avewright/proofwiki-math huggingface dataset:
Theorem: Finite Connected Simple Graph is Tree iff Size is One Less than Order/Sufficient Condition

Let $T$ be a finite connected simple graph of order $n$.
Let the size of $T$ be $n - 1$.
Then $T$ is a (finite) tree.

Proof:
Let $T$ is a connected simple graph of order $n$ with size $n - 1$.
From Finite Connected Simple Graph with Size One Less than Order has no Circuits:
:$T$ has no circuits.
Hence $T$ is a tree by definition.
{{qed}}
Theorem: 1+1 = 2

Define $0$ as the unique element in the set $P \setminus \map s P$, where:
:$P$ is the Peano Structure
:$\map s P$ is the image of the mapping $s$ defined in Peano structure
:$\setminus$ denotes the set difference.
Then:
:$1 + 1 = 2$
where:
:$1 := \map s 0$
:$2 := \map s 1 = \map s {\map s 0}$
:$+$ denotes addition
:$=$ denotes equality
:$\map s n$ denotes the successor mapping.

Proof:
$1$ is defined {{hypothesis}} as $\map s 0$ and $2$ as $\map s {\map s 0}$.
Hence the statement to be proven becomes:
:$\map s 0 + \map s 0 = \map s {\map s 0}$
Thus:
{{begin-eqn}}
{{eqn | q = \forall m, n \in P
      | l = m + \map s n
      | r = \map s {m + n}
      | c = {{Defof|Addition in Peano Structure}}
}}
{{eqn | ll= \leadsto
      | q = \forall m \in P
      | l = m + \map s 0
      | r = \map s {m + 0}
      | c = as $0 \in P$
}}
{{eqn | n = 1
      | ll= \leadsto
      | l = \map s 0 + \map s 0
      | r = \map s {\map s 0 + 0}
      | c = as $\map s 0 \in P$
}}
{{eqn-intertext|Then:}}
{{eqn | q = \forall m \in P
      | l = m + 0
      | r = m
      | c = {{Defof|Addition in Peano Structure}}
}}
{{eqn | ll= \leadsto
      | l = \map s 0 + 0
      | r = \map s 0
      | c = as $\map s 0 \in P$
}}
{{eqn | l = \map s {\map s 0 + 0}
      | r = \map s {\map s 0}
      | c = taking the successor on both sides of $(2)$
}}
{{eqn | ll= \leadsto
      | l = \map s 0 + \map s 0
      | r = \map s {\map s 0}
      | c = substituting for $\map s {\map s 0 + 0}$ from $(1)$
}}
{{end-eqn}}
{{qed}}
Theorem: 1+1 = 2

Define $0$ as the unique element in the set $P \setminus \map s P$, where:
:$P$ is the Peano Structure
:$\map s P$ is the image of the mapping $s$ defined in Peano structure
:$\setminus$ denotes the set difference.
Then:
:$1 + 1 = 2$
where:
:$1 := \map s 0$
:$2 := \map s 1 = \map s {\map s 0}$
:$+$ denotes addition
:$=$ denotes equality
:$\map s n$ denotes the successor mapping.

Proof:
Defining $1$ as $\map s 0$ and $2$ as $\map s {\map s 0}$, the statement to be proven becomes:
:$\map s 0 + \map s 0 = \map s {\map s 0}$
By the definition of addition:
:$\forall m \in P: \forall n \in P: m + \map s n = \map s {m + n}$
Letting $m = \map s 0$ and $n = 0$:
{{begin-eqn}}
{{eqn | n = 1
      | l = \map s 0 + \map s 0
      | r = \map s {\map s 0 + 0}
}}
{{end-eqn}}
By the definition of addition:
:$\forall m: m + 0 = m$
Letting $m = \map s 0$:
:$\map s 0 + 0 = \map s 0$
Taking the successor of both sides: 
{{begin-eqn}}
{{eqn | n = 2
      | l = \map s {\map s 0 + 0}
      | r = \map s {\map s 0}
}}
{{end-eqn}}
Applying Equality is Transitive to $(1)$ and $(2)$ we have:
:$\map s 0 + \map s 0 = \map s {\map s 0}$
Hence the result.
{{qed}}
Theorem: Kruskal's Algorithm produces Minimum Spanning Tree

Kruskal's Algorithm produces a minimum spanning tree.

