nyuuzyou/wikis huggingface dataset:
Parallel Law for Extremal Length

== Theorem ==
Let $X$ be a Riemann surface.
Let $\Gamma_1, \Gamma_2$ be families of rectifiable curves (or, more generally, families of disjoint unions of rectifiable curves) on $X$.
Let $\Gamma_1$ and $\Gamma_2$ be disjoint, in the sense that:
:there exist disjoint Borel subsets $A_1, A_2 \subseteq X$ such that:
::for any $\gamma_1 \in \Gamma_1$ and $\gamma_2 \in \Gamma_2$, we have $\gamma_1 \subseteq A_1$ and $\gamma_2 \subseteq A_2$.
Let $\Gamma$ be a third curve family, with the property that every element $\Gamma_1$ and every element of $\Gamma_2$ contains some element of $\Gamma$.
Then the extremal length of $\Gamma$ satisfies:
:$\dfrac 1 {\lambda \left({\Gamma}\right)} \ge \dfrac 1 {\lambda \left({\Gamma_1}\right)} + \dfrac 1 {\lambda \left({\Gamma_2}\right)}$
== Proof ==
The assumption means that every element of $\Gamma_1 \cup \Gamma_2$ contains some element of $\Gamma$.
Hence:
== Sources ==
* 
Category:Geometric Function Theory
Parallel Lines which intercept Equal Segments on Transversals

== Theorem ==
Let $3$ or more parallel lines intersect equal line segments on one transversal.
Then those same $3$ or more parallel lines intersect equal line segments on every transversal.
== Proof ==
== Sources ==
* : Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.25$
Category:Parallel Lines
Parallel Relationship is Transitive

== Theorem ==
Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $x, y, z \in S : x \ne y, x \ne z, y \ne z$.
If $x$ is parallel to $y$ and $y$ is parallel to $z$ then $x$ is parallel to $z$.
== Proof ==
Let $x$ be parallel to $y$ and $y$ be parallel to $z$.
By definition of parallel:
:$\set x$, $\set y$, $\set z$ are independent subsets
:$\set {x, y}$, $\set {y, z}$ are dependent subsets
To show that $x$ is parallel to $z$ it remains to show that:
:$\set {x, z}$ is dependent
 $\set {x, z}$ is independent.
By matroid axiom $(\text I 3)$:
:$\exists w \in \set{x, z} \setminus \set y : \set{w, y} \in \mathscr I$
By definition of the doubleton:
:$\set{x, y} \in \mathscr I \lor \set{z, y} \in \mathscr I$
This contradicts the assumption that $\set {x, y}$, $\set {y, z}$ are dependent subsets.
It follows that:
:$\set {x, z}$ is dependent
== Sources ==
* Chapter $1.$ $\S 4.$ Loops and parallel elements
Category:Matroid Parallelism
Parallel Transport Determines Connection

== Theorem ==
Let $M$ be a smooth manifold with or without boundary.
Let $\nabla$ be a connection in $TM$. 
Let $X$ and $Y$ be smooth vector fields on $M$ and $X_p$, $Y_p$ their values at $p \in M$.
Let $\gamma : I \to M$ be a smooth curve such that:
:$\map \gamma 0 = p$
:$\map {\gamma'} 0 = X_p$
Let $\nabla_X Y$ be the covariant derivative of $Y$ along $X$.
Let $P^\gamma_{h_0 h_1}$ be the parallel transport map along $\gamma$.
Then:
:$\ds \forall p \in M : \valueat{\nabla_X Y}p = \lim_{h \mathop \to 0} \frac {P^\gamma_{h0} Y_{\map \gamma h} - Y_p}{h}$
== Proof ==
== Sources ==
* : $\S 4$: Connections. Parallel Transport
Category:Parallel Transports
Category:Covariant Derivatives
Category:Connections
Parallel Transport Determines Covariant Differentiation

