I’m having trouble with a past Geometry exam question that I’m trying to solve. The problem gives me that $A$,$B$,$X$ and $Y$ are collinear. If $(ABXY)=-1$ show the following: $$\frac{2}{\overline{XY}}=\frac{2}{\overline{XB}}+\frac{1}{\overline{XA}}$$

My first thought was that $(ABXY)=-1$ tells me that $X$ and $Y$ are harmonic conjugates in respect to $A$ and $B$. Therefore, if I draw $X$ and $Y$, one will be placed between $A$ and $B$ and the other outside of $AB$ (and correct me if I’m wrong but it doesn’t matter which because either way the ratio wouldn’t change).

Then I took the following: $$\frac{(ABX)}{(ABY)}=-1 \Rightarrow \frac{\frac{\overline{AX}}{\overline{XB}}}{\frac{\overline{AY}}{\overline{YB}}}=-1$$

I started changing everything to anything I could that resembled what I’m asked to prove, but all I ended up with was $\overline{XA}=\overline{XB}$ which must be wrong considering the direction (?).

I tried going the other way too, by taking what I’m asked to prove and produce something that stands based on what I’m given but it’s not working out for me either. What am I supposed to do? Are any thoughts I’ve had so far wrong?

  • 1
    $\begingroup$ It would be useful if you explain what $(ABXY)$ means. $\endgroup$
    – user
    Commented Feb 13, 2022 at 20:02
  • $\begingroup$ It is known as the cross-ratio or double ratio. I have provided it but in a lacking manner; it is $$(ABXY)=\frac{(ABX)}{(ABY)}$$ which I’ve given above. ☺️ $\endgroup$
    – Tita
    Commented Feb 13, 2022 at 20:11
  • $\begingroup$ Are you sure it is, $$\frac{2}{\overline{XY}}=\frac{\color{red}2}{\overline{XB}}+\frac{1}{\overline{XA}}?$$ $\endgroup$
    – ACB
    Commented Feb 14, 2022 at 5:43
  • $\begingroup$ Btw, your notation is unclear to me. $\endgroup$
    – ACB
    Commented Feb 14, 2022 at 9:12
  • $\begingroup$ I’m unsure if it’s correct, but that’s exactly what’s written in the exam paper. 🙁 $\endgroup$
    – Tita
    Commented Feb 15, 2022 at 15:27

1 Answer 1


It is known that for all collinear points $A$, $B$, $C$, $D$ $$(ACBD)=1-(ABCD),$$ therefore now $$(AXBY)=2.$$ It means that $$\frac{AB}{BX}\cdot\frac{YX}{AY}=2,$$ which leads to $$\frac{2}{XY}=\frac{AB}{XB\cdot AY}.$$ Here $$\frac{AB}{XB\cdot AY}=\frac{AY+YB}{XB\cdot AY}=\frac{1}{XB}+\frac{YB}{XB\cdot AY}.$$ From your relation (where you explained the original relation $(ABXY)=-1$) $$\frac{YB}{AY}=\frac{XB}{XA}$$ follows, therefore $$\frac{YB}{XB\cdot AY}=\frac{1}{XA}.$$ So, summarized, $$\frac{2}{XY}=\frac{1}{XB}+\frac{1}{XA}.$$

Edit: Proof of the first property of the cross-ratio

$$(ACBD)=\frac{AB}{BC}\cdot\frac{DC}{AD}=\frac{CB-CA}{BC}\cdot\frac{AC-AD}{AD}=\frac{(CA-CB)(AC-AD)}{CB\cdot AD}=$$ $$\frac{CA\cdot AC-CB\cdot AC-CA\cdot AD+CB\cdot AD}{CB\cdot AD}=\frac{AC(CA+BC+AD)+CB\cdot AD}{CB\cdot AD}=$$ $$=\frac{AC\cdot BD}{CB\cdot AD}+1=1-\frac{AC\cdot DB}{CB\cdot AD}=1-(ABCD).$$

  • $\begingroup$ The final result is not as same as the one OP has written there. I have mentioned that in a comment but they haven't replied. Btw, what does (ABXY) mean? $\endgroup$
    – ACB
    Commented Feb 14, 2022 at 18:28
  • $\begingroup$ It means the cross-ratio of the four points: en.wikipedia.org/wiki/Cross-ratio $\endgroup$
    – Sz_Z
    Commented Feb 14, 2022 at 20:34
  • $\begingroup$ Hi, thanks for answering. I’ve never seen the first property you’ve used here before, do you know if I can find these properties somewhere? $\endgroup$
    – Tita
    Commented Feb 15, 2022 at 15:37
  • 1
    $\begingroup$ @Tita: These properties of the cross-ratio are listed in the wikipedia link above. However, the one I used is not so hard to prove, so I edited my answer and added a proof of it. $\endgroup$
    – Sz_Z
    Commented Feb 15, 2022 at 18:10
  • $\begingroup$ Thank you very much! I’ll assume it was a typo on behalf of the professor since your answer looks just fine. $\endgroup$
    – Tita
    Commented Feb 15, 2022 at 18:14

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