# Why is Euclidean geometry scale-invariant?

In Euclidean geometry, I frequently use concepts related to invariance under scaling. For example, I know that if two squares have different side lengths, the ratio of their side lengths is the square root of the ratio of their areas.

I became curious about the justification of this principle recently when I was asked to explain pi, the ratio of circumference to diameter of a circle, to a young student. It occurred to me half way through the explanation that I do not know a reason why pi exists; I'm not sure why all circles have the same ratio.

I haven't studied non-Euclidean geometry, but I presume this scale-invariance is a special feature of Euclidean geometry. On the surface of a sphere, for example, the sum of the angles of a triangle is related to the ratio of the area of the triangle to the surface area of the sphere, whereas Euclidean triangles have the same sum of angles regardless of size.

I know that Euclidean geometry is distinguished from other geometries by the parallel postulate, the statement that for any line and any point not on the line, there exists a unique line parallel to the original and through the point.

My question is, "How can I go from this parallel postulate and other relevant features of geometry to understanding why Euclidean geometry has no innate length scale involved?"

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Do you want an explanation entirely in terms of Euclid's axioms (which sounds rather tedious) or would you be okay with a more modern treatment (using inner products, etc.)? –  Qiaochu Yuan Feb 21 '11 at 20:17
In hyperbolic geometry for instance, the sum of the angles of a triangle is less than 180°. Interestingly, the difference between the sum and 180° is what is called the defect and the area of the triangle is proportional to the inversse of the curvature of the hyperbolic space times the defect of the triangle. –  Raskolnikov Feb 21 '11 at 20:19
@Qiaochu I know some linear algebra and a little abstract algebra from studying physics, so yes a more modern treatment would be interesting, but depending on what tools it uses I might need a reference to understand the background. –  Mark Eichenlaub Feb 21 '11 at 20:20
A super-simple answer is that curvature is not scale-invariant. A curved object, when you look at it at smaller and smaller scales tends to look flat. So in some sense Euclidean geometry is designed to be flat. –  Ryan Budney Feb 22 '11 at 0:38
@Alexei: the flat torus of course refers to $\mathbb{R}^2 / \mathbb{Z}^2$. –  Willie Wong Feb 28 '11 at 14:06

Similarity is a special feature of Euclidean geometry. It follows from the Euclidean parallel postulate:

For each line and each point not on the line, there is a unique second line, parallel to the first, that contains the point.

As a matter of fact, the following claim, called Wallis's postulate, is equivalent to the parallel postulate:

Given any triangle $\triangle ABC$ and any scale factor, r, there exists a triangle $\triangle DEF$ similar to $\triangle ABC$ where r is the ratio of the lengths of corresponding sides.

There is a list of 15 claims all logically equivalent to the Euclidean parallel postulate on the Wikipedia Parallel Postulate article. One of which, not surprisingly, is the Pythagorean theorem.

Perhaps the following will provide some insight.

Assume

AAA Similarity Theorem: If two triangles have all three pairs of angles congruent, then the triangles are similar.

the proof of which does not require the parallel postulate but is rather involved. Nonetheless, with it, we can show:

Similar Triangle Construction Theorem: If $\triangle ABC$ is triangle and $\overline{DE}$ any segment, then there is a point $F$ such that $\triangle ABC$ is similar to $\triangle DEF$.

Proof: We can construct a unique ray $\overrightarrow{DP}$ so that $\angle EDP \cong \angle A$. We can similarly construct a unique ray $\overrightarrow{EQ}$ on the same side of $\overleftrightarrow{DE}$ so that $\angle DEQ \cong \angle B$. Since the angle measures of the constructed angles, $\angle EDP$ and $\angle DEQ$, sum to less than $180$, by Euclid's fifth postulate (equivalent to Euclidean parallel postulate), the two rays must intersect at some point $F$ forming a triangle $\triangle DEF$. By the angle-sum theorem (also logically equivalent to the Euclidean parallel postulate), $\angle C \cong \angle F$. Now by the AAA similarity theorem, the two triangles are similar. $\square$

Then given any scale factor $r$, We can construct a segment $\overline {DE}$ so that $$r=\frac{DE}{AB}.$$

From this, Wallis's postulate follows straight forwardly.

Conversely, let us assume Wallis's postulate and from it derive the Euclidean parallel postulate.

Proof of EPP from Wallis's Postulate: Suppose we have a line $\ell$ and a point $P$ not on the line. We can drop a perpendicular from $P$ to line $\ell$ and let the perpendicular intersect $\ell$ at point $F$. Then construct a new line $m$ at $P$ at $90$ degrees to our perpendicular. This new line will be parallel $\ell$. If it were not, $m$ and $\ell$ would intersect forming a triangle which would violate the exterior angle theorem.

It remains to show the line $m$ is unique. Suppose there were a different line $n$ also through $P$. We must show $n$ is not parallel, that is, line $n$ intersects $\ell$.

