aarjaneiro/planetmath huggingface dataset:
angle sum identity

angle sum identity

It is desired to prove the identities

sin

⁡

(

θ

+

ϕ

)

=

sin

⁡

θ

⁢

cos

⁡

ϕ

+

cos

⁡

θ

⁢

sin

⁡

ϕ

and

cos

⁡

(

θ

+

ϕ

)

=

cos

⁡

θ

⁢

cos

⁡

ϕ

-

sin

⁡

θ

⁢

sin

⁡

ϕ

Consider the figure

where we have

∘

△

⁢

A

⁢

a

⁢

d

≡

△

⁢

C

⁢

c

⁢

b

∘

△

⁢

B

⁢

b

⁢

a

≡

△

⁢

D

⁢

d

⁢

c

∘

a

⁢

d

=

d

⁢

c

=

1

.

Also, everything is Euclidean, and in particular, the

interior angles

of any

triangle

sum to

π

.

Call

∠

⁢

A

⁢

a

⁢

d

=

θ

and

∠

⁢

b

⁢

a

⁢

B

=

ϕ

.
From the triangle , we have

∠

⁢

A

⁢

d

⁢

a

=

π

2

-

θ

and

∠

⁢

D

⁢

d

⁢

c

=

π

2

-

ϕ

, while the degenerate
angle

∠

⁢

A

⁢

d

⁢

D

=

π

, so that

∠

⁢

a

⁢

d

⁢

c

=

θ

+

ϕ

We have, therefore, that the area of the pink

parallelogram

is

sin

⁡

(

θ

+

ϕ

)

. On the other hand, we can rearrange things thus:

In this figure we see an equal pink area, but it is composed of two pieces, of areas

sin

⁡

ϕ

⁢

cos

⁡

θ

and

cos

⁡

ϕ

⁢

sin

⁡

θ

. Adding,
we have

sin

⁡

(

θ

+

ϕ

)

=

sin

⁡

ϕ

⁢

cos

⁡

θ

+

cos

⁡

ϕ

⁢

sin

⁡

θ

which gives us the first.
From definitions, it then also follows that

sin

⁡

(

θ

+

π

/

2

)

=

cos

⁡

(

θ

)

, and

sin

⁡

(

θ

+

π

)

=

-

sin

⁡

(

θ

)

.
Writing

cos

⁡

(

θ

+

ϕ

)

=

sin

⁡

(

θ

+

ϕ

+

π

/

2

)

=

sin

⁡

(

θ

)

⁢

cos

⁡

(

ϕ

+

π

/

2

)

+

cos

⁡

(

θ

)

⁢

sin

⁡

(

ϕ

+

π

/

2

)

=

sin

⁡

(

θ

)

⁢

sin

⁡

(

ϕ

+

π

)

+

cos

⁡

(

θ

)

⁢

cos

⁡

(

ϕ

)

=

cos

⁡

θ

⁢

cos

⁡

ϕ

-

sin

⁡

θ

⁢

sin

⁡

ϕ

Title

angle sum identity

Canonical name

AngleSumIdentity

Date of creation

2013-03-22 12:50:36

Last modified on

2013-03-22 12:50:36

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

14

Author

mathcam (2727)

Entry type

Theorem

Classification

msc 51-00

Related topic

ProofOfDeMoivreIdentity

Related topic

DoubleAngleIdentity

Generated on Fri Feb 9 21:49:31 2018 by

LaTeXML
angles of an isosceles triangle

angles of an isosceles triangle

The following

theorem

holds in any

geometry

in which SAS is valid. Specifically, it holds in both

Euclidean geometry

and

hyperbolic geometry

(and therefore in

neutral geometry

) as well as in

spherical geometry

.

Theorem 1

.

The angles opposite to the

congruent

sides of an

isosceles triangle

are congruent.

Proof.

Let

triangle

△

⁢

A

⁢

B

⁢

C

be isosceles such that the legs

A

⁢

B

¯

and

A

⁢

C

¯

are congruent.

Since we have

•

A

⁢

B

¯

≅

A

⁢

C

¯

•

∠

⁢

A

≅

∠

⁢

A

by the reflexive property (

http://planetmath.org/

Reflexive

) of

≅

•

A

⁢

C

¯

≅

A

⁢

B

¯

by the

symmetric

property (

http://planetmath.org/Symmetric

) of

≅

we can use SAS to conclude that

△

⁢

A

⁢

B

⁢

C

≅

△

⁢

A

⁢

C

⁢

B

. Since corresponding parts of congruent triangles are congruent, it follows that

∠

⁢

B

≅

∠

⁢

C

.
∎

In geometries in which SAS and ASA are both valid, the

converse theorem

of this theorem is also true. This theorem is stated and proven in the entry determining from angles that a triangle is isosceles.

