{{interfaces | imports = Interface:Out lines | exports = Interface:Symmetric point }}

This is part of a series of modules which prove a variety of geometrical theorems starting with Tarski's axioms for geometry. We follow the formalization of Julien Narboux<ref>The formal proofs are at http://www.lix.polytechnique.fr/Labo/Julien.Narboux/tarski.html Formalization of Tarski's geometry in the Coq proof assistant and are described in Julien Narboux (2007), "http://citeseerx.ist.psu.edu/viewdoc/summary?doi= Mechanical Theorem Proving in Tarski’s Geometry", F. Botana and T. Recio (Eds.): ADG 2006, LNAI 4869, pp. 139–156</ref> which itself closely follows a treatise by Schwabhäuser, Szmielew, and Tarski.<ref>W. Schwabhäuser, W Szmielew, and A. Tarski (1983), ''Metamathematische Methoden in der Geometrie'', ISBN 0387129588</ref>

This page defines the midpoint of a line segment, proves a few basic midpoint results, and then proves a variety of results concerning reflection through a point (symmetric points).

We import the theorems of propositional logic and predicate logic, and the geometry results so far and define some variables:

Midpoint defined

A point is the midpoint of a line segment if it is between the endpoints and is equidistant from each one.

Whether a point is the midpoint does not depend on the order in which we list the endpoints.<ref>Lemma l7_2 in Narboux</ref> This is easily proved from the corresponding theorems for betweenness and congruence.

The only midpoint of an empty line segment is the endpoint of the line segment.<ref>l7_3 in Narboux</ref>

The endpoint of an empty line segment is a midpoint of that segment.<ref>Lemma l7_3_2 in Narboux</ref>

Reflection through a point

Point symmetry, or reflection through a point, refers to the process of starting with a single point (which in special cases might be called the origin or the center) and mapping every point to a point directly opposite the given point (at the same distance).

Existence of the symmetric point

Given a point A, the symmetric point of P through A is a point P′ such that A is-midpoint-of P P′. This point always exists and is unique. We first prove existence.<ref>symmetric_point_construction in Narboux, with the cosmetic change that Narboux interchanges A and P</ref> The proof just applies line segment construction to extend the line segment P A to a point p′ by a distance of P A, or in other words, ∃ p′ (between P A p′ ∧ A p′ ≡ P A).

Symmetric point uniqueness

Uniqueness is A is-midpoint-of P P1 ∧ A is-midpoint-of P P2 → P1 = P2.<ref>symmetric_point_unicity in Narboux, again exchanging A and P</ref> We first consider the P = A case, which is just a matter of applying CongruenceIdentityFromEquality

The P ≠ A case is a straightforward application of SegmentConstructionUniqueness.

Combining the two cases completes the proof of uniqueness.

One to one

Reflecting through a point is one to one.<ref>Lemma l7_9 in Narboux</ref>


Reflecting through a point preserves distances. That is, A is-midpoint-of P′ P ∧ A is-midpoint-of Q′ Q → P Q ≡ P′ Q′.<ref>Lemma l7_13 in Narboux</ref>

thumb|right|Proof that reflecting through a point preserves distance The proof is a bit involved, so we'll start with a brief outline. The P = A case is trivial, so henceforth we assume P ≠ A. Starting with the line segment P P′, we extend the two endpoints to points X and X′ (by a distance of Q A in both cases). Then we do something similar: extend both ends of Q Q′ to Y and Y′ by a distance of P A.

The next step is X′ Y′ ≡ Y X, which we get by applying five segment to baselines X A X′ and Y′ A Y and points Y′ and X.

Then we apply inner five segment to baselines Y Q A and Y′ Q′ A and points X and X′, which gives Q X ≡ Q′ X′.

Finally, applying inner five segment to baselines X P A and X′ P′ A and points Q and Q′ yields P Q ≡ P′ Q′, as desired.

P = A case

We start formalizing this proof with the P = A case. Because A is the midpoint of Q′ Q, we know that Q′ A ≡ Q A, and we can substitute P for A. We can also substitute P′ for A, as A = P′ follows from P = A and P′ A ≡ A P.

