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🌿 Timeless Geometry: Elegant Proofs in Neutral Space

Uniqueness of the Midpoint

Let \( \overline{AB} \) be a line segment. By the Ruler Postulate, we may assign coordinates such that point \( A \) is at \( 0 \) and point \( B \) is at \( d > 0 \).

The midpoint \( M \) is defined by \( AM = MB = \frac{d}{2} \). Assign \( M \) the coordinate \( \frac{d}{2} \). Then:

• \( AM = \left| \frac{d}{2} - 0 \right| = \frac{d}{2} \)
• \( MB = \left| d - \frac{d}{2} \right| = \frac{d}{2} \)

Now suppose a different midpoint \( M' \) has coordinate \( x \). Then:

\( AM' = x \), and \( M'B = d - x \).
Set \( x = d - x \Rightarrow 2x = d \Rightarrow x = \frac{d}{2} \).

Conclusion: \( M' = M \), so the midpoint is unique.

Triangle Congruence is an Equivalence Relation

Reflexivity: Every triangle \( \triangle ABC \) is congruent to itself.

Symmetry: If \( \triangle ABC \cong \triangle DEF \), then \( \triangle DEF \cong \triangle ABC \) via rigid motion reversal.

Transitivity: If \( \triangle ABC \cong \triangle DEF \) and \( \triangle DEF \cong \triangle GHI \), then \( \triangle ABC \cong \triangle GHI \) by composition of motions.

Conclusion: Triangle congruence is reflexive, symmetric, and transitive — an equivalence relation.

Pasch’s Postulate Fails in the Missing Strip Plane

In the Missing Strip Plane, remove \( \{(x, y) \in \mathbb{R}^2 \mid -1 < x < 1\} \).

Triangle \( \triangle ABC \) with \( A(-2, 0), B(2, 0), C(0, 2) \) spans the missing strip.

Let line \( L: y = 1 \). It intersects \( \overline{AB} \) at \( P(-2, 1) \), and would intersect \( \overline{BC} \) at \( Q(2,1) \).

But the line \( y = 1 \) crosses through \( -1 < x < 1 \), which has been removed. Thus, there’s no continuous path within the plane.

Conclusion: Pasch’s Postulate fails in the Missing Strip Plane.

AAS Congruence Theorem

Theorem: In neutral geometry, if two triangles have two angles and a non-included side congruent, then the triangles are congruent.

Given: \( \angle A \cong \angle D \), \( \angle B \cong \angle E \), and \( \overline{BC} \cong \overline{EF} \).

Then: \( \angle C \cong \angle F \) by angle sum property.

Now, by the ASA Postulate (two angles and the included side), \( \triangle ABC \cong \triangle DEF \).

Conclusion: The AAS condition ensures triangle congruence.

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