Title

angles of an isosceles triangle

Canonical name

AnglesOfAnIsoscelesTriangle

Date of creation

2013-03-22 17:12:06

Last modified on

2013-03-22 17:12:06

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

10

Author

Wkbj79 (1863)

Entry type

Theorem

Classification

msc 51-00

Classification

msc 51M04

Related topic

DeterminingFromAnglesThatATriangleIsIsosceles

Related topic

PonsAsinorum

Generated on Fri Feb 9 21:49:34 2018 by

LaTeXML
annulus

annulus

An

annulus

is a two-dimensional shape which can be thought of as a disc with a smaller disc removed from its center. An annulus looks like:

Note that both the inner and outer radii may take on any values, so long as the outer radius is larger than the inner.

Title

annulus

Canonical

name

Annulus

Date of creation

2013-03-22 12:26:23

Last modified on

2013-03-22 12:26:23

Owner

akrowne (2)

Last modified by

akrowne (2)

Numerical id

12

Author

akrowne (2)

Entry type

Definition

Classification

msc 51-00

Generated on Fri Feb 9 21:49:36 2018 by

LaTeXML
Apollonius’ circle

Apollonius’ circle

Apollonius’ circle

.
The locus of a point moving so that the ratio of its

distances

from two

fixed points

is fixed, is a circle.

If two circles

C

1

and

C

2

are fixed with radii

r

1

and

r

2

, then the circle of Apollonius of the two centers with ratio

r

1

/

r

2

is the circle whose

diameter

is the

segment

that the two

homothety centers

of the circles.

Title

Apollonius’ circle

Canonical name

ApolloniusCircle

Date of creation

2013-03-22 11:44:22

Last modified on

2013-03-22 11:44:22

Owner

drini (3)

Last modified by

drini (3)

Numerical id

11

Author

drini (3)

Entry type

Definition

Classification

msc 51-00

Classification

msc 35-01

Related topic

HarmonicDivision

Generated on Fri Feb 9 21:49:37 2018 by

LaTeXML
Apollonius theorem

Apollonius theorem

Let

a

,

b

,

c

the sides of a

triangle

and

m

the length of the median to the side with length

a

.
Then

b

2

+

c

2

=

2

⁢

m

2

+

a

2

2

.

If

b

=

c

(the triangle is

isosceles

), then the theorem reduces to the

Pythagorean theorem

,

m

2

+

(

a

/

2

)

2

=

b

2

.

Title

Apollonius theorem

Canonical name

ApolloniusTheorem

Date of creation

2013-03-22 11:44:10

Last modified on

2013-03-22 11:44:10

Owner

yark (2760)

Last modified by

yark (2760)

Numerical id

14

Author

yark (2760)

Entry type

Theorem

Classification

msc 51-00

Classification

msc 18-00

Classification

msc 81-00

Related topic

Triangle

Related topic

Median

Related topic

StewartsTheorem

Related topic

ProofOfStewartsTheorem

Related topic

ProofOfApolloniusTheorem2

Related topic

ParallelogramLaw

Related topic

ProofOfParallelogramLaw

Generated on Fri Feb 9 21:49:39 2018 by

LaTeXML
area of regular polygon

area of regular polygon

Theorem 1

.

Given a

regular

n

-gon (

http://planetmath.org/RegularPolygon

) with

apothem

of length

a

and

perimeter

(

http://planetmath.org/Perimeter2

)

P

, its area is

A

=

1

2

⁢

a

⁢

P

.

Proof.

Given a regular

n

-gon

R

,

line segments

can be drawn from its center to each of its vertices. This divides

R

into

n

congruent

triangles

. The area of each of these triangles is

1

2

⁢

a

⁢

s

, where

s

is the length of one of the sides of the triangle. Also note that the perimeter of

R

is

P

=

n

⁢

s

. Thus, the area

A

of

R

is

A

=

n

⁢

(

1

2

⁢

a

⁢

s

)

=

1

2

⁢

a

⁢

(

n

⁢

s

)

=

1

2

⁢

a

⁢

P

.

∎

To illustrate what is going on in the proof, a

regular hexagon

appears below with each line segment from its center to one of its vertices drawn in red and one of its apothems drawn in blue.

