MATH 3210: Euclidean and Non-Euclidean Geometry Hilbert Planes: Euclid’s Propositions (I.1)–(I.12) and Constructions with Hilbert’s Tools March 13, 2020 Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry
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# MATH 3210: Euclidean and Non-Euclidean Geometryspot.colorado.edu/~szendrei/Geom_S20/lec-03-13.pdf · Euclidean and Non-Euclidean Geometry Hilbert Planes: Euclid’s Propositions (I.1)–(I.12)

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Transcript 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).

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?

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:

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):

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):Do these statements follow from the axioms of a Hilbert plane?

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):Do these statements follow from the axioms of a Hilbert plane?

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

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):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. ...”]:

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):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. ...”]:Do these statements follow from the axioms of a Hilbert plane?

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):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. ...”]: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:

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):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. ...”]: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:− (I1): ruler, to draw the unique line through two distinct points;

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):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. ...”]: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:− (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;

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):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. ...”]: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:− (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;− (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.

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):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. ...”]: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:− (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;− (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.These three tools are referred to as Hilbert’s tools.

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):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. ...”]: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:− (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;− (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.These three tools are referred to as Hilbert’s tools.Note: (I2), (I3), (B2) allow us to pick points in certain ways;

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):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. ...”]: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:− (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;− (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.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.)

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry Propositions (I.1)–(I.6)

Euclid’s proposition In Hilbert planes:

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry Propositions (I.1)–(I.6)

Euclid’s proposition In Hilbert planes:

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

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

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

(I.2)

(I.3)

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

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

(I.2)

(I.3)

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.)

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

(I.2)

(I.3)

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.

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

(I.2)

(I.3)

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. Axiom (C6).

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

(I.2)

(I.3)

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. Axiom (C6).

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

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

(I.2)

(I.3)

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. Axiom (C6).

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

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

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

(I.2)

(I.3)

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. 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.

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

(I.2)

(I.3)

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. 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.

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry Propositions (I.7)–(I.8)

Euclid’s proposition In Hilbert planes:

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry Propositions (I.7)–(I.8)

Euclid’s proposition In Hilbert planes:

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

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry Propositions (I.7)–(I.8)

Euclid’s proposition In Hilbert planes:

(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.

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry Propositions (I.7)–(I.8)

Euclid’s proposition In Hilbert planes:

(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.

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry Propositions (I.7)–(I.8)

Euclid’s proposition In Hilbert planes:

(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.

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry Propositions (I.7)–(I.8)

Euclid’s proposition In Hilbert planes:

(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:

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry Propositions (I.7)–(I.8)

Euclid’s proposition In Hilbert planes:

(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:

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry Propositions (I.7)–(I.8)

Euclid’s proposition In Hilbert planes:

(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:(MoS) For any triangle �DEF , for any ray

−→AX , and for each side of

the line AX ,

A

X

D F

E

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry Propositions (I.7)–(I.8)

Euclid’s proposition In Hilbert planes:

(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:(MoS) For any triangle �DEF , for any ray

−→AX , and for each side of

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

A

X

D F

E

E �

F �

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry Propositions (I.7)–(I.8)

Euclid’s proposition In Hilbert planes:

(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:(MoS) For any triangle �DEF , for any ray

−→AX , and for each side of

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 .

A

X

D F

E

E �

F �

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry Propositions (I.7)–(I.8)

Euclid’s proposition In Hilbert planes:

(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:(MoS) For any triangle �DEF , for any ray

−→AX , and for each side of

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 .

A

X

D F

E

E �

F �

Proof.

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry Propositions (I.7)–(I.8)

Euclid’s proposition In Hilbert planes:

(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:(MoS) For any triangle �DEF , for any ray

−→AX , and for each side of

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 .

A

X

D F

E

E �

F �

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

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry Propositions (I.7)–(I.8)

Euclid’s proposition In Hilbert planes:

(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:(MoS) For any triangle �DEF , for any ray

−→AX , and for each side of

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 .

A

X

D F

E

E �

F �

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 .

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry Propositions (I.7)–(I.8)

Euclid’s proposition In Hilbert planes:

(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:(MoS) For any triangle �DEF , for any ray

−→AX , and for each side of

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 .

A

X

D F

E

E �

F �

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.

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry Propositions (I.7)–(I.8)

Euclid’s proposition In Hilbert planes:

(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:(MoS) For any triangle �DEF , for any ray

−→AX , and for each side of

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 .

A

X

D F

E

E �

F �

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).

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

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

A B

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

A B

C

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry 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.• If ∠CAB ∼= ∠CBA, then �ABC is isosceles, by (I.6).

A B

C

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry 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.• If ∠CAB ∼= ∠CBA, then �ABC is isosceles, by (I.6).• Suppose ∠CAB �∼= ∠CBA.

A B

C

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry 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.• If ∠CAB ∼= ∠CBA, then �ABC is isosceles, by (I.6).• Suppose ∠CAB �∼= ∠CBA.

