aarjaneiro/planetmath huggingface dataset:
prismatoid

prismatoid

A

prismatoid

is a

polyhedron

, possibly not convex, whose vertices all lie in one or the other
of two

parallel planes

.
The perpendicular distance between the two planes is called the

altitude

of the prismatoid.
The faces that lie in the parallel planes are called the

bases

of the prismatoid.
The

midsection

is the

polygon

formed by cutting the prismatoid by
a plane

parallel

to the bases halfway between them.

The volume of a prismatoid is given by the

prismoidal formula

:

V

=

1

6

⁢

h

⁢

(

B

1

+

B

2

+

4

⁢

M

)

where

h

is the altitude,

B

1

and

B

2

are the areas of the bases and

M

is the area of the midsection.

An alternate formula is :

V

=

1

4

⁢

h

⁢

(

B

1

+

3

⁢

S

)

where

S

is the area of the polygon that is formed by cutting the prismatoid
by a plane parallel to the bases but 2/3 of the

distance

from

B

1

to

B

2

.

A proof of the prismoidal formula for the case where
the prismatoid is convex is in

[

1

]

. It is also proved in

[

2

]

for any prismatoid.
The alternate formula is proved in

[

2

]

.

Some authors impose the condition that the lateral faces must be

triangles

or

trapezoids

. However, this condition is unnecessary since it is easily shown
to hold.

References

1

A. Day Bradley, Prismatoid,

Prismoid

, Generalized Prismoid,

The American Math. Monthly,

86

, (1979), 486-490.

2

G.B. Halsted,

Rational

Geometry

: A textbook for the Science of Space. Based on
Hilbert’s Foundations

, second edition, John Wiley and Sons, New York, 1907

Title

prismatoid

Canonical name

Prismatoid

Date of creation

2013-03-22 17:12:03

Last modified on

2013-03-22 17:12:03

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

10

Author

Mathprof (13753)

Entry type

Definition

Classification

msc 51-00

Related topic

SimpsonsRule

Related topic

Volume2

Related topic

TruncatedCone

Defines

altitude

Defines

bases

Defines

prismoidal formula

Generated on Fri Feb 9 21:52:45 2018 by

LaTeXML
proof of Apollonius theorem

proof of Apollonius theorem

Let

b

=

C

⁢

A

¯

,

a

=

B

⁢

C

¯

,

c

=

A

⁢

B

¯

, and

m

=

A

⁢

M

¯

. Let

∠

⁢

C

⁢

M

⁢

A

=

θ

, so that

∠

⁢

B

⁢

M

⁢

A

=

π

-

θ

.

By the

law of cosines

,

b

2

=

m

2

+

a

2

4

-

a

⁢

m

⁢

cos

⁡

θ

and

c

2

=

m

2

+

a

2

4

-

a

⁢

m

⁢

cos

⁡

(

π

-

θ

)

=

m

2

+

a

2

4

+

a

⁢

m

⁢

cos

⁡

θ

, and adding gives

b

2

+

c

2

=

2

⁢

m

2

+

a

2

2

.

QED

Title

proof of Apollonius theorem

Canonical name

ProofOfApolloniusTheorem

Date of creation

2013-03-22 12:41:16

Last modified on

2013-03-22 12:41:16

Owner

quincynoodles (312)

Last modified by

quincynoodles (312)

Numerical id

4

Author

quincynoodles (312)

Entry type

Proof

Classification

msc 51-00

Related topic

ProofOfApolloniusTheorem2

Related topic

LengthsOfTriangleMedians

Generated on Fri Feb 9 21:52:47 2018 by

LaTeXML
proof of Apollonius theorem

proof of Apollonius theorem

Let

m

be a median of the

triangle

, as shown in the figure.

By Stewart’s theorem we have

a

⁢

(

m

2

+

(

a

2

)

2

)

=

b

2

⁢

(

a

2

)

+

c

2

⁢

(

a

2

)

and thus

m

2

+

(

a

2

)

2

=

b

2

+

c

2

2

.

Multiplying both sides by

2

gives

2

⁢

m

2

+

a

2

2

=

b

2

+

c

2

.

