avewright/proofwiki-math huggingface dataset:
Theorem: Steiner-Lehmus Theorem

Let $ABC$ be a triangle.
Denote the lengths of the angle bisectors through the vertices $A$ and $B$ by $\omega_\alpha$ and $\omega_\beta$.
Let $\omega_\alpha = \omega_\beta$.
Then $ABC$ is an isosceles triangle.
:250px

Proof:
Let $a$, $b$, and $c$ be the sides opposite $A$, $B$ and $C$ respectively.
By Length of Angle Bisector, $\omega_\alpha, \omega_\beta$ are given by:
:$\omega_\alpha^2 = \dfrac {b c} {\paren {b + c}^2} \paren {\paren {b + c}^2 - a^2}$
:$\omega_\beta^2 = \dfrac {a c} {\paren {a + c}^2} \paren {\paren {a + c}^2 - b^2}$
Equating $\omega_\alpha^2$ with $\omega_\beta^2$:
{{begin-eqn}}
{{eqn | l = \dfrac {b c} {\paren {b + c}^2} \paren {\paren {b + c}^2 - a^2}
      | r = \dfrac {a c} {\paren {a + c}^2} \paren {\paren {a + c}^2 - b^2}
}}
{{eqn | ll= \leadsto
      | l = b c \paren {a + c}^2 \paren {\paren {b + c}^2 - a^2}
      | r = a c \paren {b + c}^2 \paren {\paren {a + c}^2 - b^2}
}}
{{eqn | ll= \leadsto
      | l = b c \paren {a + c}^2 \paren {b + c + a} \paren {b + c - a}
      | r = a c \paren {b + c}^2 \paren {a + c + b} \paren {a + c - b}
}}
{{eqn | ll= \leadsto
      | l = b \paren {a + c}^2 \paren {b + c - a}
      | r = a \paren {b + c}^2 \paren {a + c - b}
      | c = as $a + b + c > 0$
}}
{{eqn | ll= \leadsto
      | l = a^2 b^2 + 2 a b^2 c + b^2 c^2 + a^2 b c + 2 a b c^2 + b c^3 - a^3 b - 2 a^2 b c - a b c^2
      | r = a^2 b^2 + 2 a^2 b c + a^2 c^2 + a b^2 c + 2 a b c^2 + a c^3 - a b^3 - 2 a b^2 c - a b c^2
}}
{{eqn | ll= \leadsto
      | l = 2 a b^2 c + b^2 c^2 - a^2 b c + b c^3 - a^3 b
      | r = 2 a^2 b c + a^2 c^2 - a b^2 c + a c^3 - a b^3
}}
{{eqn | ll= \leadsto
      | l = 3 a b^2 c - 3 a^2 b c + b^2 c^2 + b c^3 - a^3 b - a^2 c^2 - a c^3 + a b^3
      | r = 0
}}
{{eqn | ll= \leadsto
      | l = 3 a b c \paren {b - a} + c^2 \paren {b^2 - a^2} + c^3 \paren {b - a} + a b \paren {b^2 - a^2}
      | r = 0
}}
{{eqn | ll= \leadsto
      | l = \paren {b - a} \paren {3 a b c + a b \paren {a + b} + c^3 + c^2 \paren {a + b} }
      | r = 0
}}
{{eqn | ll= \leadsto
      | l = b - a
      | r = 0
      | c = as $a, b, c > 0$
}}
{{eqn | ll= \leadsto
      | l = a
      | r = b
}}
{{end-eqn}}
Therefore $ABC$ is an isosceles triangle.
{{qed}}
Theorem: Steiner-Lehmus Theorem

Let $ABC$ be a triangle.
Denote the lengths of the angle bisectors through the vertices $A$ and $B$ by $\omega_\alpha$ and $\omega_\beta$.
Let $\omega_\alpha = \omega_\beta$.
Then $ABC$ is an isosceles triangle.
:250px

