hcju/proofwikiqa huggingface dataset:
$1$ is defined as $s \left({0}\right)$ and $2$ as $s \left({s \left({0}\right)}\right)$. Therefore the statement to be proven becomes: :$s \left({0}\right) + s \left({0}\right) = s \left({s \left({0}\right)}\right)$ Thus: {{begin-eqn}} {{eqn | ll= \forall n \in P:       | l = m + s \left({n}\right)       | r = s \left({m + n}\right)       | c = Definition of Addition }} {{eqn |lll= \implies       | l = m + s \left({0}\right)       | r = s \left({m+0}\right) }} {{eqn |lll= \implies       | ll= \forall m \in P:       | l = m + s \left({0}\right)       | r = s \left({m + 0}\right) }} {{eqn | n = 1       |lll= \implies       | l = s \left({0}\right) + s \left({0}\right)       | r = s \left({s \left({0}\right) + 0}\right) }} {{eqn | ll= \forall m \in P:       | l = m + 0       | r = m       | c = Definition of Addition }} {{eqn | n = 2       |lll= \implies       | l = s \left({0}\right) + 0       | r = s \left({0}\right) }} {{eqn | l = s \left({s \left({0}\right) + 0}\right)       | r = s \left({s \left({0}\right)}\right)       | c = taking the successor on both sides of $(2)$ }} {{eqn | n = 3       |lll= \implies       | l = s \left({0}\right) + s \left({0}\right)       | r = s \left({s \left({0}\right) + 0}\right)       | c = from $(1)$ }} {{eqn | r = s \left({s \left({0}\right)}\right)       | c = from $(2)$ }} {{end-eqn}} {{qed}}
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}}
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 $\displaystyle \map f H = \sum_{e \mathop \in H} \map f e$. Let $N$ be of order $p$. Let $T$ be a spanning tree of $N$, created using Kruskal's Algorithm. Then from Size of Tree 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: :$\displaystyle \map f t = \sum_{i \mathop = 1}^{p - 1} \map f {e_i}$ 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}}
=== 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 $\left\{{u, v}\right\}$, $\left\{{v, w}\right\}$ and $\left\{{u, w}\right\}$ were adjacent. Then there would be a cycle in $G$: $\left({u, v, w, u}\right)$ and so $G$ by definition would not be a tree. So at least one pair $\left\{{u, v}\right\}$, $\left\{{v, w}\right\}$ and $\left\{{u, w}\right\}$ is not adjacent. {{WLOG}} suppose this pair is $\left\{{u, v}\right\}$. 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 Size of Tree 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}} === Sufficient Condition === $K_2$, the complete graph of order $2$, consists of $2$ vertices connected together by a single edge. From Complete Graph is Regular it follows that $K_2$ is a non-edgeless regular graph. From Circuit has Three Edges or More it follows that $K_2$ contains no circuits. It follows by definition that $K_2$ is a tree. {{qed}}
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}}
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}}
=== Necessary Condition === {{proof wanted}} === Sufficient Condition === Let $G = \struct {V, E} = C_n$ be the cycle graph of order $n$. Let $V = \set {v_1, v_2, \ldots, v_n}$ Let $u, v \in C_n$. Then both $u, v$ are on the cycle $C = \tuple {u, u_1, u_2, \ldots, u_p, v, v_1, v_2, \ldots, v_q, u}$ where $p + q + 2 = n$. By definition of cycle graph this is the only cycle in $G$. Let $e$ be the edge $u u_1$ of $G$ (or, if $u$ and $v$ are adjacent, let $e = u v$). From Size of Cycle Graph equals Order, there are exactly $n$ edges in $G$. Then the edge deletion $G - e$ has $n - 1$ edges. By Size of Tree is One Less than Order it follows that $G - e$ is a tree. Thus by Path in Tree is Unique there is exactly $1$ path from $u$ to $v$ in $G - e$, that is: :$P_1 = \tuple {u, v_q, v_{q - 1}, \ldots, v_2, v_1, v}$ By replacing $e$, there exist a second path from $u$ to $v$ in $G$: :$P_2 = \tuple {u, u_1, u_2, \ldots, u_{p - 1}, u_p, v}$ or if $u$ and $v$ are adjacent: :$P_2 = \tuple {u, v}$ {{qed}}
