MATH 3210: Euclidean and Non-Euclidean...

MATH 3210:Euclidean and Non-Euclidean Geometry

Hilbert Planes:

Euclid’s Propositions (I.1)–(I.12) andConstructions with Hilbert’s Tools

March 13, 2020

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry

Interpreting Euclid’s Propositions in Hilbert Planes

We work in a Hilbert plane, i.e., in a geometry satisfying the axioms (I1)–(I3),(B1)–(B4), and (C1)–(C6).

Which propositions from Book I of the Elements are valid in any Hilbert plane?

What does ‘valid’ mean for Euclid’s two types of propositions:

• Universal statements [“For all..., if ... then ...”], e.g., (I.4)–(I.8):

• Universal statements [“For all..., if ... then ...”], e.g., (I.4)–(I.8):Do these statements follow from the axioms of a Hilbert plane?

• Constructions, e.g. (I.1)–(I.3), (I.9)–(I.12):

� If interpreted as existential statements [“Given ..., there exists... s.t. ...”]:

� If interpreted as existential statements [“Given ..., there exists... s.t. ...”]:Do these statements follow from the axioms of a Hilbert plane?

� May also be interpreted as construction problems if we think of applications ofaxioms (I1), (C1), (C4) as construction tools:

� May also be interpreted as construction problems if we think of applications ofaxioms (I1), (C1), (C4) as construction tools:− (I1): ruler, to draw the unique line through two distinct points;

� May also be interpreted as construction problems if we think of applications ofaxioms (I1), (C1), (C4) as construction tools:− (I1): ruler, to draw the unique line through two distinct points;− (C1): transporter of line segments, to get the unique point D on a ray

−→CX

such that CD is congruent to a given segment;

−→CX

such that CD is congruent to a given segment;− (C4): transporter of angles, to get the unique ray

−→DE on a specified side of a

line DF such that ∠EDF is congruent to a given angle.

−→CX

line DF such that ∠EDF is congruent to a given angle.These three tools are referred to as Hilbert’s tools.

−→CX

line DF such that ∠EDF is congruent to a given angle.These three tools are referred to as Hilbert’s tools.Note: (I2), (I3), (B2) allow us to pick points in certain ways;

−→CX

line DF such that ∠EDF is congruent to a given angle.These three tools are referred to as Hilbert’s tools.Note: (I2), (I3), (B2) allow us to pick points in certain ways;

Counting steps: Each use of a tool is one step.(Picking points, marking points of intersection are not separate steps.)

Propositions (I.1)–(I.6)

Euclid’s proposition In Hilbert planes:

(I.1) To construct an equilateral triangleon a given segment.

There are Hilbert planes inwhich there is no equilateraltriangle on some segments.

To draw a segment equal to a givensegment at a given point.To cut off a smaller segment froma larger segment.

Replaced by axiom (C1).Construction can be done withHilbert’s tools.(Transportation of segments.)

(I.4) SAS criterion for congruence of triangles.

(I.4) SAS criterion for congruence of triangles. Axiom (C6).

(I.5) The base angles of an isoscelestriangle are equal.

True. Every step of Euclid’sproof can be justified.

(I.6) If the base angles of a triangle areequal, the triangle is isosceles.

True. Last step of Euclid’s proofrequires proper justification.

(I.7)It is not possible to put two triangleswith equal sides on the same sideof a segment.

True. Steps of Euclid’s proofcan be justified.

(I.8) SSS criterion for congruence oftriangles.

True. New proof needed, becauseEuclid used Method of Superpos.

The Method of Superposition works in all Hilbert planes:

Theorem. The following holds in every Hilbert plane:

Theorem. The following holds in every Hilbert plane:(MoS) For any triangle �DEF , for any ray

−→AX , and for each side of

the line AX ,

the line AX , there exists a triangle �AE �F � such that

the line AX , there exists a triangle �AE �F � such that(i) �AE �F � ∼= �DEF ,(ii) E � is on the ray

−→AX , and

(iii) F � is on the specified side of line AX .

−→AX , and

Proof.

−→AX , and

Proof.• Use (C1) to get E � such that (ii) and AE � ∼= DE .

−→AX , and

Proof.• Use (C1) to get E � such that (ii) and AE � ∼= DE .• Use (C4) and (C1) to get F � s.t. (iii), ∠E �AF � ∼= ∠EDF , and AF � ∼= DF .

−→AX , and

Proof.• Use (C1) to get E � such that (ii) and AE � ∼= DE .• Use (C4) and (C1) to get F � s.t. (iii), ∠E �AF � ∼= ∠EDF , and AF � ∼= DF .• Apply (C6) [SAS] to conclude that (i) holds.

−→AX , and

Proof.• Use (C1) to get E � such that (ii) and AE � ∼= DE .• Use (C4) and (C1) to get F � s.t. (iii), ∠E �AF � ∼= ∠EDF , and AF � ∼= DF .• Apply (C6) [SAS] to conclude that (i) holds.

Now (I.8) follows either using (I.7) or Hilbert’s alternative proof (see book).

A Replacement For (I.1)

In (I.9)–(I.12) Euclid uses (I.1). To carry over these propositions to Hilbert planes, wewill have to replace (I.1) by a weaker statement that is true in all Hilbert planes.