QED

Title

proof of Apollonius theorem

Canonical name

ProofOfApolloniusTheorem1

Date of creation

2013-03-22 12:41:20

Last modified on

2013-03-22 12:41:20

Owner

drini (3)

Last modified by

drini (3)

Numerical id

5

Author

drini (3)

Entry type

Proof

Classification

msc 51-00

Related topic

Median

Related topic

Cevian

Related topic

ApolloniusTheorem

Related topic

StewartsTheorem

Related topic

ProofOfStewartsTheorem

Related topic

ProofOfApolloniusTheorem

Generated on Fri Feb 9 21:52:48 2018 by

LaTeXML
proof of Brahmagupta’s formula

proof of Brahmagupta’s formula

We shall prove that the area of a

cyclic quadrilateral

with sides

p

,

q

,

r

,

s

is given by

(

T

-

p

)

⁢

(

T

-

q

)

⁢

(

T

-

r

)

⁢

(

T

-

s

)

where

T

=

p

+

q

+

r

+

s

2

.

Area of the cyclic quadrilateral = Area of

△

⁢

A

⁢

D

⁢

B

+

Area of

△

⁢

B

⁢

D

⁢

C

.

=

1

2

⁢

p

⁢

q

⁢

sin

⁡

A

+

1

2

⁢

r

⁢

s

⁢

sin

⁡

C

But since

A

⁢

B

⁢

C

⁢

D

is a cyclic quadrilateral,

∠

⁢

D

⁢

A

⁢

B

=

180

∘

-

∠

⁢

D

⁢

C

⁢

B

.

Hence

sin

⁡

A

=

sin

⁡

C

.

Therefore area now is

A

⁢

r

⁢

e

⁢

a

=

1

2

⁢

p

⁢

q

⁢

sin

⁡

A

+

1

2

⁢

r

⁢

s

⁢

sin

⁡

A

(

A

⁢

r

⁢

e

⁢

a

)

2

=

1

4

⁢

sin

2

⁡

A

⁢

(

p

⁢

q

+

r

⁢

s

)

2

4

⁢

(

A

⁢

r

⁢

e

⁢

a

)

2

=

(

1

-

cos

2

⁡

A

)

⁢

(

p

⁢

q

+

r

⁢

s

)

2

4

⁢

(

A

⁢

r

⁢

e

⁢

a

)

2

=

(

p

⁢

q

+

r

⁢

s

)

2

-

c

⁢

o

⁢

s

2

⁢

A

⁢

(

p

⁢

q

+

r

⁢

s

)

2

Applying

cosines law

for

△

⁢

A

⁢

D

⁢

B

and

△

⁢

B

⁢

D

⁢

C

and equating the

expressions

for side

D

⁢

B

,

we have

p

2

+

q

2

-

2

⁢

p

⁢

q

⁢

cos

⁡

A

=

r

2

+

s

2

-

2

⁢

r

⁢

s

⁢

cos

⁡

C

Substituting

cos

⁡

C

=

-

cos

⁡

A

(since angles

A

and

C

are suppplementary) and rearranging, we have

2

⁢

cos

⁡

A

⁢

(

p

⁢

q

+

r

⁢

s

)

=

p

2

+

q

2

-

r

2

-

s

2

substituting this in the equation for area,

4

⁢

(

A

⁢

r

⁢

e

⁢

a

)

2

=

(

p

⁢

q

+

r

⁢

s

)

2

-

1

4

⁢

(

p

2

+

q

2

-

r

2

-

s

2

)

2

16

⁢

(

A

⁢

r

⁢

e

⁢

a

)

2

=

4

⁢

(

p

⁢

q

+

r

⁢

s

)

2

-

(

p

2

+

q

2

-

r

2

-

s

2

)

2

which is of the form

a

2

-

b

2

and hence can be written in the form

(

a

+

b

)

⁢

(

a

-

b

)

as

(

2

⁢

(

p

⁢

q

+

r

⁢

s

)

+

p

2

+

q

2

-

r

2

-

s

2

)

⁢

(

2

⁢

(

p

⁢

q

+

r

⁢

s

)

-

p

2

-

q

2

+

r

2

+

s

2

)

=

(

(

p

+

q

)

2

-

(

r

-

s

)

2

)

⁢

(

(

r

+

s

)

2

-

(

p

-

q

)

2

)

=

(

p

+

q

+

r

-

s

)

⁢

(

p

+

q

+

s

-

r

)

⁢

(

p

+

r

+

s

-

q

)

⁢

(

q

+

r

+

s

-

p

)

Introducing

T

=

p

+

q

+

r

+

s

2

,

16

⁢

(

A

⁢

r

⁢

e

⁢

a

)

2

=

16

⁢

(

T

-

p

)