Proof:
Let $N = \struct {V, E, f}$ be an undirected network.
Let $H$ be a subgraph of the underlying graph $G = \struct {V, E}$ of $N$.
Let $\ds \map f H = \sum_{e \mathop \in H} \map f e$.
Let $N$ be of order $p$.
Let $T$ be a spanning tree of $N$, created using Kruskal's Algorithm.
Then from Finite Connected Simple Graph is Tree iff Size is One Less than Order, $T$ has $p - 1$ edges.
Let the edges of $T$ be ordered $e_1, e_2, \ldots, e_{p - 1}$ according to the construction of Kruskal's Algorithm.
Then:
:$\ds \map f t = \sum_{i \mathop = 1}^{p - 1} \map f {e_i}$
Let $T_0$ be a minimum spanning tree of $N$.
Suppose that $T_0$ and $T$ are not equal.
Then $T$ has one or more edges which are not in $T_0$.
Using the ordering on the edges of $T$ created above, let $e_i$ be the first edge of $T$ not in $T_0$.
By construction, $1 \le i \le p-1$.
Let $e_i$ be added to $T_0$, to obtain the graph $G_0$.
Let $e_i = u v$ where $u, v \in V$ are vertices of $N$
Then there exists a $u-v$ path $P$ in $T_0$.
Together with $e_i$ this makes a cycle $C$ in $G_0$.
Since $T$ contains no cycles, there must be an edge $e_0$ of $C$ which is not in $T$.
Thus the edge deletion $T'_0 = G_0 - e_0$ is also a spanning tree of $N$, and:
:$\map f {T'_0} = \map f {T_0} + \map f {e_i} - \map f {e_0}$
But as $T_0$ is a minimum spanning tree:
:$\map f {T_0} \le \map f {T'_0}$
So:
:$\map f {T_0} - \map f {T'_0} \le 0$
But since:
:$\map f {T_0} - \map f {T'_0} = \map f {e_0} - \map f {e_i}$
it follows that:
$\map f {e_0} \le \map f {e_i}$
However, the manner in which $T$ was constructed, $e_i$ is an edge of smallest value which can be added to $\set {e_1, e_2, \ldots, e_{i - 1} }$ without producing a cycle.
Also, if $e_0$ is added to $\set {e_1, e_2, \ldots, e_{i - 1} }$, again no cycle is produced.
Therefore $\map f {e_i} = \map f {e_0}$ and so $\map f {T'_0} = \map f {T_0}$.
Thus $T'_0$ is a minimum spanning tree which has one more edge in common with $T$ than $T_0$, that is, $e_i$.
The procedure can be repeated as many times as there are edges in $T_0$ which are not in $T$.
Finally the spanning tree $T$ is arrived at, which by the above is a minimum spanning tree.
Hence the result.
{{qed}}
Theorem: Regular Graph is Tree iff Complete Graph of Order 2

Let $G$ be a non-edgeless regular graph.
Then $G$ is a tree {{iff}} $G$ is $K_2$, the complete graph of order $2$.

Proof:
=== Necessary Condition ===
Let $G$ be a non-edgeless regular graph which is also a tree.
From Finite Tree has Leaf Nodes it follows that $G$ has at least two vertices of degree $1$.
Therefore, for $G$ to be regular it need to be $1$-regular.
Suppose $G$ has $3$ or more vertices.
Let $u, v, w$ be such vertices of $G$.
Suppose each of $\set {u, v}$, $\set {v, w}$ and $\set {u, w}$ were adjacent.
Then there would be a cycle in $G$: $\tuple {u, v, w, u}$ and so $G$ by definition would not be a tree.
So at least one pair $\set {u, v}$, $\set {v, w}$ and $\set {u, w}$ is not adjacent.
{{WLOG}} suppose this pair is $\set {u, v}$.
By definition, a tree is connected.
Therefore there exists a path $P$ between $u$ and $v$.
As $u$ and $v$ are not adjacent, there exists another vertex $y$ of $G$ through which $P$ passes.
In order for this to be the case, the degree of $y$ needs to be greater than $1$.
It follows that $G$ can not be $1$-regular.
So to be regular, $G$ can contain only $2$ vertices.
By Finite Connected Simple Graph is Tree iff Size is One Less than Order it follows that $G$ contains $2$ vertices connected together by a single edge.
By definition, this is the complete graph of order $2$.
{{qed|lemma}}
Theorem: Perimeter of Circle