By Full Angle measures 2 Pi Radians, a full angle measures $2 \pi$ radians. By definition of degree of arc, a full angle measures $360$ degrees. Thus $1$ radian is given by: :$1 \radians = \dfrac {360 \degrees} {2 \pi} = \dfrac {180 \degrees} {\pi}$ {{qed}}
By Full Angle measures 2 Pi Radians, a full angle measures $2 \pi$ radians. By definition of degree of arc, a full angle measures $360$ degrees. Thus $1$ degree of arc is given by: :$1 \degrees = \dfrac {2 \pi} {360} = \dfrac {\pi} {180}$ {{qed}}
: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}}
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}}
: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}}
: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}}
From Rectangle is Parallelogram, $ABCD$ is a parallelogram. The result then follows from a direct application of Perimeter of Parallelogram. {{qed}}
: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}}
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}}
:600px Extend line $AB$ to $E$ by length $a$. Extend line $DC$ to $F$ by length $b$. Then $BECF$ is another trapezoid whose parallel sides are of lengths $a$ and $b$ and whose height is $h$. Also, $AEFD$ is a parallelogram which comprises the two trapezoids $ABCD$ and $BECF$. So $\Box ABCD + \Box BECF = \Box AEFD$ and $\Box ABCD = \Box BECF$. $AEFD$ 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}}
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}}
: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 $AB$ 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 $n$ of 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}}
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}}
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 1iqhdq9qiwy16kc734vrhp6ppaw4sog
From Perimeter of Circle, the perimeter of $\mathcal C$ is $2 \pi r$. From Full Angle measures $2 \pi$ Radians, the angle within $\mathcal C$ 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 $\mathcal C$ within the sector $BAC$ is therefore $2 \pi r \times \dfrac \theta {2 \pi}$. Hence the result. {{qed}}
:410px Let $\mathcal A$ be the area of $\triangle ABC$. From Area of Triangle in Terms of Inradius: :$\mathcal A = r s$ where $s = \dfrac {a + b + c} 2$ is the semiperimeter of $\triangle ABC$. From Heron's Formula: :$\mathcal A = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$ where $s = \dfrac {a + b + c} 2$ is the semiperimeter of $\triangle ABC$. Hence the result: :$r = \dfrac {\sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} } } s$ {{qed}}
:320px Let $\mathcal A$ be the area of $\triangle ABC$. From Area of Triangle in Terms of Circumradius: :$\mathcal A = \dfrac {a b c} {4 R}$ From Heron's Formula: :$\mathcal A = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$ where $s = \dfrac {a + b + c} 2$ is the semiperimeter of $\triangle ABC$. Hence the result: :$R = \dfrac {abc} {4 \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} } }$ {{qed}}