⁢

(

T

-

q

)

⁢

(

T

-

r

)

⁢

(

T

-

s

)

Taking square root, we get

A

⁢

r

⁢

e

⁢

a

=

(

T

-

p

)

⁢

(

T

-

q

)

⁢

(

T

-

r

)

⁢

(

T

-

s

)

Title

proof of Brahmagupta’s formula

Canonical

name

ProofOfBrahmaguptasFormula

Date of creation

2013-03-22 13:09:14

Last modified on

2013-03-22 13:09:14

Owner

giri (919)

Last modified by

giri (919)

Numerical id

6

Author

giri (919)

Entry type

Proof

Classification

msc 51-00

Generated on Fri Feb 9 21:52:52 2018 by

LaTeXML
proof of butterfly theorem

proof of butterfly theorem

Given that

M

is the

midpoint

of a chord

P

⁢

Q

of a circle and

A

⁢

B

and

C

⁢

D

are two other chords

passing through

M

, we will prove that

M

is the midpoint of

X

⁢

Y

,

where

X

and

Y

are the points where

A

⁢

D

and

B

⁢

C

cut

P

⁢

Q

respectively.

Let

O

be the center of the circle. Since

O

⁢

M

is

perpendicular

to

X

⁢

Y

(the line from the center of the circle to the midpoint of a chord is perpendicular to the chord), to show that

X

⁢

M

=

M

⁢

Y

,

we have to prove that

∠

⁢

X

⁢

O

⁢

M

=

∠

⁢

Y

⁢

O

⁢

M

.

Drop perpendiculars

O

⁢

K

and

O

⁢

N

from

O

onto

A

⁢

D

and

B

⁢

C

, respectively.
Obviously,

K

is the midpoint of

A

⁢

D

and

N

is the midpoint of

B

⁢

C

. Further,

∠

⁢

D

⁢

A

⁢

B

=

∠

⁢

D

⁢

C

⁢

B

and

∠

⁢

A

⁢

D

⁢

C

=

∠

⁢

A

⁢

B

⁢

C

as angles subtending equal arcs.
Hence

triangles

A

⁢

D

⁢

M

and

C

⁢

B

⁢

M

are

similar

and hence

A

⁢

D

A

⁢

M

=

B

⁢

C

C

⁢

M

or

A

⁢

K

K

⁢

M

=

C

⁢

N

N

⁢

M

In other words, in triangles

A

⁢

K

⁢

M

and

C

⁢

N

⁢

M

,

two pairs of sides are proportional. Also the angles between the corresponding sides are equal. We infer that the triangles

A

⁢

K

⁢

M

and

C

⁢

N

⁢

M

are similar.
Hence

∠

⁢

A

⁢

K

⁢

M

=

∠

⁢

C

⁢

N

⁢

M

.

Now we find that

quadrilaterals

O

⁢

K

⁢

X

⁢

M

and

O

⁢

N

⁢

Y

⁢

M

both have a pair of opposite

straight angles

.
This implies that they are both

cyclic quadrilaterals

.

In

O

⁢

K

⁢

X

⁢

M

,

we have

∠

⁢

A

⁢

K

⁢

M

=

∠

⁢

X

⁢

O

⁢

M

and in

O

⁢

N

⁢

Y

⁢

M

,

we have

∠

⁢

C

⁢

N

⁢

M

=

∠

⁢

Y

⁢

O

⁢

M

.

From these two, we get

∠

⁢

X

⁢

O

⁢

M

=

∠

⁢

Y

⁢

O

⁢

M

.

Therefore

M

is the midpoint of

X

⁢

Y

.

Title

proof of butterfly theorem

Canonical name

ProofOfButterflyTheorem

Date of creation

2013-03-22 13:10:09

Last modified on

2013-03-22 13:10:09

Owner

drini (3)

Last modified by

drini (3)

Numerical id

6

Author

drini (3)

Entry type

Proof

Classification

msc 51-00

Generated on Fri Feb 9 21:52:54 2018 by

LaTeXML
proof of double angle identity

proof of double angle identity

Sine:

sin

⁡

(

2

⁢

a

)

=

sin

⁡

(

a

+

a

)

=

sin

⁡

(

a

)

⁢

cos

⁡

(

a

)

+

cos

⁡

(

a

)

⁢

sin

⁡

(

a

)

=

2

⁢

sin

⁡

(

a

)

⁢

cos

⁡

(

a

)

.