== Theorem ==
Let $M$ be a smooth manifold with or without boundary.
Let $TM$ be the tangent bundle of $M$.
Let $\nabla$ be a connection in $TM$.
Let $I \subseteq \R$ be an open real interval.
Let $\gamma : I \to M$ be a smooth curve.
Let $P_{t_1 t_0}^\gamma$ be the parallel transport map along $\gamma$.
Let $D_t$ be the covariant derivative along $\gamma$.
Suppose that $V$ is a smooth vector field along $\gamma$.
Then:
:$\ds \forall t_0 \in I : \map {D_t V} {t_0} = \lim_{t_1 \to t_0} \frac {P^\gamma_{t_1 t_0} \map V {t_1} - \map V {t_0} }{t_1 - t_0}$
== Proof ==
== Sources ==
* : $\S 4$: Connections. Parallel Transport
Category:Parallel Transports
Category:Covariant Derivatives
Parallel Transversal Theorem

== Theorem ==
== Proof ==
Let $\triangle ABC$ be a triangle.
Let $DE$ be drawn parallel to the side $BC$.
We need to show that $BD : DA = CE : EA$.
:
Let $BE$ and $CD$ be joined.
From Triangles with Equal Base and Same Height have Equal Area the area of $\triangle BDE$ is the same as the area of $\triangle CDE$.
From Ratios of Equal Magnitudes:
:$\triangle BDE : \triangle ADE = \triangle CDE : \triangle ADE$
From Areas of Triangles and Parallelograms Proportional to Base:
:$\triangle BDE : \triangle ADE = BD : DA$
By the same reasoning:
:$\triangle CDE : \triangle ADE = CE : EA$
From Equality of Ratios is Transitive:
:$BD : DA = CE : EA$
Now let the sides $AB, AC$ of $\triangle ABC$ be cut proportionally, so that $BD : DA = CE : EA$.
Join $DE$. 
We need to show that $DE \parallel BC$.
We use the same construction as above.
From Areas of Triangles and Parallelograms Proportional to Base we have that:
:$BD : DA = \triangle BDE : \triangle ADE$
:$CE : EA = \triangle CDE : \triangle ADE$
From Equality of Ratios is Transitive:
:$\triangle BDE : \triangle ADE = \triangle CDE : \triangle ADE$
So from Magnitudes with Same Ratios are Equal, the area of $\triangle BDE$ is the same as the area of $\triangle CDE$.
But these triangles are on the same base $DE$.
So from Equal Sized Triangles on Same Base have Same Height, it follows that $DE$ and $BC$ are parallel.
== Also known as ==
== Also see ==
* Midline Theorem
== Sources ==
* : Book $\text{VI}$. Propositions
* 
* 
Category:Parallel Transversal Theorem
Category:Triangles
Category:Named Theorems
Parallelepiped cut by Plane Parallel to Opposite Planes