Title

area of regular polygon

Canonical name

AreaOfRegularPolygon

Date of creation

2013-03-22 17:11:06

Last modified on

2013-03-22 17:11:06

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

6

Author

Wkbj79 (1863)

Entry type

Theorem

Classification

msc 51-00

Generated on Fri Feb 9 21:49:41 2018 by

LaTeXML
bound on area of right triangle

bound on area of right triangle

We may bound the area of a

right triangle

in terms of its

perimeter

.
The derivation of this bound is a good exercise in constrained
optimization using

Lagrange multipliers

.

Theorem 1

.

If a right triangle has perimeter

P

, then its area is bounded as

A

≤

3

-

2

⁢

2

4

⁢

P

2

with equality when one has an

isosceles right triangle

.

Proof.

Suppose a triangle has legs of length

x

and

y

. Then its

hypotenuse

has length

x

2

+

y

2

, so the perimeter is given as

P

=

x

+

y

+

x

2

+

y

2

.

The area, of course, is

A

=

1

2

⁢

x

⁢

y

.

We want to maximize

A

subject to the constraint that

P

be constant.
This means that the

gradient

of

A

will be proportional to the gradient
of

P

. That is to say, for some constant

λ

, we will have

∂

⁡

A

∂

⁡

x

=

λ

⁢

∂

⁡

P

∂

⁡

x

∂

⁡

A

∂

⁡

y

=

λ

⁢

∂

⁡

P

∂

⁡

y

Together with the constraint, these form a system of three equations
for the three quantities

x

,

y

, and

λ

. Writing them out
explicitly,

1

2

⁢

y

=

λ

⁢

(

1

+

x

x

2

+

y

2

)

1

2

⁢

x

=

λ

⁢

(

1

+

y

x

2

+

y

2

)

P

=

x

+

y

+

x

2

+

y

2

Not that we cannot have

λ

=

0

because that would mean that all
sides of our triangle would have zero length. Hence, we may eliminate

λ

between the first two equations to obtain

x

⁢

(

1

+

x

x

2

+

y

2

)

=

y

⁢

(

1

+

y

x

2

+

y

2

)

,

which may be manipulated to yield

(

x

-

y

)

⁢

(

1

+

x

+

y

x

2

+

y

2

)

=

0

.

We have two case to consider — either the first factor or the second
factor may equal zero. If the second factor equals zero,

1

+

x

+

y

x

2

+

y

2

=

0

,

move the “1” to the other side of the equation and cross-multiply
to obtain

x

+

y

=

-

x

2

+

y

2

.

Since we want

x

≥

0

and

y

≥

0

but the right-hand side is
non-positive, the only option would be to have a trianagle of
zero area.
The other possibility was to have the second factor equal zero,
which would give

x

-

y

=

0

.

In this case,

x

equals

y

. Imposing this condition on the constraint,
we see that

P

=

(

2

+

2

)

⁢

x

,

so we have the solution

x

=

P

2

+

2

=

2

-

2

2

⁢

P

y

=

P

2

+

2

=

2

-

2

2

⁢

P

.

∎

Title

bound on area of right triangle

Canonical

name

BoundOnAreaOfRightTriangle

Date of creation

2013-03-22 16:30:41

Last modified on

2013-03-22 16:30:41

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

17

Author

rspuzio (6075)

Entry type

Theorem

Classification

msc 51-00

Generated on Fri Feb 9 21:49:44 2018 by

LaTeXML
Brahmagupta’s formula

Brahmagupta’s formula

If a

cyclic quadrilateral

has sides

p

,

q

,

r

,

s

then its area is given by

(

T

-

p

)

⁢

(

T

-

q

)

⁢

(

T

-

r

)

⁢

(

T

-

s

)

where

T

=

p

+

q

+

r

+

s

2

.

Note that if

s

→

0

, Heron’s formula is recovered.

Title

Brahmagupta’s formula

Canonical name

BrahmaguptasFormula

Date of creation

2013-03-22 11:44:19

Last modified on

2013-03-22 11:44:19

Owner

drini (3)

Last modified by

drini (3)

Numerical id

11

Author

drini (3)

Entry type

Theorem

Classification

msc 51-00

Classification

msc 81T70

Classification

msc 81T25

Classification

msc 81T20

Classification

msc 81T75

Classification

msc 81T05

Classification

msc 11-02

Classification

msc 05-02

Classification

msc 05C30

Related topic

CyclicQuadrilateral

Related topic

HeronsFormula

Generated on Fri Feb 9 21:49:45 2018 by

LaTeXML
Brianchon theorem

Brianchon theorem

If an

hexagon

A

⁢

B

⁢

C

⁢

D

⁢

E

⁢

F

(not necessarily convex) is

inscribed

into
a conic (in particular into a circle), then the three

diagonals

A

⁢

D

,

B

⁢

E

,

C

⁢

F

are

concurrent

. This theorem is the dual of

Pascal
line

theorem. (C. Brianchon, 1806)

Proof:

It follows by

duality

.