Proof:
Let $a$, $b$, and $c$ be the sides opposite $A$, $B$ and $C$ respectively.
By Length of Angle Bisector, $\omega_\alpha, \omega_\beta$ are given by:
{{begin-eqn}}
{{eqn | l = \omega_\alpha^2
      | r = b c \paren {1 - \dfrac {a^2} {\paren {b + c}^2} }
}}
{{eqn | l = \omega_\beta^2
      | r = a c \paren {1 - \dfrac {b^2} {\paren {a + c}^2} }
}}
{{end-eqn}}
Equating $\omega_\alpha^2$ with $\omega_\beta^2$ yields:
{{begin-eqn}}
{{eqn | l = b c - \dfrac {b c a^2} {\paren {b + c}^2}
      | r = a c - \dfrac {a c b^2} {\paren {a + c}^2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = c \paren {b - a}
      | r = \dfrac {b c a^2} {\paren {b + c}^2} - \dfrac {a c b^2} {\paren {a + c}^2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = c \paren {b - a} \paren {b + c}^2 \paren {a + c}^2
      | r = b c a^2 \paren {a + c}^2 - a c b^2 \paren {b + c}^2
      | c = 
}}
{{eqn | n = 1
      | r = c a \paren {a \paren {a + c}^2 - b\paren {b + c}^2}
      | c = 
}}
{{end-eqn}}
Substituting $b$ for $a$ in $(1)$ proves that $b = a$ is a solution for $(1)$.
We still have to show that $b = a$ is the only solution for $(1)$.
{{AimForCont}} $b \ne a$.
{{WLOG}}, suppose $b > a$.
Because $c \paren {b - a} \paren {b + c}^2 \paren {a + c}^2 > 0$, the {{LHS}} of $(1)$ is positive.
We have {{hypothesis}} that:
:$a$, $b$, and $c$ are positive real numbers
and {{hypothesis}}:
:$a < b$
By Real Number Ordering is Compatible with Multiplication and since $a > 0$:
:$a^2 < a \cdot b$
For the same reason and since $b > 0$:
:$a \cdot b < b^2$
By the Transitive Law:
:$a^2 < b^2$
Then:
{{begin-eqn}}
{{eqn | l = a^3
      | o = <
      | r = b^3
      | c = {{mutatis}}
}}
{{eqn | ll= \leadsto
      | l = a^2 - b^2
      | o = <
      | r = 0
      | c = 
}}
{{eqn | l = a^3 - b^3
      | o = <
      | r = 0
      | c = 
}}
{{end-eqn}}
Since $c a > 0$, the {{RHS}} of $(1)$ will be negative {{iff}}:
:$a \paren {a + c}^2 - b \paren {b + c}^2 < 0$
We rearrange as follows:
{{begin-eqn}}
{{eqn | l = a \paren {a + c}^2 - b \paren {b + c}^2
      | r = a \paren {a^2 + 2 a c + c^2} - b \paren {b^2 + 2 b c + c^2}
}}
{{eqn | r = a^3 + 2 a^2 c + a c^2 - b^3 - 2 b^2 c - b c^2
}}
{{eqn | n = 2
      | r = \paren {a^3 - b^3} + 2 c \paren {a^2 - b^2} + c^2 \paren {a - b}
}}
{{end-eqn}}
From above, each term in parenthesis in $(2)$ is negative.
As $c > 0$ we have:
{{begin-eqn}}
{{eqn | l = 2 c
      | o = >
      | r = 0
}}
{{eqn | l = c^2
      | o = >
      | r = 0 
}}
{{end-eqn}}
Hence each term in $(2)$ is negative.
Hence their sum is also negative.
Therefore, the {{RHS}} of $(1)$ is negative.
Hence we have reached a contradiction.
{{qed|lemma}}
Similarly, we also get a contradiction by assuming $b < a$.
This assumption yields a negative {{LHS}} and a positive {{RHS}} for $(1)$.
Therefore, by Proof by Contradiction, $b = a$ is the only solution for $(1)$.
Therefore $ABC$ is an isosceles triangle.
{{qed}}
Theorem: Steiner-Lehmus Theorem

Let $ABC$ be a triangle.
Denote the lengths of the angle bisectors through the vertices $A$ and $B$ by $\omega_\alpha$ and $\omega_\beta$.
Let $\omega_\alpha = \omega_\beta$.
Then $ABC$ is an isosceles triangle.
:250px