Cosine:

cos

⁡

(

2

⁢

a

)

=

cos

⁡

(

a

+

a

)

=

cos

⁡

(

a

)

⁢

cos

⁡

(

a

)

+

sin

⁡

(

a

)

⁢

sin

⁡

(

a

)

=

cos

2

⁡

(

a

)

-

sin

2

⁡

(

a

)

.

By using the identity

sin

2

⁡

(

a

)

+

cos

2

⁡

(

a

)

=

1

we can change the expression above into the

alternate forms

cos

⁡

(

2

⁢

a

)

=

2

⁢

cos

2

⁡

(

a

)

-

1

=

1

-

2

⁢

sin

2

⁡

(

a

)

.

Tangent

:

tan

⁡

(

2

⁢

a

)

=

tan

⁡

(

a

+

a

)

=

tan

⁡

(

a

)

+

tan

⁡

(

a

)

1

-

tan

⁡

(

a

)

⁢

tan

⁡

(

a

)

=

2

⁢

tan

⁡

(

a

)

1

-

tan

2

⁡

(

a

)

.

Title

proof of double angle identity

Canonical name

ProofOfDoubleAngleIdentity

Date of creation

2013-03-22 12:50:30

Last modified on

2013-03-22 12:50:30

Owner

drini (3)

Last modified by

drini (3)

Numerical id

4

Author

drini (3)

Entry type

Proof

Classification

msc 51-00

Generated on Fri Feb 9 21:52:56 2018 by

LaTeXML
proof of Erdös-Anning Theorem

proof of Erdös-Anning Theorem

Let

A

,

B

and

C

be three non-collinear points. For an additional point

P

consider the

triangle

A

⁢

B

⁢

P

. By using the triangle inequality for the sides

P

⁢

B

and

P

⁢

A

we find

-

|

A

⁢

B

|

≤

|

P

⁢

B

|

-

|

P

⁢

A

|

≤

|

A

⁢

B

|

. Likewise, for triangle

B

⁢

C

⁢

P

we get

-

|

B

⁢

C

|

≤

|

P

⁢

B

|

-

|

P

⁢

C

|

≤

|

B

⁢

C

|

. Geometrically, this means the point

P

lies on two

hyperbola

with

A

and

B

or

B

and

C

respectively as foci. Since all the lengths involved here are by

assumption

integer there are only

2

⁢

|

A

⁢

B

|

+

1

possibilities for

|

P

⁢

B

|

-

|

P

⁢

A

|

and

2

⁢

|

B

⁢

C

|

+

1

possibilities for

|

P

⁢

B

|

-

|

P

⁢

C

|

. These hyperbola are distinct since they don’t have the same

major axis

. So for each pair of hyperbola we can have at most

4

points of

intersection

and there can be no more than

4

⁢

(

2

⁢

|

A

⁢

B

|

+

1

)

⁢

(

2

⁢

|

B

⁢

C

|

+

1

)

points satisfying the conditions.

Title

proof of Erdös-Anning

Theorem

Canonical name

ProofOfErdosAnningTheorem

Date of creation

2013-03-22 13:19:11

Last modified on

2013-03-22 13:19:11

Owner

lieven (1075)

Last modified by

lieven (1075)

Numerical id

4

Author

lieven (1075)

Entry type

Proof

Classification

msc 51-00

Generated on Fri Feb 9 21:52:58 2018 by

LaTeXML
proof of Heron’s formula

proof of Heron’s formula

Let

α

be the angle between the sides

b

and

c

, then we get from the

cosines law

:

cos

⁡

α

=

b

2

+

c

2

-

a

2

2

⁢

b

⁢

c

.

Using the equation

sin

⁡

α

=

1

-

cos

2

⁡

α

we get:

sin

⁡

α

=

-

a

4

-

b

4

-

c

4

+

2

⁢

b

2

⁢

c

2

+

2

⁢

a

2

⁢

b

2

+

2

⁢

a

2

⁢

c

2

2

⁢

b

⁢

c

.

Now we know:

Δ

=

1

2

⁢

b

⁢

c

⁢

sin

⁡

α

.

So we get:

Δ

=

1

4

⁢

-

a

4

-

b

4

-

c

4

+

2

⁢

b

2

⁢

c

2

+

2

⁢

a

2

⁢

b

2

+

2

⁢

a

2

⁢

c

2

=

1

4

⁢

(

a

+

b

+

c

)

⁢

(

b

+

c

-

a

)

⁢

(

a

+

c

-

b

)

⁢

(

a

+

b

-

c

)

=

s

⁢

(

s

-

a

)

⁢

(

s

-

b

)

⁢

(

s

-

c

)

.