The perimeter $C$ of a circle with radius $r$ is given by:
:$C = 2 \pi r$

Proof:
By definition, the perimeter is the length of the circumference of the circle.
Let $C$ be the perimeter.
Then:
{{begin-eqn}}
{{eqn | l = \pi
      | r = \frac C {2 r}
      | c = {{Defof|Pi|$\pi$ (pi)}}
}}
{{eqn | ll= \leadsto
      | l = C
      | r = 2 \pi r
}}
{{end-eqn}}
{{qed}}
Theorem: Euler's Criterion/Quadratic Residue

Let $p$ be an odd prime.
Let $a \not \equiv 0 \pmod p$.
Then:
{{begin-eqn}}
{{eqn | l = a^{\frac {p-1} 2}
      | o = \equiv
      | r = 1
      | rr= \pmod p
      | c = {{iff}} $a$ is a quadratic residue of $p$
}}
{{eqn | l=a ^{\frac {p-1} 2}
      | o = \equiv
      | r = -1
      | rr= \pmod p
      | c = {{iff}} $a$ is a quadratic non-residue of $p$.
}}
{{end-eqn}}

Proof:
Trivially, any $a \not \equiv 0 \pmod p$ is either a quadratic residue or a quadratic non-residue, modulo $p$.
Therefore, it suffices to check the sufficient condition for both of the equations (i.e., the ''if'' parts from the ''iff''s).
So let $a$ be a quadratic non-residue of $p$.
Also, let $b \in \set {1, 2, \ldots, p - 1}$.
The congruence $b x \equiv a \pmod p$ has (modulo $p$) a unique solution $b'$ by Solution of Linear Congruence.
Note that $b' \not\equiv b$, because otherwise we would have $b^2 \equiv a \pmod p$ and $a$ would be a quadratic residue of $p$.
It follows that the residue classes $\set {1, 2, \ldots, p - 1}$ modulo $p$ fall into $\dfrac {p - 1} 2$ pairs $b, b'$ such that $b b' \equiv a \pmod p$.
Therefore, we have:
:$\paren {p - 1}! = 1 \times 2 \times \cdots \times \paren {p - 1} \equiv a \times a \times \cdots \times a \equiv a^{\frac {p - 1} 2} \pmod p$
From Wilson's Theorem, we also have:
:$\paren {p - 1}! \equiv -1 \pmod p$
And so, for any quadratic non-residue of $p$:
:$a^{\frac {p - 1} 2} \equiv -1 \pmod p$
Subsequently, let $a$ be a quadratic residue of $p$.
By definition of a quadratic residue, the congruence $x^2 \equiv a \pmod p$ has a solution $x$.
Suppose also $y$ is a solution. Then we have:
{{begin-eqn}}
{{eqn | l = x^2 - y^2
      | o = \equiv
      | r = 0
      | rr= \pmod p
}}
{{eqn | ll= \leadstoandfrom
      | l = \paren {x + y} \paren {x - y}
      | o = \equiv
      | r = 0
      | rr= \pmod p
}}
{{end-eqn}}
So either $x + y \equiv 0 \pmod p$ or $x - y \equiv 0 \pmod p$ from Product is Zero Divisor means Zero Divisor.
It follows that when $c \pmod p$ is one solution, $-c \pmod p$ is, too.
Also, these solutions are distinct as $p$ is odd.
Furthermore, we conclude that these are the only two solutions.
Now, remove $c$ and $p - c$ from $\set {1, 2, \ldots, p - 1}$.
The remaining integers fall, modulo $p$, into $\dfrac {p - 3} 2$ pairs $b, b'$ such that $b b' \equiv a \pmod p$ by Solution of Linear Congruence.
Therefore, we can compute the following:
{{begin-eqn}}
{{eqn | l = \paren {p - 1}!
      | r = 1 \times 2 \times \cdots \times c \times \cdots \times \paren {p - c} \times \cdots \times \paren {p - 1}
}}
{{eqn | o = \equiv
      | r = a \times a \times \cdots \times a \times c \times \paren {-c}
      | rr= \pmod p
}}
{{eqn | o = \equiv
      | r = a^{\frac {p - 3} 2} \paren {-c^2}
      | rr= \pmod p
}}
{{eqn | o = \equiv
      | r = a^{\frac {p - 3} 2} \paren {-a}
      | rr= \pmod p
}}
{{eqn | o = \equiv
      | r = -a^{\frac {p - 1} 2}
      | rr= \pmod p
}}
{{end-eqn}}
Again applying Wilson's Theorem, we conclude:
:$-a^{\frac {p-1} 2} \equiv - 1 \pmod p$
The assertion for quadratic residues $a$ of $p$ follows.
{{qed}}
Theorem: Euler's Criterion/Quadratic Residue