Proof:
=== {{Lemma|Steiner-Lehmus Theorem|1}} ===
{{:Steiner-Lehmus Theorem/Lemma 1}}{{qed|lemma}}
=== {{Lemma|Steiner-Lehmus Theorem|2}} ===
{{:Steiner-Lehmus Theorem/Lemma 2}}{{qed|lemma}}
400px
Let $\triangle ABC$ be a triangle.
Let $\angle ABC$ be bisected by $BM$
Let $\angle ACB$ be bisected by $CN$.
Let $BM = CN$.
Suppose $\angle ABC < \angle ACB$.
By {{Lemma|Steiner-Lehmus Theorem|2}}:
:$CN < BM$
This is a contradiction.
{{qed|lemma}}
Suppose $\angle ABC < \angle ACB$.
By {{Lemma|Steiner-Lehmus Theorem|2}}:
:$CN > BM$
This is a contradiction.
{{qed|lemma}}
Since $\angle ABC$ can be neither less than nor greater than $\angle ACB$, $\angle ABC = \angle ACB$.
By Triangle with Two Equal Angles is Isosceles:
:$\triangle ABC$ is an isosceles triangle.
The result follows.
{{qed}}
Theorem: Triangle Right-Angle-Hypotenuse-Side Congruence

If two right triangles have:
:their hypotenuses equal
:another of their respective sides equal
they will also have:
:their third sides equal
:the remaining two angles equal to their respective remaining angles.

Proof:
Let $\triangle ABC$ and $\triangle DEF$ be two triangles having sides $AB = DE$ and $AC = DF$, and with $\angle ABC = \angle DEF = 90^\circ$.
By Pythagoras' Theorem:
:$BC = \sqrt {AB^2 + AC^2}$
and:
:$EF = \sqrt {DE^2 + DF^2}$
:$\therefore BC = \sqrt {AB^2 + AC^2} = \sqrt {DE^2 + DF^2} = EF$
The part that the remaining two angles are equal to their respective remaining angles follows from Triangle Side-Side-Side Congruence.
{{qed}}
Theorem: Triangle Right-Angle-Hypotenuse-Side Congruence

If two right triangles have:
:their hypotenuses equal
:another of their respective sides equal
they will also have:
:their third sides equal
:the remaining two angles equal to their respective remaining angles.

Proof:
350px
Let $\triangle ADB$ and $\triangle ADC$ both be right triangles.
Let them have equal hypotenuse and one leg ($AD$) equal.
{{hypothesis}}:
:$AB = AC$
:$\angle ADB = \angle ADC = \ $ one right angle
:$AD$ is shared
So the two triangles can be drawn as shown with $BD$ and $DC$ joined at $D$.
By addition:
:$\angle BDA + \angle CDA = \angle BDC = \ $ two right angles
By Two Angles making Two Right Angles make Straight Line:
:$BDC$ is one straight line
:the points $BDC$ are collinear
So:
:$\triangle ABC$ is a triangle.
By definition of isosceles triangle:
:$\triangle ABC$ is isosceles.
Since $AD \perp BDC$, by definition of perpendicular bisector:
:$BD = DC$
By Triangle Side-Angle-Side Congruence:
:$\triangle ABD \cong \triangle ACD$
{{qed}}
Theorem: Internal Angles of Regular Polygon

The size $A$ of each internal angle of a regular $n$-gon is given by:
:$A = \dfrac {\paren {n - 2} 180 \degrees} n$

Proof:
From Sum of Internal Angles of Polygon, we have that the sum $S$ of all internal angles of a $n$-gon is:
:$S = \paren {n - 2} 180 \degrees$
From the definition of a regular polygon, all the internal angles of a regular polygon are equal.
Therefore, the size $A$ of each internal angle of a regular polygon with $n$ sides is:
:$A = \dfrac {\paren {n - 2} 180 \degrees} n$
{{qed}}
Theorem: Sum of Internal Angles of Polygon

The sum $S$ of all internal angles of a polygon with $n$ sides is given by the formula $S = \paren {n - 2} 180 \degrees$.