This is Heron’s formula.

□

Title

proof of Heron’s formula

Canonical name

ProofOfHeronsFormula

Date of creation

2013-03-22 12:41:38

Last modified on

2013-03-22 12:41:38

Owner

mathwizard (128)

Last modified by

mathwizard (128)

Numerical id

5

Author

mathwizard (128)

Entry type

Proof

Classification

msc 51-00

Generated on Fri Feb 9 21:53:01 2018 by

LaTeXML
proof of Mollweide’s equations

proof of Mollweide’s equations

We transform the equation

(

a

+

b

)

⁢

sin

⁡

γ

2

=

c

⁢

cos

⁡

(

α

-

β

2

)

to

a

⁢

cos

⁡

(

α

2

+

β

2

)

+

b

⁢

cos

⁡

(

α

2

+

β

2

)

=

c

⁢

cos

⁡

α

2

⁢

cos

⁡

β

2

+

c

⁢

sin

⁡

α

2

⁢

sin

⁡

β

2

,

using the fact that

γ

=

π

-

α

-

β

. The left hand side can be further expanded, so that we get:

a

⁢

(

cos

⁡

α

2

⁢

cos

⁡

β

2

-

sin

⁡

α

2

⁢

sin

⁡

β

2

)

+

b

⁢

(

cos

⁡

α

2

⁢

cos

⁡

β

2

-

sin

⁡

α

2

⁢

sin

⁡

β

2

)

=

c

⁢

cos

⁡

α

2

⁢

cos

⁡

β

2

+

c

⁢

sin

⁡

α

2

⁢

sin

⁡

β

2

.

Collecting terms we get:

(

a

+

b

-

c

)

⁢

cos

⁡

α

2

⁢

cos

⁡

β

2

-

(

a

+

b

+

c

)

⁢

sin

⁡

α

2

⁢

sin

⁡

β

2

=

0

.

Using

s

:=

a

+

b

+

c

2

and using the equations

sin

⁡

α

2

=

(

s

-

b

)

⁢

(

s

-

c

)

b

⁢

c

cos

⁡

β

2

=

s

⁢

(

s

-

a

)

b

⁢

c

we get:

2

⁢

s

⁢

(

s

-

c

)

c

⁢

(

s

-

a

)

⁢

(

s

-

b

)

a

⁢

b

-

2

⁢

s

⁢

(

s

-

c

)

c

⁢

(

s

-

a

)

(

s

-

b

)

)

a

⁢

b

=

0

,

which is obviously true. So we can prove the first equation by going backwards. The second equation can be proved in quite the same way.

Title

proof of Mollweide’s equations

Canonical name

ProofOfMollweidesEquations

Date of creation

2013-03-22 12:50:10

Last modified on

2013-03-22 12:50:10

Owner

mathwizard (128)

Last modified by

mathwizard (128)

Numerical id

5

Author

mathwizard (128)

Entry type

Proof

Classification

msc 51-00

Generated on Fri Feb 9 21:53:03 2018 by

LaTeXML
proof of parallelogram law

proof of parallelogram law

The proof follows directly from

Apollonius theorem

noticing that each

diagonal

is a median for the

triangles

in which

parallelogram

is split by the other diagonal.
Also, the diagonals bisect each other.

Therefore, Apollonius theorem implies

2

⁢

(

d

1

2

)

2

+

(

d

2

2

)

2

=

u

2

+

v

2

.

Multiplying both sides by

2

and simplification leads to the desired expression.

Title

proof of parallelogram law

Canonical name

ProofOfParallelogramLaw

Date of creation

2013-03-22 12:41:29

Last modified on

2013-03-22 12:41:29

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

5

Author

Mathprof (13753)

Entry type

Proof

Classification

msc 51-00

Related topic

ApolloniusTheorem

Related topic

Median

Related topic

ProofOfParallelogramLaw2

Generated on Fri Feb 9 21:53:05 2018 by

LaTeXML
proof of Ptolemy’s inequality

proof of Ptolemy’s inequality

Looking at the quadrilateral

A

⁢

B

⁢

C

⁢

D

we construct a point

E

, such that the

triangles

A

⁢

C

⁢

D

and

A

⁢

E

⁢

B

are

similar

(

∠

⁢

A

⁢

B

⁢

E

=

∠

⁢

C

⁢

D

⁢

A

and

∠

⁢

B

⁢

A

⁢

E

=

∠

⁢

C

⁢

A

⁢

D

).