Let $p$ be an odd prime.
Let $a \not \equiv 0 \pmod p$.
Then:
{{begin-eqn}}
{{eqn | l = a^{\frac {p-1} 2}
      | o = \equiv
      | r = 1
      | rr= \pmod p
      | c = {{iff}} $a$ is a quadratic residue of $p$
}}
{{eqn | l=a ^{\frac {p-1} 2}
      | o = \equiv
      | r = -1
      | rr= \pmod p
      | c = {{iff}} $a$ is a quadratic non-residue of $p$.
}}
{{end-eqn}}

Proof:
First note that the square roots of $1$ are $1, -1 \pmod p$.
Also, we have that $a^{p - 1} \equiv 1 \pmod p$ by Fermat's Little Theorem.
Combining these two observations, we find:
:$a^{\frac {p - 1} 2} \equiv 1 \text{ or } -1 \pmod p$
The theorem is therefore equivalent to stating that $a$ is a quadratic residue modulo $p$ {{iff}} $a^{\frac{p - 1} 2} \equiv 1 \pmod p$.
Namely, considering the above, we see this also implies that all quadratic non-residues will be congruent to $-1 \pmod p$.
We prove each direction of the equivalent statement separately:
=== Sufficient Condition ===
Assume $a$ is a quadratic residue modulo $p$.
We pick $k$ such that $k^2 \equiv a \pmod p$.
Then by Congruence of Powers and Fermat's Little Theorem:
:$a^{\frac{p-1} 2} \equiv k^{p-1} \equiv 1 \pmod p$
=== Necessary Condition ===
Now assume $a^{\frac{p - 1} 2} \equiv 1 \pmod p$.
Then let $y$ be a primitive root modulo $p$, so that $a$ can be written as $y^j$.
In particular:
:$y^{j \frac {p - 1} 2} \equiv 1 \pmod p$
From the definition of $y$, it has order $p-1$.
It follows that $p - 1 \divides j \dfrac {p - 1} 2$ from Element to Power of Multiple of Order is Identity.
We conclude that $j$ is necessarily an even integer, and denote $j' = \dfrac j 2$.
Let $k$ be such that $k \equiv y^{j'} \pmod p$.
By construction, we have:
:$k^2 \equiv y^{2 j'} \equiv y^j \equiv a \pmod p$
Hence $a$ is a quadratic residue modulo $p$.
{{qed}}
Theorem: Two Paths between Vertices in Cycle Graph

Let $G$ be a simple graph.
Let $u, v$ be vertices in $G$ such that $u \ne v$.
Then:
:for any two vertices $u, v$ in $G$ such that $u \ne v$ there exists exactly two paths between $u$ and $v$
{{iff}}:
:$G$ is a cycle graph.

Proof:
=== Necessary Condition ===
{{proof wanted}}
Theorem: Value of Radian in Degrees

The value of a radian in degrees is given by:
:$1 \radians = \dfrac {180 \degrees} {\pi} \approx 57 \cdotp 29577 \, 95130 \ 82320 \, 87679 \, 8154 \ldots \degrees$ 
{{OEIS|A072097}}

Proof:
By Measurement of Full Angle, a full angle measures $2 \pi$ radians.
By definition of degree of angle, a full angle measures $360$ degrees.
Thus $1$ radian is given by:
:$1 \radians = \dfrac {360 \degrees} {2 \pi} = \dfrac {180 \degrees} {\pi}$
{{qed}}
Theorem: Value of Degree in Radians

The value of a degree in radians is given by:
:$1 \degrees = \dfrac {\pi} {180} \radians \approx 0 \cdotp 01745 \, 32925 \, 19943 \, 29576 \, 9236 \ldots \radians$
{{OEIS|A019685}}

Proof:
By Measurement of Full Angle, a full angle measures $2 \pi$ radians.
By definition of degree of angle, a full angle measures $360$ degrees.
Thus $1$ degree of angle is given by:
:$1 \degrees = \dfrac {2 \pi} {360} = \dfrac \pi {180}$
{{qed}}
Theorem: Difference of Two Squares/Geometric Proof 1