Proof:
The Polygon Triangulation Theorem shows that there exists a triangulation of the polygon that consists of $n - 2$ triangles.
The sides of these triangles are sides and chords of the polygon, where the chords lie completely in the interior of $P$.
Hence the vertices of the triangles are vertices of the polygon.
Sum of Angles of Triangle equals Two Right Angles shows that the sum of the internal angles of a triangle is $180 \degrees$.
As the triangulation covers the polygon, the sum of the internal angles of the vertices of the triangles in the triangulation is equal to $S$.
So:
:$S = \paren {n - 2} 180 \degrees$
{{qed}}
Theorem: Sum of Internal Angles of Polygon

The sum $S$ of all internal angles of a polygon with $n$ sides is given by the formula $S = \paren {n - 2} 180 \degrees$.

Proof:
This proof assumes that the polygon is convex.
Name a vertex as $A_1$, go clockwise and name the vertices as $A_2, A_3, \ldots, A_n$.
By joining $A_1$ to every vertex except $A_2$ and $A_n$, one can form $\paren {n - 2}$ triangles in a fan triangulation of the convex polygon.
From Sum of Angles of Triangle equals Two Right Angles, the sum of the internal angles of a  triangle is $180 \degrees$.
Therefore, the sum of internal angles of a polygon with $n$ sides is $\paren {n - 2} 180 \degrees$.
{{qed}}
Theorem: Five Platonic Solids

There exist exactly five platonic solids:
{{begin-itemize}}
{{item|(1):|the regular tetrahedron}}
{{item|(2):|the cube}}
{{item|(3):|the regular octahedron}}
{{item|(4):|the regular dodecahedron}}
{{item|(5):|the regular icosahedron.}}
{{end-itemize}}
{{:Euclid:Proposition/XIII/18/Endnote}}

Proof:
Consider a convex regular polyhedron $P$.
Let $m$ be the number of sides of each of the regular polygons that form the faces of $P$.
Let $n$ be the number of those polygons which meet at each vertex of $P$.
From Internal Angles of Regular Polygon, the internal angles of each face of $P$ measure $180^\circ - \dfrac {360^\circ} m$.
The sum of the internal angles must be less than $360^\circ$.
So:
{{begin-eqn}}
{{eqn | l = n \paren {180^\circ - \dfrac {360^\circ} m}
      | o = <
      | r = 360^\circ
      | c = 
}}
{{eqn | ll= \leadsto
      | l = n \paren {1 - \dfrac 2 m}
      | o = <
      | r = 2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = n \paren {m - 2}
      | o = <
      | r = 2m
      | c = 
}}
{{eqn | ll= \leadsto
      | l = n \paren {m - 2} - 2 \paren {m - 2}
      | o = <
      | r = 2m - 2 \paren {m - 2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {m - 2} \paren {n - 2}
      | o = <
      | r = 4
      | c = 
}}
{{end-eqn}}
But $m$ and $n$ are both greater than $2$.
So:
:if $m = 3$, $n$ can only be $3$, $4$ or $5$
:if $m = 4$, $n$ can only be $3$
:if $m = 5$, $n$ can only be $3$
and $m$ cannot be greater than $3$.
There are $5$ possibilities in all.
Therefore all platonic solids have been accounted for.
{{qed}}
Theorem: Space in which All Convergent Sequences have Unique Limit not necessarily Hausdorff

Let $T = \struct{S, \tau}$ be a topological space.
Let $T$ be such that all convergent sequences have a unique limit.
Then it is not necessarily the case that $T$ is a Hausdorff space.

Proof:
Let $T = \struct{\R, \tau}$ be the set of real numbers $\R$ with the countable complement topology.
From Countable Complement Space is not $T_2$, $T$ is not a Hausdorff space.
Suppose $\sequence{x_n}$ is a sequence in $\R$ which converges to $x$.
Then $C = \set{x_n: x_n \ne x}$ is closed in $T$ because it is countable.
So $X \setminus C$ is a neighborhood of $x$.
This means there is some $N \in \N$ such that:
:$\forall n > N: x_n \in X \setminus C$
That is, $x_n = x$ for large $n$.
This means that if $x_n \to y$ then $y = x$, proving limits in $T$ are unique.
{{qed}}
Category:Countable Complement Topologies
Category:Hausdorff Spaces
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Theorem: Geometric Sequence with Coprime Extremes is in Lowest Terms

Let $G_n = \sequence {a_0, a_1, \ldots, a_n}$ be a geometric sequence of integers.
Let:
:$a_0 \perp a_n$
where $\perp$ denotes coprimality.
Then $G_n$ is in its lowest terms.