Title

Brianchon theorem

Canonical name

BrianchonTheorem

Date of creation

2013-03-22 14:04:29

Last modified on

2013-03-22 14:04:29

Owner

vmoraru (1243)

Last modified by

vmoraru (1243)

Numerical id

7

Author

vmoraru (1243)

Entry type

Theorem

Classification

msc 51-00

Generated on Fri Feb 9 21:49:47 2018 by

LaTeXML
butterfly theorem

butterfly theorem

Let

M

be the

midpoint

of a chord

P

⁢

Q

of a circle, through which two other chords

A

⁢

B

and

C

⁢

D

are drawn. If

A

⁢

D

intersects

P

⁢

Q

at

X

and

C

⁢

B

intersects

P

⁢

Q

at

Y

,then

M

is also the midpoint of

X

⁢

Y

.

The theorem gets its name from the shape of the figure, which resembles a butterfly.

Title

butterfly theorem

Canonical name

ButterflyTheorem

Date of creation

2013-03-22 13:10:06

Last modified on

2013-03-22 13:10:06

Owner

giri (919)

Last modified by

giri (919)

Numerical id

8

Author

giri (919)

Entry type

Theorem

Classification

msc 51-00

Generated on Fri Feb 9 21:49:49 2018 by

LaTeXML
CPCTC

CPCTC

“CPCTC” is an acronym which stands for “corresponding parts of

congruent

(

http://planetmath.org/Congruent2

)

triangles

(

http://planetmath.org/Triangle

) are congruent.” In other , if

△

⁢

A

⁢

B

⁢

C

≅

△

⁢

D

⁢

E

⁢

F

,

(1)

then all of the following are true:

•

∠

⁢

A

≅

∠

⁢

D

;

•

A

⁢

B

¯

≅

D

⁢

E

¯

;

•

∠

⁢

B

≅

∠

⁢

E

;

•

B

⁢

C

¯

≅

E

⁢

F

¯

;

•

∠

⁢

C

≅

∠

⁢

F

;

•

A

⁢

C

¯

≅

D

⁢

F

¯

.

Note.

The notation in (1) indicates that the

isomorphism

in question maps

A

↦

D

,

B

↦

E

,  and

C

↦

F

.

Title

CPCTC

Canonical name

CPCTC

Date of creation

2013-03-22 17:12:09

Last modified on

2013-03-22 17:12:09

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

7

Author

Wkbj79 (1863)

Entry type

Definition

Classification

msc 51-00

Synonym

corresponding parts of congruent triangles are congruent

Generated on Fri Feb 9 21:49:51 2018 by

LaTeXML
Carnot circles

Carnot circles

If

A

⁢

B

⁢

C

is a

triangle

, and

H

is the

orthocenter

, then we have
three circles so that every circle contains two vertices and
the orthocenter. The three circles are called the

Carnot
circles

.

Title

Carnot circles

Canonical name

CarnotCircles

Date of creation

2013-03-22 14:05:24

Last modified on

2013-03-22 14:05:24

Owner

vmoraru (1243)

Last modified by

vmoraru (1243)

Numerical id

6

Author

vmoraru (1243)

Entry type

Definition

Classification

msc 51-00

Generated on Fri Feb 9 21:49:53 2018 by

LaTeXML
centroid

centroid

The

centroid

of a

triangle

(also called

center of gravity

of
the triangle) is the point where the three medians intersect each other.

In the figure,

A

⁢

A

′

,

B

⁢

B

′

and

C

⁢

C

′

are medians and

G

is the centroid of

A

⁢

B

⁢

C

.
The centroid

G

has the

property

that divides the medians in the ratio

2

:

1

, that is

A

⁢

G

=

2

⁢

G

⁢

A

′

B

⁢

G

=

2

⁢

G

⁢

B

′

C

⁢

G

=

2

⁢

G

⁢

C

′

.

