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In the hyperbolic plane, suppose $L$ is a line with points $A, B, C$ on the same side of line $L$. Assume further points $P, Q, R$ are on line $L$ such that segments $AP$, $BQ$ and $CR$ are perpendicular to $L$ and the three lengths $AP$, $BQ$ and $CR$ are all equal. Prove $A, B, C$ are not collinear.

Suppose they were collinear. Then $ACRQ$ is a rectangle since all the angles are $90$ degrees due to being perpendicular to the line Rectangles do not exist outside of Euclidean geometry since the interior angle sum is less then $360$ degrees.

Would that be sufficient to show its a rectangle? Or is there a better approach to take. Or is there any direct proof ideas that could work

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  • $\begingroup$ $\square ACRQ$ has right angles along $L$, but it's not obvious that collinearity of $A$, $B$, $C$ forces it to have more right angles. $\endgroup$ – Blue Apr 18 '18 at 20:56
  • $\begingroup$ How else could I show its a rectangle then? $\endgroup$ – HighSchool15 Apr 18 '18 at 21:42
  • $\begingroup$ Don't you want ACRP instead? ABQP and BCRQ are rectangles too. $\endgroup$ – Somos Apr 18 '18 at 22:22
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We may assume $Q$ is between $P$ and $R$.

image

Claim. In $\square ABQP$, $\angle PAB \cong \angle QBA$.

As noted in the question, the angle-sum of a hyperbolic quadrilateral is less than four right angles, so the Claim's angles must each be less than one right angle.

Likewise, in $\square CRQB$, $\angle RCB \cong \angle QBC$, and these angles are each less than a right angle.

Then $\angle ABQ + \angle QBC$ is less than two right angles ---that is, the angles are not supplementary--- so that $\overline{AB}$ and $\overline{BC}$ cannot be collinear. $\square$


Proof of Claim. $$\triangle APQ \stackrel{\text{SAS}}{\cong} \triangle BQP \;\implies\; \overline{AQ}\cong\overline{BP} \;\implies\;\triangle PAB \stackrel{\text{SSS}}{\cong}\triangle QBA \;\implies\;\angle A\cong\angle B$$

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I don’t see that the new angles you’ve created by drawing a line from $A$ to $C$ have to be right angles.

Here’s a rough argument that you may like, and may want to clean up:

Lemma. Let $\ell$ be a line in the hyperbolic plane, and $X,Y$ two points on the same side of $\ell$ and equidistant from $\ell$. Let $X',Y'$ be the points on $\ell$ that are gotten by dropping perpendiculars from $X$ and $Y$, respectively. Then in the quadrilateral $XYY'X'$, the angles at $X$ and $Y$ are equal (and less than $\pi/2$).

I think that to prove this, one may argue from symmetry somehow by exchanging $X$ and $Y$ as well as $X',Y'$.

Now let’s apply to your situation, assuming that your point $Q$ is between $P$ and $R$, and using your to-be-disproven hypothesis that $A,B,C$ are collinear.

In the quadrilateral $ABQP$, the angle $\angle ABQ$ is less than $\pi/2$ and in the quadrilateral $CBQR$, the angle $\angle CBQ$ is also less than $\pi/2$. But these angles have to be supplementary, since $B$ has been assumed to be on the segment $\overline{AC}$. No good

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