Proof:
Let $G_n = \sequence {a_0, a_1, \ldots, a_n}$ be natural numbers in geometric sequence such that $a_0 \perp a_n$.
{{AimForCont}} there were to exist another set of natural numbers in geometric sequence:
:$G\,'_n = \sequence {b_0, b_1, \cdots, b_n}$
with the same common ratio where:
:$\forall k \in \N_{\le n}: a_k > b_k$
From {{EuclidPropLink|book = VII|prop = 14|title = Proportion of Numbers is Transitive}}:
:$a_0 : a_n = b_0 : b_n$
But by hypothesis:
:$a_0 \perp a_n$
and so from:
:{{EuclidPropLink|prop = 21|title = Coprime Numbers form Fraction in Lowest Terms}}
and:
:{{EuclidPropLink|prop = 20|title = Ratios of Fractions in Lowest Terms}}
it follows that:
:$a_0 \divides b_0$
However, this contradicts the assumption that $b_0 < a_0$.
Therefore $a_0, a_1, \cdots, a_n$ are the least of those with the same common ratio.
{{qed}}
Theorem: Geometric Sequence with Coprime Extremes is in Lowest Terms

Let $G_n = \sequence {a_0, a_1, \ldots, a_n}$ be a geometric sequence of integers.
Let:
:$a_0 \perp a_n$
where $\perp$ denotes coprimality.
Then $G_n$ is in its lowest terms.

Proof:
Let $G_n = \sequence {a_0, a_1, \ldots, a_n}$ be natural numbers in geometric sequence such that $a_0 \perp a_n$.
{{AimForCont}} $G\,'_n = \sequence {b_0, b_1, \cdots, b_n}$ be another set of natural numbers in geometric sequence with the same common ratio where:
:$\forall k \in \N_{\le n}: a_k > b_k$
By definition of geometric sequence:
:$a_0 = r^n a_n$
:$b_0 = r^n b_n$
Hence:
:$\dfrac {a_0} {a_n} = \dfrac {b_0} {b_n}$
By hypothesis:
: $a_0 \perp a_n$
Thus $\dfrac {a_0} {a_n}$ is in canonical form.
From Canonical Form of Rational Number is Unique it follows that $a_0$ and $a_n$ are the only two integers fulfilling these conditions.
Thus:
:$\forall p, q: \dfrac p q = \dfrac {a_0} {a_n}: a_0 \le p, a_n \le q$
But it was supposed that:
:$b_0 < a_0$
:$b_n < a_n$
From this contradiction it follows that there can be no such $b_0, \ldots, b_n$.
Hence the result.
{{qed}}
Theorem: Measurements of Common Angles/Straight Angle

The measurement of a straight angle is $180 \degrees$ or $\pi$ radians.

Proof:
From Measurement of Full Angle, a full rotation is defined to be $360 \degrees$ or $2 \pi$ radians.
Since lines are straight, it therefore follows that from any point on a line, the angle between one side of the line and the other is one half of a full rotation.
Therefore, the measurement of a straight angle is:
:$\dfrac {360 \degrees} 2 = 180 \degrees$
or:
:$\dfrac {2 \pi} 2 = \pi$
{{qed}}
Theorem: Measurements of Common Angles/Right Angle

The measurement of a '''right angle''' is $\dfrac {180 \degrees} 2 = 90 \degrees$ or $\dfrac \pi 2$.

Proof:
A right angle is equal to one half of a straight angle.
From Measurement of Straight Angle it follows that the measurement of a right angle is $\dfrac {180 \degrees} 2 = 90 \degrees$ or $\dfrac \pi 2$.
{{qed}}
Theorem: Measurements of Common Angles/Full Angle

A full angle is equal to $360 \degrees$ or $2 \pi$ radians.