== Theorem ==
== Proof ==
:
Let the parallelepiped $ABCD$ be cut by the plane $FG$ which is parallel to the opposite planes $RA$ and $DH$.
It is to be demonstrated that the ratio of the base $AEFV$ to the base $EHCF$ equals the ratio of the parallelepiped $ABFU$ to the parallelepiped $EGCD$.
Let $AH$ be produced in each direction.
Let any number of straight lines $AK, KL$ equal to $AE$ be made.
Let any number of straight lines $HM, MN$ equal to $EH$ be made.
Let the parallelograms $LP, KV, HW, MS$ be completed.
Let the parallelepipeds $LQ, KR, DM, MT$ be completed.
We have that the straight lines $LK, KA, AE$ are equal.
Therefore the parallelograms $LP, KV, AF$ are equal.
From :
:the parallelograms $LX, KQ, AR$ are equal.
For the same reason, the parallelograms $EC, HW, MS$ are equal.
Therefore the parallelepipeds $HG, HI, NT$ are equal.
Further, the parallelograms $DH, MY, MT$ are equal.
Therefore, in the parallelepipeds $LQ, KR, AU$, three planes equal three planes.
But the three planes equal the three planes opposite.
Therefore the three parallelepipeds $LQ, KR, AU$ are equal to one another.
For the same reason, the three parallelepipeds $ED, DM, MT$ are equal to one another.
Therefore, whatever multiple the base $LF$ is of the base $AF$, the same multiple also is the parallelepiped $LU$ of the parallelepiped $AU$.
For the same reason, whatever multiple the base $NF$ is of the base $FH$, the same multiple also is the parallelepiped $NU$ of the parallelepiped $HU$.
And if the base $LF$ equals the base $NF$, the parallelepiped $LU$ equals the parallelepiped $NU$.
If the base $LF$ exceeds the base $NF$, the parallelepiped $LU$ exceeds the parallelepiped $NU$.
If the base $LF$ falls short of the base $NF$, the parallelepiped $LU$ falls short of the parallelepiped $NU$.
We have four magnitudes:
:the two bases $AF$ and $FH$
and:
:the two parallelepipeds $AU$ and $HU$.
Equimultiples have been taken of the base $AF$ and the parallelepiped $AU$: the base $LF$ and the parallelepiped $LU$.
and equimultiples have been taken of the base $HF$ and the parallelepiped $HU$: the base $NF$ and the parallelepiped $NU$.
It has been proved that:
: if the base $LF$ exceeds the base $FN$, the parallelepiped $LU$ exceeds the parallelepiped $NU$
:if the base $LF$ equals the base $NF$, the parallelepiped $LU$ equals the parallelepiped $NU$
:if the base $LF$ falls short of the base $NF$, the parallelepiped $LU$ falls short of the parallelepiped $NU$.
Therefore, as the base $AF$ is to the base $FH$, so the parallelepiped $AU$ is to the parallelepiped $UH$.
== Sources ==
* : Book $\text{XI}$. Propositions
Category:Parallelepipeds
Parallelepiped cut by Plane through Diagonals of Opposite Planes is Bisected

== Theorem ==
== Proof ==
:
Let the parallelepiped $AB$ be cut by the plane $CDEF$ through the diagonals $CF$ and $DE$ of opposite planes.
It is to be demonstrated that the parallelepiped $AB$ is bisected by the plane $CDEF$
From :
:$\triangle CGF = \triangle CFB$
and:
:$\triangle ADE = \triangle DEH$
while:
:the parallelogram $CA$ is equal to the parallelogram $EB$
and:
:the parallelogram $GE$ is equal to the parallelogram $CH$
as they are opposite.
Therefore from :
:the prism contained by $\triangle CGF$ and $\triangle ADE$ and the three parallelograms $GE, AC, AE$
equals:
:the prism contained by $\triangle CFB$ and $\triangle DEH$ and the three parallelograms $CH, BE, CE$.
Hence the result.
== Sources ==
* : Book $\text{XI}$. Propositions
Category:Parallelepipeds
Parallelepiped formed from Three Proportional Lines equal to Equilateral Parallelepiped with Equal Angles to it formed on Mean

== Theorem ==
== Proof ==
:
Let $A, B, C$ be three straight lines in proportion:
:$A : B = B : C$
Let a solid angle be constructed at $E$ contained by the plane angles $\angle DEG, \angle GEF, \angle FED$.
Let each of the straight lines $DE, GE, EF$ be made equal to $B$.
Let the parallelepiped $EK$ be completed.
Let the straight line $LM$ be made equal to $A$.
Let a solid angle be constructed at $L$ contained by the plane angles $\angle NLO, \angle OLM, \angle MLN$.
Let the straight line $LO$ be made equal to $B$.
Let the straight line $LN$ be made equal to $C$.
We have that:
:$A : B = B : C$
while:
:$A = LM$
:$B = LO = ED = EF$
:$C = LN$
Therefore:
:$LM = EF = DE : LN$
Thus the sides about the equal angles $\angle NLM, \angle DEF$ are reciprocally proportional.
Therefore from :
:the parallelogram $MN$ equals the parallelogram $DF$.
It follows from :
:the perpendiculars from $G$ and $O$ to the planes containing $\angle NLM$ and $\angle DEF$ are equal.
Therefore the parallelepipeds $LH$ and $EK$ are of equal height.
So from :
:$LH = EK$.
We have that:
:$LH$ is the parallelepiped formed out of $A, B, C$
:$EK$ is the parallelepiped formed out of $B$.
Hence the result.
== Sources ==
* : Book $\text{XI}$. Propositions
Category:Parallelepipeds
Parallelepipeds of Same Height have Volume Proportional to Bases