This means that:

A

⁢

E

A

⁢

C

=

A

⁢

B

A

⁢

D

=

B

⁢

E

D

⁢

C

,

from which follows that

B

⁢

E

=

A

⁢

B

⋅

D

⁢

C

A

⁢

D

.

Also because

∠

⁢

E

⁢

A

⁢

C

=

∠

⁢

B

⁢

A

⁢

D

and

A

⁢

D

A

⁢

C

=

A

⁢

B

A

⁢

E

the triangles

E

⁢

A

⁢

C

and

B

⁢

A

⁢

D

are similar. So we get:

E

⁢

C

=

A

⁢

C

⋅

D

⁢

B

A

⁢

D

.

Now if

A

⁢

B

⁢

C

⁢

D

is cyclic we get

∠

⁢

A

⁢

B

⁢

E

+

∠

⁢

C

⁢

B

⁢

A

=

∠

⁢

A

⁢

D

⁢

C

+

∠

⁢

C

⁢

B

⁢

A

=

180

∘

.

This means that the points

C

,

B

and

E

are on one line and thus:

E

⁢

C

=

E

⁢

B

+

B

⁢

C

Now we can use the

formulas

we already found to get:

A

⁢

C

⋅

D

⁢

B

A

⁢

D

=

A

⁢

B

⋅

D

⁢

C

A

⁢

D

+

B

⁢

C

.

Multiplication

with

A

⁢

D

gives:

A

⁢

C

⋅

D

⁢

B

=

A

⁢

B

⋅

D

⁢

C

+

B

⁢

C

⋅

A

⁢

D

.

Now we look at the case that

A

⁢

B

⁢

C

⁢

D

is not cyclic. Then

∠

⁢

A

⁢

B

⁢

E

+

∠

⁢

C

⁢

B

⁢

A

=

∠

⁢

A

⁢

D

⁢

C

+

∠

⁢

C

⁢

B

⁢

A

≠

180

∘

,

so the points

E

,

B

and

C

form a triangle and from the

triangle inequality

we know:

E

⁢

C

<

E

⁢

B

+

B

⁢

C

.

Again we use our formulas to get:

A

⁢

C

⋅

D

⁢

B

A

⁢

D

<

A

⁢

B

⋅

D

⁢

C

A

⁢

D

+

B

⁢

C

.

From this we get:

A

⁢

C

⋅

D

⁢

B

<

A

⁢

B

⋅

D

⁢

C

+

B

⁢

C

⋅

A

⁢

D

.

Putting this together we get Ptolomy’s

inequality

:

A

⁢

C

⋅

D

⁢

B

≤

A

⁢

B

⋅

D

⁢

C

+

B

⁢

C

⋅

A

⁢

D

,

with equality iff

A

⁢

B

⁢

C

⁢

D

is cyclic.

Title

proof of Ptolemy’s inequality

Canonical name

ProofOfPtolemysInequality

Date of creation

2013-03-22 12:46:09

Last modified on

2013-03-22 12:46:09

Owner

mathwizard (128)

Last modified by

mathwizard (128)

Numerical id

5

Author

mathwizard (128)

Entry type

Proof

Classification

msc 51-00

Generated on Fri Feb 9 21:53:07 2018 by

LaTeXML
proof of Ptolemy’s theorem

proof of Ptolemy’s theorem

Let

A

⁢

B

⁢

C

⁢

D

be a cyclic quadrialteral. We will prove that

A

⁢

C

⋅

B

⁢

D

=

A

⁢

B

⋅

C

⁢

D

+

B

⁢

C

⋅

D

⁢

A

.

Find a point

E

on

B

⁢

D

such that

∠

⁢

B

⁢

C

⁢

A

=

∠

⁢

E

⁢

C

⁢

D

. Since

∠

⁢

B

⁢

A

⁢

C

=

∠

⁢

B

⁢

D

⁢

C

for opening the same arc, we have

triangle

similarity

△

⁢

A

⁢

B

⁢

C

∼

△

⁢

D

⁢

E

⁢

C

and so

A

⁢

B

D

⁢

E

=

C

⁢

A

C

⁢

D

which implies

A

⁢

C

⋅

E

⁢

D

=

A

⁢

B

⋅

C

⁢

D

.

