# In neutral geometry construct a transversal through two parallel lines and a point between them

Is this possible to do?

Here is what I have tried. Draw a line that is perpendicular to either of the parallel lines that go through the point between the parallel lines, but then there is no way to guarantee that the two perpendiculars are really the same line.

I am not 100% sure if this is even possible but I am also having trouble envisioning what the counter example would look like. Even in non-Euclidean geometry, it is difficult for me to envision what a situation would be where you a point contained between two lines and are not able to draw a line that intersects both lines and the points.

Any input would be greatly appreciated!

## 1 Answer

This is not true in hyperbolic geometry. It is easy to see a counterexample in Poincare half-plane model (https://en.wikipedia.org/wiki/Poincar%C3%A9_half-plane_model). There, the points are the points of the upper half-plane $$\{(x,y)\in\mathbb R^2\mid y>0\}$$, and we have two types of lines: straight lines orthogonal to $$x$$-axis, so sets given by $$\{(x_0,y)\mid y>0\}$$ for any $$x_0\in\mathbb R$$ are lines, and upper parts of the circles with centers on $$x$$-axis, i.e. sets given by $$\{(x,y)\mid (x-x_0)^2+y^2=r^2,y>0\}$$ for $$x_0\in\mathbb R$$ and $$r>0$$ are lines.

Consider lines $$p$$ and $$q$$ given by: $$p=\{(x,y)\mid x^2+y^2=1,y>0\}$$ and $$q=\{(x,y)\mid (x-2)^2+y^2=1,y>0\}$$; they don't intersect. Points between $$p$$ and $$q$$ are all points outside of these circles. Take e.g. $$M=(0,3)$$. It is easy to see that lines $$QM$$ don't intersect $$p$$ when $$Q\in q$$.

Also, you can easily see an counterexample in Beltrami-Klein model (https://en.wikipedia.org/wiki/Beltrami%E2%80%93Klein_model).