== Theorem ==
== Proof ==
:
Let $AB$ and $CD$ be parallelepipeds of the same height.
Let $AE$ be the base of $AB$ and $CF$ be the base of $AB$.
Using :
:Let the parallelogram $FH$ equal in area to $AE$ be applied to $FG$.
Let the parallelepiped $GK$ be constructed with the same height as $CD$ on the base $FH$.
Then from :
:the parallelepiped $AB$ equals the parallelepiped $GH$.
We have that the parallelepiped $CK$ is cut by the plane $DG$ which is parallel to opposite planes.
Therefore from :
:the parallelepiped $CD$ is to the parallelepiped $DH$ as the base $CF$ is to the base $FH$.
But:
:the base $FH$ equals the base $AE$
and:
:the parallelepiped $GK$ equals the parallelepiped $AB$.
Therefore, as the base $AE$ is to the base $CF$, the parallelepiped $AB$ is to the parallelepiped $CD$.
== Sources ==
* : Book $\text{XI}$. Propositions
Category:Parallelepipeds
Parallelepipeds on Same Base and Same Height whose Extremities are not on Same Lines are Equal in Volume

== Theorem ==
== Proof ==
:
Let $CM$ and $CN$ be parallelepipeds on the same base $AB$ and of the same height.
Let the endpoints of their vertical sides:
:$AF, AG, LM, LN, CD, CE, BH, BK$
be not on the same straight lines.
It is to be demonstrated that the parallelepiped $CM$ is equal to the parallelepiped $CN$.
Let $NK$ and $DH$ be produced, and meet each other at $R$.
Let $FM$ and $GE$ be produced to $P$ and $Q$.
Let $AO, LP, CQ, BR$ be joined.
We have that $CM$ and $CP$ are on the same parallelogram $ABCL$ and of the same height, and the endpoints of their vertical sides:
:$AF, AO, LM, LP, CD, CQ, BH, BR$
are on the same straight lines $FP$ and $FR$.
So from :
:the parallelepiped $CM$, of which the parallelogram $ABCL$ is the base and $FDHM$ the opposite
is equal to
:the parallelepiped $CP$, of which the parallelogram $ABCL$ is the base and $OQRP$ the opposite.
But we also have that $CP$ and $CN$ are on the same parallelogram $ABCL$ and of the same height, and the endpoints of their vertical sides:
:$AG, AO, LN, LP, CE, CQ, BK, BR$
are on the same straight lines $GQ$ and $NR$.
So from :
:the parallelepiped $CN$, of which the parallelogram $ABCL$ is the base and $ACBL$ the opposite
is equal to
:the parallelepiped $CP$, of which the parallelogram $ABCL$ is the base and $OQRP$ the opposite.
Hence the parallelepiped $CM$ equals the parallelepiped $CN$.
== Sources ==
* : Book $\text{XI}$. Propositions
Category:Parallelepipeds
Parallelepipeds on Same Base and Same Height whose Extremities are on Same Lines are Equal in Volume

