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 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.)

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 $$1,2$$.

• You defined $P$ as the intersection of $A_1B_2$ and $A_2B_2$, which seems like a typo. Did you mean $P=A_1B_1\cap A_2B_2$? Jan 29, 2021 at 5:08
• Yes, corrected, thanks! Jan 29, 2021 at 11:48