Proof:
By definition, $1$ '''radian''' is the angle which sweeps out an arc on a circle whose length is the radius $r$ of the circle.
From Perimeter of Circle, the length of the circumference of a circle of radius $r$ is equal to $2 \pi r$.
Therefore, $1$ radian sweeps out $\dfrac 1 {2 \pi}$ of a circle.
It follows that $2 \pi$ radians sweeps out the entire circle, or one full angle.
By definition of degree of angle, a full rotation is $360 \degrees$.
Therefore, a full angle is $360 \degrees$ or $2 \pi$.
{{qed}}
Theorem: Measurements of Common Angles/Acute Angle

An acute angle measures $\theta$, where:
:$0 \degrees < \theta < 90 \degrees$
or:
:$0 < \theta < \dfrac \pi 2$

Proof:
An acute angle is defined to be an angle whose measure is between that of a zero angle and a right angle.
A zero angle measures $0$ by definition, and a right angle measures $90 \degrees$ or $\dfrac \pi 2$.
Hence the result.
{{qed}}
Theorem: Measurements of Common Angles/Obtuse Angle

An '''obtuse angle''' measures $\theta$, where:
:$90 \degrees < \theta < 180 \degrees$
or:
:$\dfrac \pi 2 < \theta < \pi$

Proof:
An obtuse angle is defined to be an angle whose measure is between that of a right angle and a straight angle.
A right angle measures $90 \degrees$ or $\dfrac \pi 2$ and a straight angle measures $180 \degrees$ or $\pi$.
Hence the result.
{{qed}}
Theorem: Measurements of Common Angles/Reflex Angle

A reflex angle measures $\theta$, where:
:$180 \degrees < \theta < 360 \degrees$
or:
:$\pi < \theta < 2 \pi$

Proof:
A reflex angle is defined to be an angle whose measure is between that of a straight angle and a full angle.
A straight angle measures $180 \degrees$ or $\pi$ and a full angle measures $360 \degrees$ or $2 \pi$.
Hence the result.
{{qed}}
Theorem: Supplementary Interior Angles implies Parallel Lines

Given two infinite straight lines which are cut by a transversal, if the interior angles on the same side of the transversal are supplementary, then the lines are parallel.

Proof:
:200px
Let $AB$ and $CD$ be infinite straight lines.
Let $EF$ be a transversal that cuts them.
Let at least one pair of interior angles on the same side of the transversal be supplementary.
{{WLOG}}, let those interior angles be $\angle BGH$ and $\angle DHG$.
So, by definition, $\angle DHG + \angle BGH$ equals two right angles.
Also, from Two Angles on Straight Line make Two Right Angles, $\angle AGH + \angle BGH$ equals two right angles.
Then from Euclid's first and third common notion and Euclid's fourth postulate:
:$\angle AGH = \angle DHG$
Finally, by Equal Alternate Angles implies Parallel Lines:
:$AB \parallel CD$
{{qed}}
{{Euclid Note|28|I|{{EuclidNoteConverse|prop = 29|title = Parallelism implies Supplementary Interior Angles|part = third}}|part = second}}
Theorem: Equal Corresponding Angles implies Parallel Lines

Given two infinite straight lines which are cut by a transversal, if the corresponding angles are equal, then the lines are parallel.

Proof:
:200px
Let $AB$ and $CD$ be infinite straight lines.
Let $EF$ be a transversal that cuts them.
Let at least one pair of corresponding angles be equal.
{{WLOG}}, let $\angle EGB = \angle GHD$.
By the Vertical Angle Theorem:
:$\angle GHD = \angle EGB = \angle AGH$
Thus by Equal Alternate Angles implies Parallel Lines:
:$AB \parallel CD$
{{qed}}
{{Euclid Note|28|I|{{EuclidNoteConverse|prop = 29|title = Parallelism implies Equal Corresponding Angles|part = second}}|part = first}}
Theorem: Equivalence of Definitions of Tangent Vector

Let $M$ be a smooth manifold.
Let $m \in M$ be a point. 
Let $V$ be an open neighborhood of $m$. 
Let $\map {C^\infty} {V, \R}$ be defined as the set of all smooth mappings $f: V \to \R$.
{{TFAE|def = Tangent Vector}}