== Theorem ==
== Proof ==
:
Let $CM$ and $CN$ be parallelepipeds on the same bases and of the same height.
Let the endpoints of their vertical sides:
:$AG, AF, LM, LN, CD, CE, BH, BK$
be on the same straight lines $FN$ and $DK$.
It is to be demonstrated that the parallelepiped $CM$ is equal to the parallelepiped $CN$.
We have that each of the figures $CH$ and $CK$ is a parallelogram.
Thus from :
:$CB$ equals each of the straight lines $DH$ and $EK$.
Thus $DH = EK$.
Let $EH$ be subtracted from both $DH$ and $EK$.
Therefore their remainders $DE$ and $HK$ are equal.
So from:
:
and:
:
it follows that:
:$\triangle DCE = \triangle HBK$
and from :
:the parallelogram $DG$ equals the parallelogram $HN$.
For the same reason:
:$\triangle AFG = \triangle MLN$
But the parallelogram $CF$ equals the parallelogram $BM$, as they are opposite.
Therefore:
:the prism contained by $\triangle AFG$ and $\triangle DCE$ and the three parallelograms $AD, DG, CG$
equals:
:the prism contained by $\triangle MLN$ and $\triangle HBK$ and the three parallelograms $BM, HN, BN$.
Let there be added to each the solid of which the parallelogram $AB$ is its base and $GEHM$ its opposite.
Therefore the whole parallelepiped $CM$ is equal to the whole parallelepiped $CN$.
== Sources ==
* : Book $\text{XI}$. Propositions
Category:Parallelepipeds
Parallelism implies Equal Alternate Angles

== Theorem ==
Given two infinite straight lines which are cut by a transversal, if the lines are parallel, then the alternate angles are equal.
== Proof ==
:
Let $AB$ and $CD$ be parallel infinite straight lines.
Let $EF$ be a transversal that cuts them.
 the alternate angles are not equal.
Then one of the pair $\angle AGH$ and $\angle GHD$ must be greater.
, let $\angle AGH$ be greater.
From Two Angles on Straight Line make Two Right Angles, $\angle AGH + \angle BGH$ equal two right angles.
So $\angle GHD + \angle BGH$ is less than two right angles.
By Euclid's fifth postulate, lines extended infinitely from angles less than two right angles must meet.
But the lines are parallel.
So by definition the lines do not intersect.
From this contradiction it follows that the alternate angles are be equal.
== Sources ==
* : Book $\text{I}$. Propositions
* 
* 
Category:Transversals (Geometry)
Category:Parallel Lines
Parallelism implies Equal Alternate Angles, Corresponding Angles, and Supplementary Interior Angles

== Theorem ==
Worded in more contemporary language:
=== Parallelism implies Equal Alternate Angles ===
=== Parallelism implies Equal Corresponding Angles ===
=== Parallelism implies Supplementary Interior Angles ===
== Sources ==
* : Book $\text{I}$. Propositions
* : Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.6$
Category:Transversals (Geometry)
Category:Parallel Lines
Category:Supplementary Angles
Parallelism implies Equal Corresponding Angles

== Theorem ==
Given two infinite straight lines which are cut by a transversal, if the lines are parallel, then the corresponding angles are equal.
== Proof ==
:
Let $AB$ and $CD$ be parallel infinite straight lines.
Let $EF$ be a transversal that cuts them.
From Parallelism implies Equal Alternate Angles:
:$\angle AGH = \angle DHG$
By the Vertical Angle Theorem:
:$\angle EGB = \angle AGH = \angle DHG$
== Sources ==
* : Book $\text{I}$. Propositions
* 
* 
Category:Transversals (Geometry)
Category:Parallel Lines
Parallelism implies Supplementary Interior Angles

== Theorem ==
Given two infinite straight lines which are cut by a transversal, if the lines are parallel, then the interior angles on the same side of the transversal are supplementary.
== Proof ==
:
Let $AB$ and $CD$ be parallel infinite straight lines.
Let $EF$ be a transversal that cuts them.
From Parallelism implies Equal Corresponding Angles and Euclid's second common notion:
:$\angle EGB + \angle BGH = \angle DHG + \angle BGH$
From Two Angles on Straight Line make Two Right Angles, $\angle EGB + \angle BGH$ equals two right angles.
So by definition, $\angle BGH$ and $\angle DHG$ are supplementary.
== Sources ==
* : Book $\text{I}$. Propositions
Category:Transversals (Geometry)
Category:Parallel Lines
Category:Supplementary Angles
Parallelism is Equivalence Relation

