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.

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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.)


1 Answer 1


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.


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 $1,2$.

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

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