If $n$ is our perpendicular $\overleftrightarrow{PF}$ obviously it intersects $\ell$. So if $n$ is not $\overleftrightarrow{PF}$, lets choose a point $Q$ of $n$ between lines $\ell$ and $m$ but not on $\overleftrightarrow{PF}$. Let $R$ be the point of intersection of a perpendicular dropped from $Q$ onto line $\overleftrightarrow{PF}$. Let $r=\frac{PF}{PR}$. Wallis's theorem assures us that there is somewhere a triangle $\triangle ABC$ that is similar to $\triangle PQR$ with scale factor $r$. We can construct a new triangle $\triangle PSF$ on the segment $\overline {PF}$ congruent to $\triangle ABC$ by placing $S$ on $\ell$ at distance $BC$ from $F$ and on the same side of $\overleftrightarrow{PF}$ as $Q$. This triangle is congruent to $\triangle ABC$ by the side-angle-side congruence theorem and so is similar to $\triangle PQR$. So the angles $\angle FPS$ and $\angle RPQ$ are the same angle. Hence the lines $n$, $\overleftrightarrow{PQ}$, and $\overleftrightarrow{PS}$ are the same line, and $n$ does indeed intersect $\ell$. $\square$

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You give a statement of the AAA Similarity Theorem but usually people take as the definition of similarity the statement: Two triangles whose corresponding angles are congruent are called similar. In the hyperbolic plane (multiple parallels) one can prove that if two triangles are similar then they are congruent. –  Joseph Malkevitch Feb 22 '11 at 5:09
@Joseph: The definition that I've used is: Two triangles △ABC and △DEF are similar under a correspondence A↔D, B↔E, and C↔F iff $\angle A \cong \angle D$,$\quad \angle B \cong \angle E$, $\: \angle C \cong \angle F$ and there is a positive number r such that $$r=\frac{AB}{DE}=\frac{BC}{EF}=\frac{CA}{FD}.$$ Using the reals with geometry for distance, angle measure and proportion, following G. D. Birkhoff, seems more natural than Hilbert's numberless approach. I suspect if the foundations of the reals was in better shape when he did his geometry work, Hilbert might have taken the same approach. –  Eric Nitardy Feb 22 '11 at 7:05
@Eric Yes, there are other approaches to axiomatics other than Hilbert's approach. The version of Birkhoff's that I am most familiar with appears in R. Millman and G. Parker's book, Geometry: A Metric Approach with Models. Loosely speaking Hilbert wanted to preserve a "synthetic" view. –  Joseph Malkevitch Feb 22 '11 at 14:53
@Eric (continued) On page 219 of (first edition, 1981) Millman and Parker write: "We shall define similar triangles to be triangles with corresponding angles congruent." Take a look at Millman and Parker's Theorem 9.1.9. A neutral geometry satisfies EPP (parallel postulate) if and only if there are non-congruent triangles ABC and DEF with angles A and D congruent, B and E congruent and C and F congruent. –  Joseph Malkevitch Feb 22 '11 at 14:54
@Joseph: The definitions and development I know best is from an unpublished manuscript by John M. Lee titled "Axiomatic Geometry". It is based in part on Birkhoff and in part on the School Mathematic Study Group postulates. For the past few weeks, I've been reading "Euclidean and Non-Euclidean Geometries" by Marvin J. Greenberg. I prefer John Lee's approach so far. I'll put Millman and Parker on my reading list. –  Eric Nitardy Feb 22 '11 at 16:28

I am currently studying Perspective and Projective geometry in my math history class and I have learned a bit about the differences in geometries from it... Perspective and Projective geometry differ from Euclidean geometry in that they use descriptive properties and Euclidean geometry follows metric properties. This means that, in Euclidean geometry, you must maintain *distance/length, ratios of distance/length, angles, etc.* Therefore, scaling an object (similar triangles for example) would be invariant. Also, in Perspective/Projective geometry, distances and ratios of distance and angles are not maintained like in Euclidean geometry which is why parallel lines meet at a point! Unlike Euclidean geometry where parallel lines NEVER MEET.

I'm not sure if this is making any sense, but this is what I've learned so far from studying Projective geometry. Hope it helps...

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Maybe this reason isn't "deep" enough, but it's because the Euclidean distance formula (the Pythagorean identity) is scale invariant:

$$a^2 + b^2 = c^2$$

$$(Xa)^2 + (Xb)^2 = X^2(a^2+b^2) = (Xc)^2$$

whereas the distance formulas in spherical geometry:

$$\cos a \cos b = \cos c$$

and hyperbolic geometry:

$$\cosh a \cosh b = \cosh c$$

lack this property.

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Interesting point. Then all I need to do is understand how why the Pythagorean theorem is special to Euclidean geometry. The most famous proofs that pop to mind for me involve things like similar triangles, though, and so using them would be circular reasoning. –  Mark Eichenlaub Feb 21 '11 at 20:24