Title

centroid

Canonical

name

Centroid

Date of creation

2013-03-22 11:55:44

Last modified on

2013-03-22 11:55:44

Owner

drini (3)

Last modified by

drini (3)

Numerical id

11

Author

drini (3)

Entry type

Definition

Classification

msc 51-00

Synonym

barycenter

Synonym

center of gravity

Related topic

Median

Related topic

Orthocenter

Related topic

Triangle

Related topic

EulerLine

Related topic

CevasTheorem

Related topic

CenterOfATriangle

Related topic

LemoinePoint

Related topic

GergonneTriangle

Related topic

TrigonometricVersionOfCevasTheorem

Generated on Fri Feb 9 21:49:54 2018 by

LaTeXML
chord

chord

A chord is the

line segment

joining two points on a curve.
Usually it is used to refer to a line segment whose

end points

lie on a circle.

Title

chord

Canonical name

Chord

Date of creation

2013-03-22 13:08:51

Last modified on

2013-03-22 13:08:51

Owner

giri (919)

Last modified by

giri (919)

Numerical id

4

Author

giri (919)

Entry type

Definition

Classification

msc 51-00

Generated on Fri Feb 9 21:49:56 2018 by

LaTeXML
circumcircle

circumcircle

For any

triangle

A

⁢

B

⁢

C

there is always
a circle

passing through

its three vertices.

Such circle is called a

circumcircle

.
Its radius is the

circumradius

,
and its center is the

circumcenter

.
The circumcenter lies at the

intersection

of the

perpendicular bisectors

of the sides of the triangle.

Since the perpendicular bisector of a segment
is the locus of points at the same

distance

from the segment

endpoints

,
the points on the perpendicular bisector of

A

⁢

B

are equidistant to

A

and

B

.
The points in the perpendicular bisector of

B

⁢

C

are equidistant to

B

and

C

,
and thus the intersection point

O

is at the same distance from

A

,

B

and

C

.

In a more general setting, if

P

is any

polygon

,
its circumcircle would be a circle passing through all vertices,
and circumradius and circumcenter are defined similarly.
However, unlike triangles, circumcircles need not to exist for any polygon.
For

instance

, a non-rectangular

parallelogram

has no circumcircle,
for no circle passes through the four vertices.
A

quadrilateral

that does possess a circumcircle
is called a

cyclic quadrilateral

.

Title

circumcircle

Canonical

name

Circumcircle

Date of creation

2013-03-22 15:00:32

Last modified on

2013-03-22 15:00:32

Owner

yark (2760)

Last modified by

yark (2760)

Numerical id

8

Author

yark (2760)

Entry type

Definition

Classification

msc 51-00

Related topic

Triangle

Related topic

CyclicQuadrilateral

Related topic

SimsonsLine

Defines

circumcenter

Defines

circumcentre

Defines

circumradius

Generated on Fri Feb 9 21:50:04 2018 by

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classification of Platonic solids

classification of Platonic solids

Proposition.

.

The

regular tetrahedron

, regular octahedron, regular icosahedron, cube, and regular dodecahedron
are the only

Platonic solids

.

Proof.

Each vertex of a Platonic solid is

incident

with at least three
faces. The

interior angles

incident with that vertex must sum to
less than

2

⁢

π

, for otherwise the solid would be flat at that
vertex. Since all faces of the solid have the same number of sides,
this implies bounds on the number of faces which could meet at a
vertex.

The interior angle of an

equilateral triangle

has measure

π

3

,
so a Platonic solid could only have three, four, or five

triangles

meeting at each vertex. By

similar

reasoning, a Platonic solid could
only have three squares or three

pentagons

meeting at each vertex. But
the interior angle of a

regular hexagon

has measure

2

⁢

π

3

.
To avoid flatness a solid with

hexagons

as faces would thus have
to have only two faces meeting at each vertex, which is impossible. For

polygons

with more sides it only gets worse.

Since a Platonic solid is uniquely determined by the number and kind of
faces meeting at each vertex, there are at most five Platonic solids,
with the numbers and kinds of faces listed above. But
these correspond to the five known Platonic solids. Hence there are
exactly five Platonic solids.
∎

Title

classification of Platonic solids

Canonical

name

ClassificationOfPlatonicSolids

Date of creation

2013-03-22 15:53:07

Last modified on

2013-03-22 15:53:07

Owner

mps (409)

Last modified by

mps (409)

Numerical id

5

Author

mps (409)

Entry type

Result

Classification

msc 51-00

Generated on Fri Feb 9 21:50:05 2018 by

LaTeXML
collinear

collinear

In a , points are said to be

collinear

if they all lie on a

straight line

.