== Theorem ==
Let $S$ be the set of straight lines in the plane.
For $l_1, l_2 \in S$, let $l_1 \parallel l_2$ denote that $l_1$ is parallel to $l_2$.
Then $\parallel$ is an equivalence relation on $S$.
== Proof ==
Checking in turn each of the criteria for equivalence:
=== Reflexivity ===
=== Symmetry ===
=== Transitivity ===
$\parallel$ has been shown to be reflexive, symmetric and transitive.
Hence by definition it is an equivalence relation.
== Sources ==
* : Introduction to Set Theory: $1$. Elementary Operations on Sets
* : Appendix $\text{A}.3$: Equivalence Relations: $(3)$
* 
* 
Category:Parallel Lines
Category:Examples of Equivalence Relations
Parallelism is Equivalence Relation/Transitivity

== Theorem ==
Let $S$ be the set of straight lines in the plane.
For $l_1, l_2 \in S$, let $l_1 \parallel l_2$ denote that $l_1$ is parallel to $l_2$.
Then $\parallel$ is a transitive relation on $S$.
== Proof ==
From Parallelism is Transitive Relation:
:$l_1 \parallel l_2$ and $l_2 \parallel l_3$ implies $l_1 \parallel l_3$.
Thus $\parallel$ is seen to be transitive.
== Sources ==
* : Appendix $\text{A}.2$: Cartesian Products and Relations: Problem Set $\text{A}.2$: $7 \ \text{(a)}$
Category:Parallel Lines
Category:Examples of Transitive Relations
Parallelism is Reflexive Relation

== Theorem ==
Let $S$ be the set of straight lines in the plane.
For $l_1, l_2 \in S$, let $l_1 \parallel l_2$ denote that $l_1$ is parallel to $l_2$.
Then $\parallel$ is a reflexive relation on $S$.
== Proof ==
By definition of parallel lines, the contemporary definition is for a straight line to be declared parallel to itself.
Hence for a straight line $l$:
:$l \parallel l$
Thus $\parallel$ is seen to be reflexive.
== Also see ==
* Parallelism is Symmetric Relation
* Parallelism is Transitive Relation
* Parallelism is Equivalence Relation
== Sources ==
* : Introduction to Set Theory: $1$. Elementary Operations on Sets
* : Appendix $\text{A}.2$: Cartesian Products and Relations: Problem Set $\text{A}.2$: $7 \ \text{(a)}$
Category:Parallel Lines
Category:Examples of Reflexive Relations
Parallelism is Symmetric Relation

== Theorem ==
Let $S$ be the set of straight lines in the plane.
For $l_1, l_2 \in S$, let $l_1 \parallel l_2$ denote that $l_1$ is parallel to $l_2$.
Then $\parallel$ is a symmetric relation on $S$.
== Proof ==
Let $l_1 \parallel l_2$.
By definition of parallel lines, $l_1$ does not meet $l_2$ when produced indefinitely.
Hence $l_2$ similarly does not meet $l_1$ when produced indefinitely.
That is:
:$l_2 \parallel l_1$
Thus $\parallel$ is seen to be symmetric.
== Also see ==
* Parallelism is Reflexive Relation
* Parallelism is Transitive Relation
* Parallelism is Equivalence Relation
== Sources ==
* : Introduction to Set Theory: $1$. Elementary Operations on Sets
* : Appendix $\text{A}.2$: Cartesian Products and Relations: Problem Set $\text{A}.2$: $7 \ \text{(a)}$
Category:Parallel Lines
Category:Examples of Symmetric Relations
Parallelism is Transitive Relation