Theorem. (Existence of isosceles triangles)Given a line segment AB, there exists an isosceles triangle with baseAB; in fact, such a triangle is constructible with Hilbert’s tools.

Proof.• (I3) ⇒ there is a point C not on line AB.

Proof.• (I3) ⇒ there is a point C not on line AB.• If ∠CAB ∼= ∠CBA, then �ABC is isosceles, by (I.6).

Proof.• (I3) ⇒ there is a point C not on line AB.• If ∠CAB ∼= ∠CBA, then �ABC is isosceles, by (I.6).• Suppose ∠CAB �∼= ∠CBA.

� Prop. 9.5 ⇒ ∠CAB < ∠CBA or ∠CAB > ∠CBA; say the first.

� Prop. 9.5 ⇒ ∠CAB < ∠CBA or ∠CAB > ∠CBA; say the first.� Def. of < ⇒ there is a ray

−→BE in the interior of ∠CBA

such that ∠CAB ∼= ∠EBA.A B

such that ∠CAB ∼= ∠EBA.� Crossbar Theorem ⇒ −→

BE meets AC at a point D. A B

BE meets AC at a point D.� (I.6) ⇒ �ABD is isosceles.

Construction with Hilbert’s tools: Given AB.

Construction with Hilbert’s tools: Given AB.Pick a point C not on line AB.

Construction with Hilbert’s tools: Given AB.Pick a point C not on line AB.1. Draw line AC.

Construction with Hilbert’s tools: Given AB.Pick a point C not on line AB.1. Draw line AC.2. Draw line BC. Suppose ∠CAB < ∠CBA.

Construction with Hilbert’s tools: Given AB.Pick a point C not on line AB.1. Draw line AC.2. Draw line BC. Suppose ∠CAB < ∠CBA.3. Transport ∠CAB to ∠EBA so that E ,C are on the same side of line AB.

Construction with Hilbert’s tools: Given AB.Pick a point C not on line AB.1. Draw line AC.2. Draw line BC. Suppose ∠CAB < ∠CBA.3. Transport ∠CAB to ∠EBA so that E ,C are on the same side of line AB.Get D (where

−→BE meets AC).

(I.9)(I.10)(I.11)

To bisect an angle.To bisect a segment.To construct a perpendicular to a lineat a given point of the line.

(I.12) To drop a perpendicular from a point toa line not containing the point.

(I.9)(I.10)(I.11)

Euclid’s method yields a construc-tion with Hilbert’s tools, but eachuse of (I.1) has to be replaced byconstructing an isosceles �.

(I.9)(I.10)(I.11)

Constructible with Hilbert’s tools,but not with Euclid’s method.

(I.9)(I.10)(I.11)

The constructions can be justified the same way as before, becausethe proof of the Kite Lemma carries over to Hilbert planes. (See below.)

(I.9)(I.10)(I.11)

Construction for (I.12) with Hilbert’s tools:Given � and A /∈ �.

(I.9)(I.10)(I.11)

Construction for (I.12) with Hilbert’s tools:Given � and A /∈ �. Pick distinct points B,C on �.

(I.9)(I.10)(I.11)

Construction for (I.12) with Hilbert’s tools:Given � and A /∈ �. Pick distinct points B,C on �.1. Draw line AB. �

(I.9)(I.10)(I.11)

Construction for (I.12) with Hilbert’s tools:Given � and A /∈ �. Pick distinct points B,C on �.1. Draw line AB.2. Transport ∠ABC to ∠XBC on the opposite sides of line �.

(I.9)(I.10)(I.11)

Construction for (I.12) with Hilbert’s tools:Given � and A /∈ �. Pick distinct points B,C on �.1. Draw line AB.2. Transport ∠ABC to ∠XBC on the opposite sides of line �.3. Transport BA to BA� on

−→BX .

(I.9)(I.10)(I.11)

−→BX .

4. Draw line AA�.

(I.9)(I.10)(I.11)

−→BX .

4. Draw line AA�.

A�Justification. B,C exist by (I2).

(I.9)(I.10)(I.11)

−→BX .

4. Draw line AA�.

A�Justification. B,C exist by (I2).From the construction and (C6), �ABC ∼= �A�BC (so ABA�C is a kite).

(I.9)(I.10)(I.11)

−→BX .

4. Draw line AA�.

Justification. B,C exist by (I2).From the construction and (C6), �ABC ∼= �A�BC (so ABA�C is a kite).AA� meets � at a point D, because A,A� are on opposite sides of �.

(I.9)(I.10)(I.11)

−→BX .

4. Draw line AA�.

Justification. B,C exist by (I2).From the construction and (C6), �ABC ∼= �A�BC (so ABA�C is a kite).AA� meets � at a point D, because A,A� are on opposite sides of �.D is different from at least one of B,C; say D �= B.

(I.9)(I.10)(I.11)

−→BX .

4. Draw line AA�.

Justification. B,C exist by (I2).From the construction and (C6), �ABC ∼= �A�BC (so ABA�C is a kite).AA� meets � at a point D, because A,A� are on opposite sides of �.D is different from at least one of B,C; say D �= B.By (C6), �ABD ∼= �A�BD, therefore ∠ADB ∼= ∠A�DB, and hence AA� ⊥ BD = �.

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