Also notice that

△

⁢

A

⁢

D

⁢

C

∼

△

⁢

B

⁢

E

⁢

C

since have two pairs of equal angles. The similarity implies

A

⁢

C

B

⁢

C

=

A

⁢

D

B

⁢

E

which implies

A

⁢

C

⋅

B

⁢

E

=

B

⁢

C

⋅

D

⁢

A

.

So we finally have

A

⁢

C

⋅

B

⁢

D

=

A

⁢

C

⁢

(

B

⁢

E

+

E

⁢

D

)

=

A

⁢

B

⋅

C

⁢

D

+

B

⁢

C

⋅

D

⁢

A

.

Title

proof of Ptolemy’s theorem

Canonical name

ProofOfPtolemysTheorem

Date of creation

2013-03-22 12:38:31

Last modified on

2013-03-22 12:38:31

Owner

drini (3)

Last modified by

drini (3)

Numerical id

11

Author

drini (3)

Entry type

Proof

Classification

msc 51-00

Related topic

PtolemysTheorem

Related topic

CyclicQuadrilateral

Generated on Fri Feb 9 21:53:09 2018 by

LaTeXML
proof of Pythagorean theorem

proof of Pythagorean theorem

This is a geometrical

proof of Pythagorean theorem

. We begin with our

triangle

:

{xy}

,

(

0

,

0

)

;

(

20

,

0

)

*

*

@

-

;

(

20

,

10

)

*

*

@

-

;

(

0

,

0

)

*

*

@

-

,

(

10

,

-

2

)

*

a

,

(

23

,

6

)

*

b

,

(

10

,

7

)

*

c

Now we use the

hypotenuse

as one side of a square:

{xy}

,

(

0

,

0

)

;

(

20

,

0

)

*

*

@

-

;

(

20

,

10

)

*

*

@

-

;

(

0

,

0

)

*

*

@

-

;

(

-

10

,

20

)

*

*

@

-

;

(

10

,

30

)

*

*

@

-

;

(

20

,

10

)

*

*

@

-

,

(

10

,

-

2

)

*

a

,

(

23

,

6

)

*

b

,

(

10

,

7

)

*

c

and draw in four more identical triangles

{xy}

,

(

0

,

0

)

;

(

20

,

0

)

*

*

@

-

;

(

20

,

10

)

*

*

@

-

;

(

0

,

0

)

*

*

@

-

;

(

-

10

,

20

)

*

*

@

-

;

(

10

,

30

)

*

*

@

-

;

(

20

,

10

)

*

*

@

-

;

(

20

,

30

)

*

*

@

-

;

(

-

10

,

30

)

*

*

@

-

;

(

-

10

,

0

)

*

*

@

-

;

(

0

,

0

)

*

*

@

-

,

(

10

,

-

2

)

*

a

,

(

23

,

6

)

*

b

,

(

10

,

7

)

*

c

Now for the proof. We have a large square, with each side of length

a

+

b

, which is subdivided into one smaller square and four
triangles. The area of the large square must be equal to the combined
area of the shapes it is made out of, so we have

(

a

+

b

)

2

=

c

2

+

4

⁢

(

1

2

⁢

a

⁢

b

)

a

2

+

b

2

+

2

⁢

a

⁢

b

=

c

2

+

2

⁢

a

⁢

b

a

2

+

b

2

=

c

2

(1)

Title

proof of Pythagorean theorem

Canonical name

ProofOfPythagoreanTheorem

Date of creation

2013-03-22 11:56:36

Last modified on

2013-03-22 11:56:36

Owner

drini (3)

Last modified by

drini (3)

Numerical id

8

Author

drini (3)

Entry type

Proof

Classification

msc 51-00

Related topic

PythagorasTheorem

Generated on Fri Feb 9 21:53:11 2018 by

LaTeXML
proof of Pythagorean theorem

proof of Pythagorean theorem

Let

A

⁢

B

⁢

C

be a

right triangle

with

hypotenuse

B

⁢

C

. Draw the height

A

⁢

T

.