== Theorem ==
Parallelism between straight lines is a transitive relation.
== Proof ==
:
Let the straight lines $AB$ and $CD$ both be parallel to the straight line $EF$.
Let the straight line $GK$ be a transversal that cuts the parallel lines $AB$ and $EF$.
By Parallelism implies Equal Alternate Angles:
:$\angle AGK = \angle GHF$
By Playfair's Axiom, there is only one line that passes through $H$ that is parallel to $CD$ (namely $EF$).
Therefore the transversal $GK$ cannot be parallel to $CD$.
Hence the two lines must therefore intersect.
The straight line $GK$ also cuts the parallel lines $EF$ and $CD$.
So from Parallelism implies Equal Corresponding Angles:
:$\angle GHF = \angle GKD$.
Thus $\angle AGK = \angle GKD$.
So from Equal Alternate Angles implies Parallel Lines:
:$AB \parallel CD$
== Also see ==
* Parallelism is Reflexive Relation
* Parallelism is Symmetric Relation
* Parallelism is Equivalence Relation
== Sources ==
* : Book $\text{I}$. Propositions
* : Introduction to Set Theory: $1$. Elementary Operations on Sets
* : Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.7$
Category:Parallel Lines
Category:Examples of Transitive Relations
Parallelogram Law

== Theorem ==
Let $\mathbf a$ and $\mathbf b$ be vector quantities.
Consider a parallelogram, two of whose adjacent sides represent $\mathbf a$ and $\mathbf b$ (in magnitude and direction).
:
Then the diagonal of the parallelogram through that common point represents the magnitude and direction of $\mathbf a + \mathbf b$, the sum of $\mathbf a$ and $\mathbf b$.
== Proof ==
== Examples ==
== Sources ==
* : Chapter $\text I$. Addition and Subtraction of Vectors. Centroids: Addition and subtraction of vectors: $4$. Component and Resultant
* : Chapter $\text I$: Definitions. Elements of Vector Algebra: $3$. Addition and Subtraction of Vectors
* : Chapter $\text I$: Vector Algebra: $\S 1$.
* : Chapter $1$: Vector Algebra: $1.1$ Definition of a Vector
* : $\S 1$: $n$-Dimensional vector spaces
* : Chapter $\text {V}$: Vector Spaces: $\S 26$. Vector Spaces and Modules
* : $\text {I}$: Motion, Sound and Heat: Chapter $4$: Gravitation: Combination of Forces
* : $\S 22$: Fundamental Definitions: $3.$
* : Chapter $7$: Vector Spaces: $\S 32$. Definition of a Vector Space
* : Chapter $1$ Vector Analysis $1.1$ Definitions, Elementary Approach: $(1.2)$
* : $1$: Complex Numbers: Vector Interpretation of Complex Numbers
* 
* 
* 
* 
* 
* 
* 
* 
* 
Category:Parallelogram Law
Category:Vector Algebra
Category:Applied Mathematics
Parallelogram Law/Examples/3 Weights from Pulleys

== Example of Use of Parallelogram Law ==
Let $3$ bodies with mass be suspended by cords from pulleys like so:
:
The bodies will arrange themselves into equilibrium when the vector corresponding to the weight $\mathbf F_3$ of the middle body is equal and opposite the vector corresponding to the vector sum of the weights $\mathbf F_1$ and $\mathbf F_2$ according to the Parallelogram Law.
== Sources ==
* : Chapter $1$ Vector Analysis $1.1$ Definitions, Elementary Approach
Category:Parallelogram Law
Parallelogram Law (Inner Product Space)

== Theorem ==
Let $\struct {V, \innerprod \cdot \cdot}$ be an inner product space.
Let $\norm \cdot$ be the inner product norm of $\struct {V, \innerprod \cdot \cdot}$. 
Let $f, g \in V$ be arbitrary.
Then:
:$\norm {f + g}^2 + \norm {f - g}^2 = 2 \paren {\norm f^2 + \norm g^2}$
== Proof ==
== Also see ==
* Norm satisfying Parallelogram Law induced by Inner Product establishes a converse result: if a norm satisfies the parallelogram law, then it is induced by an inner product.
== Sources ==
* 
* : $\text{I}$ Hilbert Spaces: $\S 2.$ Orthogonality: $2.3$ Parallelogram Law
* $8.3$: Properties of the Induced Norms
Category:Inner Product Spaces