:$\forall x, y \in \R: x^2 - y^2 = \paren {x + y} \paren {x - y}$

Proof:
{{:Euclid:Proposition/II/5}}
:400px
Let $AB$ be cut into equal segments at $C$ and unequal segments at $D$.
Then the rectangle contained by $AD$ and $DB$ together with the square on $CD$ equals the square on $BC$.
(That is, let $x = AC, y = CD$. Then $\paren {x + y} \paren {x - y} + y^2 = x^2$.)
This is proved as follows.
Construct the square $CBFE$ on $CB$, and join $BE$.
Construct $DG$ parallel to $CE$ through $G$, and let $DG$ cross $BE$ at $H$.
Construct $KM$ parallel to $AB$ through $H$.
Construct $AK$ parallel to $BF$ through $A$.
From Complements of Parallelograms are Equal:
:$\Box CDHL = \Box FGHM$.
Add the square $DBMH$ to each.
Then $\Box CBML = \Box DBFG$.
But as $AC = CB$, from Parallelograms with Equal Base and Same Height have Equal Area we have that:
:$\Box ACLK = \Box CBML$
Add $\Box CDHL$ to each.
Then $\Box ADHK$ is equal in area to the gnomon $CBFGHL$.
But $\Box ADHK$ is the rectangle contained by $AD$ and $DB$, because $DB = DH$.
So the gnomon $CBFGHL$ is equal in area to the rectangle contained by $AD$ and $DB$.
Now $\Box LHGE$ is equal to the square on $CD$.
Add $\Box LHGE$ to each of the gnomon $CBFGHL$ and $\Box ADHK$.
Then the gnomon $CBFGHL$ together with $\Box LHGE$ equals the rectangle contained by $AD$ and $DB$ and the square on $CD$.
But the gnomon $CBFGHL$ together with $\Box LHGE$ is the square $CBFE$.
Hence the result.
{{qed}}
Theorem: Area of Parallelogram/Square

The area of a square equals the product of one of its bases and the associated altitude.

Proof:
:190px
From Area of Square:
:$\paren {ABCD} = a^2$
where $a$ is the length of one of the sides of the square.
The altitude of a square is the same as its base.
Hence the result.
{{Qed}}
Theorem: Area of Parallelogram/Rectangle

The area of a rectangle equals the product of one of its bases and the associated altitude.

Proof:
Let $ABCD$ be a rectangle.
:300px
Then construct the square with side length:
:$\map \Area {AB + BI}$
where $BI = BC$, as shown in the figure above.
Note that $\square CDEF$ and $\square BCHI$ are squares.
Thus:
:$\square ABCD \cong \square CHGF$
Since congruent shapes have the same area:
:$\map \Area {ABCD} = \map \Area {CHGF}$ (where $\map \Area {FXYZ}$ denotes the area of the plane figure $FXYZ$).
Let $AB = a$ and $BI = b$.
Then the area of the square $AIGE$ is equal to:
{{begin-eqn}}
{{eqn | l = \paren {a + b}^2
      | r = a^2 + 2 \map \Area {ABCD} + b^2
}}
{{eqn | l = \paren {a^2 + 2 a b + b^2}
      | r = a^2 + 2 \map \Area {ABCD} + b^2
}}
{{eqn | l = a b
      | r = \map \Area {ABCD}
}}
{{end-eqn}}
{{Qed}}
Theorem: Area of Parallelogram/Parallelogram

Let $ABCD$ be a parallelogram whose adjacent sides are of length $a$ and $b$ enclosing an angle $\theta$.
The area of $ABCD$ equals the product of one of its bases and the associated altitude:
{{begin-eqn}}
{{eqn | l = \map \Area {ABCD}
      | r = b h
      | c = 
}}
{{eqn | r = a b \sin \theta
      | c = 
}}
{{end-eqn}}
where:
:$b$ is the side of $ABCD$ which has been chosen to be the base
:$h$ is the altitude of $ABCD$ from $b$.