:400px Let $I$ be the incenter of $\triangle ABC$. Let $r$ be the inradius of $\triangle ABC$. The total area of $\triangle ABC$ is equal to the sum of the areas of the triangle formed by the vertices of $\triangle ABC$ and its incenter: :$\mathcal A = \map \Area {\triangle AIB} + \map \Area {\triangle BIC} + \map \Area {\triangle CIA}$ Let $AB$, $BC$ and $CA$ be the bases of $\triangle AIB, \triangle BIC, \triangle CIA$ respectively. The lengths of $AB$, $BC$ and $CA$ respectively are $c, a, b$. The altitude of each of these triangles is $r$. Thus from Area of Triangle in Terms of Side and Altitude: {{begin-eqn}} {{eqn | l = \map \Area {\triangle AIB}       | r = \frac {c r} 2 }} {{eqn | l = \map \Area {\triangle BIC}       | r = \frac {a r} 2 }} {{eqn | l = \map \Area {\triangle CIA}       | r = \frac {b r} 2 }} {{end-eqn}} Thus: :$\displaystyle \mathcal A = r \frac {a + b + c} 2$ That is: :$\mathcal A = r s$ where $s = \dfrac {a + b + c} 2$ is the semiperimeter of $\triangle ABC$. {{qed}}
:400px From Regular Polygon composed of Isosceles Triangles, let $\triangle OAB$ be one of the $n$ isosceles triangles that compose $P$. Then $\mathcal A$ is equal to $n$ times the area of $\triangle OAB$. Let $d$ be the length of one side of $P$. Then $d$ is the length of the base of $\triangle OAB$. Let $h$ be the altitude of $\triangle OAB$. The angle $\angle AOB$ is equal to $\dfrac {2 \pi} n$. Then: :$(1): \quad h = r \cos \dfrac \pi n$ :$(2): \quad d = 2 r \sin \dfrac \pi n$ So: {{begin-eqn}} {{eqn | l = \mathcal A       | r = n \frac {h d} 2       | c = Area of Triangle in Terms of Side and Altitude }} {{eqn | r = \frac n 2 \paren {r \cos \frac \pi n} \paren {2 r \sin \dfrac \pi n}       | c = substituting from $(1)$ and $(2)$ above }} {{eqn | r = \frac 1 2 n r^2 2 \paren {\cos \frac \pi n} \paren {\sin \dfrac \pi n}       | c = rearranging }} {{eqn | r = \frac 1 2 n r^2 \paren {\sin \frac {2 \pi} n}       | c = Double Angle Formula for Sine }} {{eqn | r = \frac 1 2 n r^2 \sin \frac {2 \pi} n       | c = simplifying }} {{end-eqn}} {{qed}}
:400px From Regular Polygon composed of Isosceles Triangles, let $\triangle OAB$ be one of the $n$ isosceles triangles that compose $P$. Then $\mathcal P$ is equal to $n$ times the base of $\triangle OAB$. Let $d$ be the length of one side of $P$. Then $d$ is the length of the base of $\triangle OAB$. The angle $\angle AOB$ is equal to $\dfrac {2 \pi} n$. Then: :$d = 2 r \sin \dfrac \pi n$ So: {{begin-eqn}} {{eqn | l = \mathcal P       | r = n d       | c =  }} {{eqn | r = n \paren {2 r \sin \dfrac \pi n}       | c = substituting from above }} {{eqn | r = 2 n r \sin \dfrac \pi n       | c = simplifying }} {{end-eqn}} {{qed}}
:400px From Regular Polygon composed of Isosceles Triangles, let $\triangle OAB$ be one of the $n$ isosceles triangles that compose $P$. Then $\AA$ is equal to $n$ times the area of $\triangle OAB$. Also, $r$ is the length of the altitude of $\triangle OAB$. Let $d$ be the length of one side of $P$. Then $d$ is the length of the base of $\triangle OAB$. The angle $\angle AOB$ is equal to $\dfrac {2 \pi} n$. Then $d = 2 r \tan \dfrac \pi n$ So: {{begin-eqn}} {{eqn | l = \AA       | r = n \frac {r d} 2       | c = Area of Triangle in Terms of Side and Altitude }} {{eqn | r = \frac n 2 r \paren {2 r \tan \dfrac \pi n}       | c = substituting from above }} {{eqn | r = n r^2 \tan \dfrac \pi n       | c = rearranging }} {{end-eqn}} {{qed}}
