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