Proof:
:425px
Let $ABCD$ be the parallelogram whose area is being sought.
Let $F$ be the foot of the altitude from $C$
Also construct the point $E$ such that $DE$ is the altitude from $D$ (see figure above).
Extend $AB$ to $F$.
Then:
{{begin-eqn}}
{{eqn | l = AD
      | o = \cong
      | r = BC
}}
{{eqn | l = \angle AED
      | o = \cong
      | r = \angle BFC
}}
{{eqn | l = DE
      | o = \cong
      | r = CF
}}
{{end-eqn}}
Thus:
:$\triangle AED \cong \triangle BFC \implies \map \Area {AED} = \map \Area {BFC}$
So:
{{begin-eqn}}
{{eqn | l = \map \Area {ABCD}
      | r = EF \cdot FC
      | c = 
}}
{{eqn | r = AB \cdot DE
      | c = 
}}
{{eqn | r = b h
      | c = 
}}
{{eqn | r = a b \sin \theta
      | c = {{Defof|Sine of Angle}}: $h = a \sin \theta$
}}
{{end-eqn}}
{{qed}}
Theorem: Perimeter of Rectangle

Let $ABCD$ be a rectangle whose side lengths are $a$ and $b$.
The perimeter of $ABCD$ is $2 a + 2 b$.

Proof:
:300px
From Rectangle is Parallelogram, $ABCD$ is a parallelogram.
By Opposite Sides and Angles of Parallelogram are Equal it follows that:
:$AB = CD$
:$BC = AD$
The perimeter of $ABCD$ is $AB + BC + CD + AD$.
But $AB = CD = a$ and $BC = AD = b$.
Hence the result.
{{qed}}
Theorem: Perimeter of Rectangle

Let $ABCD$ be a rectangle whose side lengths are $a$ and $b$.
The perimeter of $ABCD$ is $2 a + 2 b$.

Proof:
From Rectangle is Parallelogram, $ABCD$ is a parallelogram.
The result then follows from a direct application of Perimeter of Parallelogram.
{{qed}}
Theorem: Perimeter of Parallelogram

Let $ABCD$ be a parallelogram whose side lengths are $a$ and $b$.
The perimeter of $ABCD$ is $2 a + 2 b$.

Proof:
:400px
By Opposite Sides and Angles of Parallelogram are Equal it follows that:
:$AB = CD$
:$BC = AD$
The perimeter of $ABCD$ is $AB + BC + CD + AD$.
But $AB = CD = a$ and $BC = AD = b$.
Hence the result.
{{qed}}
Theorem: Perimeter of Triangle

Let $ABC$ be a triangle.
Then the perimeter $P$ of $ABC$ is given by:
:$P = a + b + c$
where $a, b, c$ are the lengths of the sides of $ABC$.

Proof:
The perimeter of a plane geometric figure is defined as the total length of the boundary.
By definition, the boundary of a triangle comprises the three sides of that triangle.
Hence the result.
{{qed}}
Theorem: Area of Trapezium

:410px
Let $ABCD$ be a trapezium:
:whose parallel sides are of lengths $a$ and $b$
and
:whose height is $h$.
Then the area of $ABCD$ is given by:
:$\Box ABCD = \dfrac {h \paren {a + b} } 2$

Proof:
:600px
Extend line $AB$ to $E$ by length $a$.
Extend line $DC$ to $F$ by length $b$.
Then $BEFC$ is another trapezium whose parallel sides are of lengths $a$ and $b$ and whose height is $h$.
Also, $AEFD$ is a parallelogram which comprises the two trapezia $ABCD$ and $BEFC$.
So $\Box ABCD + \Box BEFC = \Box AEFD$ and $\Box ABCD = \Box BEFC$.
$AEFD$ is of altitude $h$ with sides of length $a + b$.
Thus from Area of Parallelogram the area of $AEFD$ is given by:
:$\Box AEFD = h \paren {a + b}$
It follows that $\Box ABCD = \dfrac {h \paren {a + b} } 2$
{{qed}}
Theorem: Perimeter of Trapezium

:400px
Let $ABCD$ be a trapezium:
:whose parallel sides are of lengths $a$ and $b$
:whose height is $h$.
and
:whose non-parallel sides are at angles $\theta$ and $\phi$ with the parallels.
The perimeter $P$ of $ABCD$ is given by:
:$P = a + b + h \paren {\csc \theta + \csc \phi}$
where $\csc$ denotes cosecant.

