# 11.1: Neutral plane

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Let us remove Axiom V from our axiomatic system. This way we define a new object called neutral plane or absolute plane. (In a neutral plane, the Axiom V may or may not hold.)

Clearly, any theorem in neutral geometry holds in Euclidean geometry. In other words, the Euclidean plane is an example of a neutral plane. In the next chapter we will construct an example of a neutral plane that is not Euclidean.

In this book, the Axiom V was used starting from Chapter 6. Therefore all the statements before hold in neutral geometry.

It makes all the discussed results about half-planes, signs of angles, congruence conditions, perpendicular lines and reflections true in neutral geometry.

Let us give an example of a theorem in neutral geometry that admits a simpler proof in Euclidean geometry.

## Theorem $$\PageIndex{1}$$ Hypotenuse-leg congruence condition

Assume that triangle $$ABC$$ and $$A'B'C'$$ have right angles at $$C$$ and $$C'$$ respectively, $$AB = A'B'$$ and $$AC = A'C'$$. Then $$\triangle ABC \cong \triangle A'B'C'$$.

Proof

Euclidean proof. By the Pythagorean theorem $$BC = B'C'$$. Then the statement follows from the SSS congruence condition.

The proof of the Pythagorean theorem used properties of similar triangles, which in turn used Axiom V. Therefore this proof does not work in a neutral plane.

Neutral proof. Suppose that $$D$$ denotes the reflection of $$A$$ across $$(BC)$$ and $$D'$$ denotes the reflection of $$A'$$ across $$(B'C')$$. Note that

$$AD = 2 \cdot AC = 2 \cdot A'C' = A'D'$$, $$BD = BA = B'A' = B'D'.$$

By SSS congruence condition (Theorem 4.4.1), we get that $$\triangle ABD \cong \triangle A'B'D'$$.

The statement follows, since $$C$$ is the midpoint of $$[AD]$$ and $$C'$$ is the midpoint of $$[A'D']$$.

## Exercise $$\PageIndex{1}$$

Give a proof of Exercise 8.7.1 that works in the neutral plane.

Hint

Suppose that $$D$$ denotes the midpoint of $$[BC]$$. Assume $$(AD)$$ is the angle bisector at $$A$$.

Let $$A' \in [AD)$$ be the point distinct from $$A$$ such that $$AD = A'D$$. Note that $$\triangle CAD \cong \triangle BA'D$$. In particular, $$\measuredangle BAA' = \measuredangle AA'B$$. It remains to apply Theorem 4.3.1 for $$\triangle ABA'$$.

## Exercise $$\PageIndex{2}$$

Let $$ABCD$$ be an inscribed quadrangle in the neutral plane. Show that

$$\measuredangle ABC + \measuredangle CDA \equiv \measuredangle BCD + \measuredangle DAB.$$

Hint

The statement is evident if $$A, B, C$$, and $$D$$ lie on one line. In the remaining case, suppose that $$O$$ denotes the circumcenter. Apply theorem about isosceles triangle (Theorem 4.3.1) to the triangles $$AOB, BOC, COD, DOA$$.

(Note that in the Euclidean plane the statement follows from Corollary 9.3.2 and Exercise 7.4.5, but one cannot use these statements in the neutral plane.)

Note that one cannot use the Corollary 9.3.2 to solve the exercise above, since it uses Theorem 9.1.1 and Theorem 9.2.1, which in turn uses Theorem 7.4.1.

This page titled 11.1: Neutral plane is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Anton Petrunin via source content that was edited to the style and standards of the LibreTexts platform.