To understand the organization of the proof, it is based on the substitution A = P′ ∧ A = P → (Q′ A ≡ Q A ↔ Q′ P′ ≡ Q P).

Constructing X, X′, Y, and Y′

The first step for the P ≠ A case is to construct the points we will need. The first lemma constructs two points (and we'll be able to use it twice, once for x and x′ and once for y and y′).

Applying betweenness transitivity

thumb|right|X′, P′, A, P, and X are in that order and so are Y′, Q′, A, Q, and Y Visually the diagram makes it clear that the points of each of the two lines are in a certain order: in the one case X′, P′, A, P, and X and in the other case Y′, Q′, A, Q, and Y. To express this formally involves a large number of betweenness formulas, and we now prove a number of them. The first is between A P X, which follows from between P′ A P and between P′ P X.

A similar result is between A Q Y, which follows from between Q′ A Q and between Q′ Q Y.

The next betweenness result is between A Q′ Y′. Unlike the previous two, we need to apply transitivity twice (because of the way we constructed Y′). The first application applies transitivity to between Q′ A Q and between Q′ Q Y to give between Q′ A Y.

The same reasoning applied to the other line gives between A P′ X′. The first application of transitivity is to between P′ A P and between P′ P X and gives between P′ A X.


thumb|right|A X, A Y, A X′, and A Y′ are all congruent From the diagram it should be clear that the line segments A X, A Y, A X′, and A Y′ are all congruent, as they each consist of one line segment congruent to P A plus one congruent to Q A. We formalize this through a series of lemmas which apply the three segment theorem.

We have one more similar congruence, A Y ≡ A Y′, which follows from outer three segment on lines A Q Y and A Q′ Y′.

Other similar congruences can be derived from those by transitivity. Here is A X ≡ A X′

X′ Y′ ≡ Y X

All that seems fairly preparatory, and visually obvious. Now for a more substantial result: X′ Y′ ≡ Y X, which we get by applying outer five segment to baselines X A X′ and Y′ A Y and points Y′ and X. We break some of the antecedents of outer five segment into their own lemmas, and then proceed with outer five segment.

Q X ≡ Q′ X′

Now it is time to apply inner five segment to baselines Y Q A and Y′ Q′ A and points X and X′, which gives Q X ≡ Q′ X′.

P Q ≡ P′ Q′

The next step is another application of inner five segment, which proves P Q ≡ P′ Q′. It uses baselines X P A and X′ P′ A and points Q and Q′.

Finishing the proof

Combining the construction of the points and the proof of P Q ≡ P′ Q′ from those points gives us the P ≠ A case of the proof.

And finally we combine the P = A and P ≠ A cases.


Reflecting through a point preserves betweenness relationships. That is, if a point is between two points and we reflect all three points through a common point A, then the corresponding one symmetric point is between the corresponding two symmetric points. In symbols, A is-midpoint-of P P′ ∧ A is-midpoint-of Q Q′ ∧ A is-midpoint-of R R′ ∧ between P Q R → between P′ Q′ R′.<ref>l7_15 in Narboux</ref>

The proof first applies SymmetricPointDistance to each pair of points to prove the degenerate triangle congruence P Q R ≅ P′ Q′ R′. From there, it is a simple application of BetweennessDegenerateTriangle.


The reflections (through a common point) of congruent line segments are congruent. In symbols, A is-midpoint-of P P′ ∧ A is-midpoint-of Q Q′ ∧ A is-midpoint-of R R′ ∧ A is-midpoint-of S S′ ∧ P Q ≡ R S → P′ Q′ ≡ R′ S′.<ref>l7_16 in Narboux</ref>

The proof is just a matter of applying congruence transitivity to P Q ≡ P′ Q′ and R S ≡ R′ S′, which come from SymmetricPointDistance, and P Q ≡ R S (which we are given).



We now export to Interface:Symmetric point.



{{DEFAULTSORT:{{PAGENAME}}}} Category:Euclidean geometries (general) and generalizations Category:Foundations of classical theories (including reverse mathematics)

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