Proof:
=== Definition 2 implies Definition 1 ===
Let $\lambda \in \R$ and $f, g \in \map {C^\infty} {V, \R}$. 
{{begin-eqn}}
{{eqn | l = \map {X_m} {f + \lambda g}
      | r = \map {\frac \d {\d \tau} {\restriction_0} } {\map {\paren {f + \lambda g} \circ \gamma} \tau}
      | c = Definition 2
}}
{{eqn | r = \map {\frac \d {\d \tau} {\restriction_0} } {\map {f \circ \gamma} \tau} + \lambda \map {\frac \d {\d \tau} {\restriction_0} } {\map {g \circ \gamma} \tau} 
}}
{{eqn | r = \map {X_m} f + \lambda \map {X_m} g 
}}
{{end-eqn}}
Thus $X_m$ is linear. 
{{begin-eqn}}
{{eqn | l = \map {X_m} {f g}
      | r = \frac \d {\d \tau} {\restriction_0} \map {\paren {f g} \circ \gamma} \tau
      | c = Definition 2
}}
{{eqn | r = \frac \d {\d \tau} {\restriction_0} \map {f \circ \gamma} \tau \, \map {g \circ \gamma} 0 + \map {f \circ \gamma} 0 \, \frac \d {\d \tau} {\restriction_0} \map {g \circ \gamma} \tau
      | c = Product Rule for Derivatives
}}
{{eqn | r = \map {X_m} f \map g m + \map f m \map {X_m} g
      | c = Definition 2, $\map \gamma 0 = m$
}}
{{end-eqn}}
Hence $X_m$ satisfies the ''Leibniz law''. 
Thus $X_m$ satisfies Definition 1.
{{qed|lemma}}
Theorem: Parallelism implies Equal Alternate Angles

Given two infinite straight lines which are cut by a transversal, if the lines are parallel, then the alternate angles are equal.

Proof:
:200px
Let $AB$ and $CD$ be parallel infinite straight lines.
Let $EF$ be a transversal that cuts them.
{{AimForCont}} the alternate angles are not equal.
Then one of the pair $\angle AGH$ and $\angle GHD$ must be greater.
{{WLOG}}, 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.
{{qed}}
{{Euclid Note|29|I|{{EuclidNoteConverse|prop = 27|title = Equal Alternate Angles implies Parallel Lines}}<br><br>This is the first proposition of {{BookLink|The Elements|Euclid}} to make use of Euclid's fifth postulate.|part = first}}
Theorem: Parallelism implies Equal Corresponding Angles

Given two infinite straight lines which are cut by a transversal, if the lines are parallel, then the corresponding angles are equal.

Proof:
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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:
:$\angle EGB = \angle AGH = \angle DHG$
{{qed}}
{{Euclid Note|29|I|{{EuclidNoteConverse|prop = 28|title = Equal Corresponding Angles implies Parallel Lines|part = first}}|part = second}}
Theorem: Proportion is Reflexive

Proportion is a reflexive relation.
That is, every real variable is proportional to itself:
:$\forall x \in \R: x \propto x$

Proof:
Let $x$ be arbitrary.
Then $x = 1 \times x$.
The result follows from the definition of reflexivity and proportion.
{{qed}}
Category:Proportion
Category:Examples of Reflexive Relations
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Theorem: Proportion is Symmetric

Proportion is a symmetric relation.
That is:
:$\forall x, y \in \R: x \propto y \implies y \propto x$

Proof:
Let $x, y$ be arbitrary.
Let $x$ be proportional to $y$:
:$x \propto y$
Then by definition:
{{begin-eqn}}
{{eqn | q = \exists k \ne 0
      | l = x
      | r = k \times y
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y
      | r = k^{-1} \times x
      | c = 
}}
{{end-eqn}}
The result follows from the definition of symmetry and proportion.
{{qed}}
Category:Proportion
Category:Examples of Symmetric Relations
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Theorem: Proportion is Transitive

Proportion is a transitive relation.
That is:
:$\forall x, y, z \in \R: x \propto y \land y \propto z \implies x \propto z$

Proof:
Let $x, y, z$ be arbitrary.
Let $x$ be proportional to $y$ and $y$ to $z$:
:$x \propto y \land y \propto z$
Then by definition:
:$\exists j, k \ne 0: x = j \times y \land y = k \times z$
Substituting $k \times z$ for $y$:
:$x = \paren {j \times k} \times z$
so $j \times k$ is the desired constant of proportion.
The result follows from the definition of transitivity and proportion.
{{qed}}
Category:Proportion
Category:Examples of Transitive Relations
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