of the points in $(A_1,B_1;P,C_1)=-1$,) and intersect them with $Ay$. By perspectivity we get:
$$
(C_2,A;C_3,C_1)=-1\ .
$$
We isolate this relation, and state a simpler result which does not involve the points $A_1,A_2$. The same notations are used.


Proposition: Let $A,C_2;C_1,C_3$ be four points on a line in harmonic proportion. Let $B_1,B_2$ be two further points on a line through $A$. Construct $D_1,D_2$ as in the OP. Then $A,D_1,D_2$ are on a line.

Pedestrian proof:
We have:
$$
\begin{aligned}
+1 &=
\color{blue}{\frac{D_1C_2}{D_1B_1}}
\cdot \frac{B_2B_1}{B_2A}
\cdot \frac{C_1A}{C_1C_2}\ ,
\\
&\text{ Menelaus, $\Delta C_2B_1A$, secant $D_1B_2C_1$,}
\\[2mm]
+1 &=
\color{blue}{\frac{D_2C_2}{D_2B_2}}
\cdot \frac{B_1B_2}{B_1A}
\cdot \frac{C_3A}{C_3C_2}\ ,
\\
&\text{ Menelaus, $\Delta C_2B_2A$, secant $D_2B_1C_3$,}
\\[2mm]
&\text{ which gives}
\\[2mm]
\color{blue}{\frac{D_1B_1}{D_1C_2}}\cdot
\color{blue}{\frac{D_2C_2}{D_2B_2}}\cdot
\color{maroon}{\frac{AB_2}{AB_1}}
&=
\frac{B_2B_1}{B_2A}
\cdot \frac{C_1A}{C_1C_2}\ \cdot\
\frac{B_1A}{B_1B_2}
\cdot \frac{C_3C_2}{C_3A}
\ \cdot\
\color{maroon}{\frac{AB_2}{AB_1}}
\\
&=
-\frac{C_1A}{C_1C_2} \cdot \frac{C_3C_2}{C_3A}
\\
&=
-(C_1,C_3;A,C_2)
\\
&=-(-1)=+1\ . 
\end{aligned}
$$
The reciprocal Menelaus in $\Delta C_2B_1B_2$ shows now the claimed colinearity.
$\square$
Note: The statement and the proof of the isolated proposition may be simpler to follow if we use a better notation, for instance $A'$ instead of $C_2$, and the remained points using maybe only indices $1,2$.