In the following picture,

A

,

P

, and

B

are collinear.

Title

collinear

Canonical name

Collinear

Date of creation

2013-03-22 12:24:40

Last modified on

2013-03-22 12:24:40

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

17

Author

Wkbj79 (1863)

Entry type

Definition

Classification

msc 51-00

Synonym

collinearity

Related topic

PappussTheorem

Related topic

EulerLine

Related topic

MenelausTheorem

Generated on Fri Feb 9 21:50:07 2018 by

LaTeXML
compass and straightedge construction of angle bisector

compass and straightedge construction of angle bisector

One can construct the (interior)

angle bisector

of a given angle using

compass

and straightedge as follows:

1.

With one point of the compass on the vertex (

http://planetmath.org/Vertex5

) of the angle, draw an arc that

intersects

both sides (

http://planetmath.org/Side3

) of the angle.

2.

Draw an arc from each of these points of

intersection

so that the arcs intersect in the interior of the angle. The compass needs to stay open the same amount throughout this step.

3.

Draw the ray from the vertex of the angle to the intersection of the two arcs drawn during the previous step.

This construction is justified because the point determined in the second step is equidistant from the two rays and thus must lie on the angle bisector.

If you are interested in seeing the rules for compass and straightedge constructions, click on the provided.

Title

compass and straightedge construction of angle bisector

Canonical name

CompassAndStraightedgeConstructionOfAngleBisector

Date of creation

2013-03-22 17:11:09

Last modified on

2013-03-22 17:11:09

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

18

Author

Wkbj79 (1863)

Entry type

Algorithm

Classification

msc 51-00

Classification

msc 51M15

Synonym

construction of angle bisector

Generated on Fri Feb 9 21:50:10 2018 by

LaTeXML
compass and straightedge construction of parallel line

compass and straightedge construction of parallel line

Task.

Construct the line

parallel

to a given line

ℓ

and

passing through

a given point

P

which is not on

ℓ

.

Solution.

1.

Draw a circle

c

1

with center (

http://planetmath.org/Center8

)

P

and intersecting

ℓ

at two points, one of which is

A

.

2.

Draw a second circle

c

2

with center

A

and the same radius

r

as

c

1

. This circle also

intersects

ℓ

at two points, one of which is

B

.

3.

Draw a third circle

c

3

with center

B

and radius

r

.  Let

C

be the

intersection

point of

c

3

(drawn below in red) with

c

1

(drawn below in green) which lies on the same side of

ℓ

as

P

does.  The line

P

⁢

C

(drawn below in blue) is the required parallel to

ℓ

.

Note 1.

The construction is based on the fact that the

quadrilateral

P

⁢

A

⁢

B

⁢

C

is a

parallelogram

. In fact,

P

⁢

A

⁢

B

⁢

C

is a

rhombus

. The reasoning is as follows:

•

The green circle shows that

P

⁢

C

¯

and

P

⁢

A

¯

are

congruent

.

•

The black circle shows that

P

⁢

A

¯

and

A

⁢

B

¯

are congruent.

•

The red circle shows that

A

⁢

B

¯

and

B

⁢

C

¯

are congruent.

•

Since

P

⁢

A

⁢

B

⁢

C

is a quadrilateral with all sides congruent, it is a rhombus (and therefore a parallelogram).

Note 2.

It is clear that the construction only needs the

compass

, not a straightedge: In determining the point

C

, the straightedge is totally superfluous, and the points

P

and

C

determine the desired line (which thus is not necessary to actually draw!). It may be proved that all constructions with compass and straightedge are possible using only the compass.

Note 3.

Another construction of the parallel uses the fact that the

endpoints

of two congruent chords (red) in a circle determine two parallel chords:

If you are interested in seeing the rules for compass and straightedge constructions, click on the provided.

Title

compass and straightedge construction of parallel line

Canonical name

CompassAndStraightedgeConstructionOfParallelLine

Date of creation

2013-03-22 17:11:18

Last modified on

2013-03-22 17:11:18

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

24

Author

pahio (2872)

Entry type

Algorithm

Classification

msc 51-00

Classification

msc 51M15

Synonym

construction of parallel

Synonym

construction of parallel line

Related topic

ParallelPostulate

Related topic

NSectionOfLineSegmentWithCompassAndStraightedge

Generated on Fri Feb 9 21:50:12 2018 by

LaTeXML