Using the

right angles

∠

⁢

B

⁢

A

⁢

C

and

∠

⁢

A

⁢

T

⁢

B

and the fact that the sum of angles on any triangle is

180

∘

, it can be shown that

∠

⁢

B

⁢

A

⁢

T

=

∠

⁢

A

⁢

C

⁢

T

∠

⁢

T

⁢

A

⁢

C

=

∠

⁢

C

⁢

B

⁢

A

and therefore we have the following triangle

similarities

:

△

⁢

A

⁢

B

⁢

C

∼

△

⁢

T

⁢

B

⁢

A

∼

△

⁢

T

⁢

A

⁢

C

.

From those similarities, we have

A

⁢

B

B

⁢

C

=

T

⁢

B

B

⁢

A

and thus

A

⁢

B

2

=

B

⁢

C

⋅

T

⁢

B

. Also

A

⁢

C

B

⁢

C

=

T

⁢

C

A

⁢

C

and thus

A

⁢

C

2

=

B

⁢

C

⋅

T

⁢

C

. We have then

A

⁢

B

2

+

A

⁢

C

2

=

B

⁢

C

⁢

(

B

⁢

T

+

T

⁢

C

)

=

B

⁢

C

⋅

B

⁢

C

=

B

⁢

C

2

which concludes the proof.

Title

proof of Pythagorean theorem

Canonical

name

ProofOfPythagoreanTheorem1

Date of creation

2013-03-22 12:48:39

Last modified on

2013-03-22 12:48:39

Owner

drini (3)

Last modified by

drini (3)

Numerical id

8

Author

drini (3)

Entry type

Proof

Classification

msc 51-00

Generated on Fri Feb 9 21:53:13 2018 by

LaTeXML
proof of Simson’s line

proof of Simson’s line

Given a

△

⁢

A

⁢

B

⁢

C

with a point

P

on its

circumcircle

(other than

A

,

B

,

C

),
we will prove that the feet of the

perpendiculars

drawn from P to the sides

A

⁢

B

,

B

⁢

C

,

C

⁢

A

(or their prolongations) are

collinear

.

Since

P

⁢

W

is perpendicular to

B

⁢

W

and

P

⁢

U

is perpendicular to

B

⁢

U

the point

P

lies on the circumcircle of

△

⁢

B

⁢

U

⁢

W

.

By

similar

arguments

,

P

also lies on the circumcircle of

△

⁢

A

⁢

W

⁢

V

and

△

⁢

C

⁢

U

⁢

V

.

This implies that

P

⁢

U

⁢

B

⁢

W

,

P

⁢

U

⁢

C

⁢

V

and

P

⁢

V

⁢

W

⁢

A

are all

cyclic quadrilaterals

.

Since

P

⁢

U

⁢

B

⁢

W

is a cyclic quadrilateral,

∠

⁢

U

⁢

P

⁢

W

=

180

∘

-

∠

⁢

U

⁢

B

⁢

W

implies

∠

⁢

U

⁢

P

⁢

W

=

180

∘

-

∠

⁢

C

⁢

B

⁢

A

Also

C

⁢

P

⁢

A

⁢

B

is a cyclic quadrilateral, therefore

∠

⁢

C

⁢

P

⁢

A

=

180

∘

-

∠

⁢

C

⁢

B

⁢

A

(opposite angles in a cyclic quarilateral are

supplementary

).

From these two, we get

∠

⁢

U

⁢

P

⁢

W

=

∠

⁢

C

⁢

P

⁢

A

Subracting

∠

⁢

C

⁢

P

⁢

W

, we have

∠

⁢

U

⁢

P

⁢

C

=

∠

⁢

W

⁢

P

⁢

A

Now, since

P

⁢

V

⁢

W

⁢

A

is a cyclic quadrilateral,

∠

⁢

W

⁢

P

⁢

A

=

∠

⁢

W

⁢

V

⁢

A

also, since

U

⁢

P

⁢

V

⁢

C

is a cyclic quadrilateral,

∠

⁢

U

⁢

P

⁢

C

=

∠

⁢

U

⁢

V

⁢

C

Combining these two results with the previous one, we have

∠

⁢

W

⁢

V

⁢

A

=

∠

⁢

U

⁢

V

⁢

C

This implies that the points

U

,

V

,

W

are collinear.

Title

proof of Simson’s line

Canonical name

ProofOfSimsonsLine

Date of creation

2013-03-22 13:08:26

Last modified on

2013-03-22 13:08:26

Owner

giri (919)

Last modified by

giri (919)

Numerical id

9

Author

giri (919)

Entry type

Proof

Classification

msc 51-00

Generated on Fri Feb 9 21:53:24 2018 by

LaTeXML