🌐 Does the Missing Strip Plane Obey Euclid?

And can a line segment truly have only one midpoint?

Does the Missing Strip Plane Satisfy the Euclidean Parallel Postulate?

Answer: Yes — the Missing Strip Plane does satisfy the Euclidean Parallel Postulate.

Consider the plane \( \mathbb{R}^2 \) with the vertical strip \( \{(x, y) \in \mathbb{R}^2 \mid -1 < x < 1\} \) removed. Although lines are interrupted by this strip, their direction and extension behave as in Euclidean geometry.

For any point \( P \notin L \), where \( L \) is a line, there still exists exactly one line through \( P \) that does not intersect \( L \), preserving the core of Euclid's parallel postulate.

Justification:

The missing strip does not create new intersections or additional parallels. For instance, if \( L \) is the line \( x = a \) with \( |a| \geq 1 \), the strip \( -1 < x < 1 \) is irrelevant to the existence and uniqueness of a parallel through \( P \). The parallel postulate stands firm.

Prove that a Line Segment Has a Unique Midpoint

Let \( \overline{AB} \) be a line segment in neutral geometry.

By the Ruler Postulate, we may assign coordinates such that \( A \) is at 0 and \( B \) is at \( d > 0 \), where \( d \) is the length of the segment.

Define point \( M \) to have coordinate \( \frac{d}{2} \). Then:

\( AM = |0 - \frac{d}{2}| = \frac{d}{2} \) and \( MB = |d - \frac{d}{2}| = \frac{d}{2} \), so \( AM = MB \) and \( M \) is a midpoint.

Suppose another point \( M' \) is also a midpoint with coordinate \( x \). Then: \[ AM' = x,\quad M'B = d - x \] If \( AM' = M'B \), then \( x = d - x \Rightarrow 2x = d \Rightarrow x = \frac{d}{2} \).

Thus, \( M' \) coincides with \( M \), proving uniqueness.

Conclusion: In any neutral geometry — even one that dares omit Euclid's Fifth — every segment bows to symmetry. Its midpoint, poised at perfect balance, is one and only one.