Proof:
The perimeter $P$ of $ABCD$ is given by:
:$P = AB + BC + CD + AD$
where the lines are used to indicate their length.
Thus:
{{begin-eqn}}
{{eqn | n = 1
      | l = AB
      | r = b
}}
{{eqn | n = 2
      | l = CD
      | r = a
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = h
      | r = AD \sin \theta
      | c = {{Defof|Sine of Angle}}
}}
{{eqn | ll= \leadsto
      | l = AD
      | r = \frac h {\sin \theta}
      | c = 
}}
{{eqn | n = 3
      | r = h \csc \theta
      | c = Cosecant is Reciprocal of Sine
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = h
      | r = BC \sin \phi
      | c = {{Defof|Sine of Angle}}
}}
{{eqn | ll= \leadsto
      | l = BC
      | r = \frac h {\sin \phi}
      | c = 
}}
{{eqn | n = 4
      | r = h \csc \phi
      | c = Cosecant is Reciprocal of Sine
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = P
      | r = AB + BC + CD + AD
}}
{{eqn | r = b + h \csc \phi + a + h \csc \theta
      | c = from $(1)$, $(2)$, $(3)$ and $(4)$ 
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Area of Regular Polygon

Let $P$ be a regular $n$-sided polygon whose side length is $b$.
Then the area of $P$ is given by:
:$\Box P = \dfrac 1 4 n b^2 \cot \dfrac \pi n$
where $\cot$ denotes cotangent.

Proof:
:400px
Let $H$ be the center of the regular $n$-sided polygon $P$.
Let one of its sides be $AB$.
Consider the triangle $\triangle ABH$.
As $P$ is regular and $H$ is the center, $AH = BH$ and so $\triangle ABH$ is isosceles.
Thus $b = AB$ is the base of $\triangle ABH$.
Let $h = GH$ be its altitude.
See the diagram.
Then:
{{begin-eqn}}
{{eqn | l = \triangle ABH
      | r = \frac {b h} 2
      | c = Area of Triangle in Terms of Side and Altitude
}}
{{eqn | r = \frac b 2 \frac b 2 \cot \alpha
      | c = {{Defof|Cotangent of Angle}}
}}
{{eqn | r = \frac {b^2} 4 \cot \frac \pi n
      | c = $\alpha$ is half the apex of $\triangle ABH$, and $n$ of such apices fit into the full circle of $2 \pi$
}}
{{end-eqn}}
The full polygon $P$ is made up of $n$ such triangles, each of which has the same area as $\triangle ABH$.
Hence:
:$\Box P = \dfrac 1 4 n b^2 \cot \dfrac \pi n$
{{qed}}
Theorem: Perimeter of Regular Polygon

Let $P$ be a regular $n$-sided polygon whose side length is $b$.
Then the perimeter $L$ of $P$ is given by:
:$L = n b$

Proof:
By definition, an $n$-sided polygon has $n$ sides.
By definition, a regular polygon has sides all the same length.
By definition, the perimeter of a polygon is the total length of all its sides.
Thus $P$ has $n$ sides all of length $b$.
Hence $L = n b$.
{{qed}}
Theorem: Commensurability is Transitive

Let $a$, $b$, $c$ be three real numbers.
Let $a$ and $b$ be commensurable, and $b$ and $c$ be commensurable.
Then $a$ and $c$ are commensurable.

Proof:
From the definition of commensurablility:
:$\dfrac a b, \dfrac b c \in \Q$
where $\Q$ denotes the set of all rational numbers.
From Rational Multiplication is Closed:
:$\dfrac a b \times \dfrac b c \in \Q$
Cancelling $b$, we have:
:$\dfrac a c \in \Q$
Hence the result.
{{qed}}
Category:Number Theory
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Theorem: Arc Length of Sector

Let $\CC = ABC$ be a circle whose center is $A$ and with radii $AB$ and $AC$.
Let $BAC$ be the sector of $\CC$ whose angle between $AB$ and $AC$ is $\theta$.
:300px
Then the length $s$ of arc $BC$ is given by:
:$s = r \theta$
where:
:$r = AB$ is the length of the radius of the circle
:$\theta$ is measured in radians.

Proof:
From Perimeter of Circle, the perimeter of $\CC$ is $2 \pi r$.
From Measurement of Full Angle, the angle within $\CC$ is $2 \pi$.
{{explain|Why is the density of the arc length uniform? i.e. why does equal rotation sweeps out equal arc length?}}
The fraction of the perimeter of $\CC$ within the sector $BAC$ is therefore $2 \pi r \times \dfrac \theta {2 \pi}$.
Hence the result.
{{qed}}