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

A B

C

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry 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.• If ∠CAB ∼= ∠CBA, then �ABC is isosceles, by (I.6).• Suppose ∠CAB �∼= ∠CBA.

� 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

C

E

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry 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.• If ∠CAB ∼= ∠CBA, then �ABC is isosceles, by (I.6).• Suppose ∠CAB �∼= ∠CBA.

� 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.� Crossbar Theorem ⇒ −→

BE meets AC at a point D. A B

C

E

D

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry 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.• If ∠CAB ∼= ∠CBA, then �ABC is isosceles, by (I.6).• Suppose ∠CAB �∼= ∠CBA.

� 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.� Crossbar Theorem ⇒ −→

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

A B

C

E

D

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry 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.• If ∠CAB ∼= ∠CBA, then �ABC is isosceles, by (I.6).• Suppose ∠CAB �∼= ∠CBA.

� 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.� Crossbar Theorem ⇒ −→

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

A B

C

E

D

Construction with Hilbert’s tools: Given AB.

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry 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.• If ∠CAB ∼= ∠CBA, then �ABC is isosceles, by (I.6).• Suppose ∠CAB �∼= ∠CBA.

� 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.� Crossbar Theorem ⇒ −→

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

A B

C

E

D

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

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry 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.• If ∠CAB ∼= ∠CBA, then �ABC is isosceles, by (I.6).• Suppose ∠CAB �∼= ∠CBA.

� 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.� Crossbar Theorem ⇒ −→

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

A B

C

E

D

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

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry 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.• If ∠CAB ∼= ∠CBA, then �ABC is isosceles, by (I.6).• Suppose ∠CAB �∼= ∠CBA.

� 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.� Crossbar Theorem ⇒ −→

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

A B

C

E

D

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.

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry 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.• If ∠CAB ∼= ∠CBA, then �ABC is isosceles, by (I.6).• Suppose ∠CAB �∼= ∠CBA.

� 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.� Crossbar Theorem ⇒ −→

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

A B

C

E

D

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.

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry 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.• If ∠CAB ∼= ∠CBA, then �ABC is isosceles, by (I.6).• Suppose ∠CAB �∼= ∠CBA.

� 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.� Crossbar Theorem ⇒ −→

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

A B

C

E

D

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).

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry Propositions (I.9)–(I.12)

Euclid’s proposition In Hilbert planes:

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry Propositions (I.9)–(I.12)

Euclid’s proposition In Hilbert planes:

(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.

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry Propositions (I.9)–(I.12)

Euclid’s proposition In Hilbert planes:

(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.

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

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

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry Propositions (I.9)–(I.12)

Euclid’s proposition In Hilbert planes:

(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.

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

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

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

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry Propositions (I.9)–(I.12)

Euclid’s proposition In Hilbert planes:

(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.

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

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

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

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

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry Propositions (I.9)–(I.12)

Euclid’s proposition In Hilbert planes:

(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.

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

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

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

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

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

A

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry Propositions (I.9)–(I.12)

Euclid’s proposition In Hilbert planes:

(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.

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

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

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

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

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

A

B C

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry Propositions (I.9)–(I.12)

Euclid’s proposition In Hilbert planes:

(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.

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

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

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

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

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

A

B C

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry Propositions (I.9)–(I.12)

Euclid’s proposition In Hilbert planes:

(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.

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

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

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

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

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 �.

A

B C

X

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry Propositions (I.9)–(I.12)

Euclid’s proposition In Hilbert planes:

(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.

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

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

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

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

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 .

A

B C

X

A�

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry Propositions (I.9)–(I.12)

Euclid’s proposition In Hilbert planes:

(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.

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

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

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

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

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 .

4. Draw line AA�.

A

B C

X

A�

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry Propositions (I.9)–(I.12)

Euclid’s proposition In Hilbert planes:

(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.

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

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

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

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

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 .

4. Draw line AA�.

A

B C

X

A�Justification. B,C exist by (I2).

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry Propositions (I.9)–(I.12)

Euclid’s proposition In Hilbert planes:

(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.

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

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

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

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

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 .

4. Draw line AA�.

A

B C

X

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

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry Propositions (I.9)–(I.12)

Euclid’s proposition In Hilbert planes:

(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.

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

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

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

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

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 .

4. Draw line AA�.

A

B C

X

A�

D

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 �.

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry Propositions (I.9)–(I.12)

Euclid’s proposition In Hilbert planes:

(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.

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

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

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

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

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 .

4. Draw line AA�.

A

B C

X

A�

D

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.

Hilbert Planes: MATH 3210: Euclidean and Non-Euclidean Geometry Propositions (I.9)–(I.12)

Euclid’s proposition In Hilbert planes:

(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.

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

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

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

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

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 .

4. Draw line AA�.

A

B C

X

A�

D

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 = �.

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

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