:400px From Regular Polygon composed of Isosceles Triangles, let $\triangle OAB$ be one of the $n$ isosceles triangles that compose $P$. Then $\mathcal A$ is equal to $n$ times the area of $\triangle OAB$. Also, $r$ is the length of the altitude of $\triangle OAB$. Let $d$ be the length of one side of $P$. Then $d$ is the length of the base of $\triangle OAB$. The angle $\angle AOB$ is equal to $\dfrac {2 \pi} n$. Then: : $d = 2 r \tan \dfrac \pi n$ So: {{begin-eqn}} {{eqn | l = \mathcal P       | r = n d       | c = Perimeter of Regular Polygon }} {{eqn | r = 2 n r \tan \dfrac \pi n       | c = substituting from above }} {{end-eqn}} {{qed}}
:300px {{begin-eqn}} {{eqn | l = \mathcal A       | r = \frac 1 2 b h       | c = Area of Triangle in Terms of Side and Altitude }} {{eqn | r = \frac 1 2 b \paren {r \cos \dfrac \theta 2}       | c = Definition of Cosine }} {{eqn | r = \frac 1 2 2 \paren {r \sin \dfrac \theta 2} \paren {r \cos \dfrac \theta 2}       | c = Definition of Sine }} {{eqn | r = \frac 1 2 r^2 \sin \theta       | c = Double Angle Formula for Sine }} {{end-eqn}} {{qed}}
A direct application of Area of Triangle in Terms of Two Sides and Angle: :$\mathcal A = \dfrac 1 2 a b \sin \theta$ where $a = b = r$. {{qed}}
:350px Let $BDCE$ be the segment $S$. Let $b$ be the length of the base of $S$. Let $BACE$ be the sector of $C$ whose angle is $\theta$. The $\AA$ is equal to the area of $BACE$ minus the area of the isosceles triangle $\triangle ABC$ whose base is $b$. Let $h$ be the altitude of $\triangle ABC$. From Area of Sector, the area of sector $BACE$ is $\dfrac 1 2 r^2 \theta$. From Area of Isosceles Triangle, the area of $\triangle ABC$ is $\dfrac 1 2 r^2 \sin \theta$. Thus: {{begin-eqn}} {{eqn | l = \AA       | r = \frac 1 2 r^2 \theta - \frac 1 2 r^2 \sin \theta       | c =  }} {{eqn | r = \frac 1 2 r^2 \paren {\theta - \sin \theta}       | c =  }} {{end-eqn}} {{qed}}
Let $K$ be an ellipse aligned in a cartesian plane in reduced form. Then from Equation of Ellipse in Reduced Form: :$\dfrac {x^2} {a^2} + \dfrac {y^2} {b^2} = 1$ Thus: :$y = \pm b \sqrt {1 - \dfrac {x^2} {a^2} }$ From the geometric interpretation of the definite integral: {{begin-eqn}} {{eqn | l = \AA       | r = b \int_{-a}^a \paren {\sqrt {1 - \dfrac {x^2} {a^2} } - \paren {-\sqrt {1 - \dfrac {x^2} {a^2} } } } \rd x }} {{eqn | r = b \int_{-a}^a 2 \sqrt {1 - \dfrac {x^2} {a^2} } \rd x }} {{end-eqn}} Let $x = a \sin \theta$ (note that we can do this because $-a \le x \le a$). Thus: :$\theta = \map \arcsin {\dfrac x a}$ and: :$\d x = a \cos \theta \rd \theta$ Then: {{begin-eqn}} {{eqn | l = \AA       | r = b \int_{\map \arcsin {\frac {-a} a} }^{\map \arcsin {\frac a a} } 2 a \sqrt {1 - \frac {\paren {a \sin \theta}^2} {a^2} } \cos \theta \rd \theta       | c = Integration by Substitution }} {{eqn | r = b \int_{-\frac {\pi} 2}^{\frac {\pi} 2} 2 a \sqrt {1 - \sin^2 \theta} \cos \theta \rd \theta }} {{eqn | r = b \int_{-\frac {\pi} 2}^{\frac {\pi} 2} 2 a \sqrt {\cos^2 \theta} \cos \theta \rd \theta       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = a b \int_{-\frac {\pi} 2}^{\frac {\pi} 2} 2 \cos^2 \theta \rd \theta }} {{eqn | r = a b \int_{-\frac {\pi} 2}^{\frac {\pi} 2} \paren {1 + \map \cos {2 \theta} } \rd \theta       | c = Double Angle Formula for Cosine: Corollary 1 }} {{eqn | r = a b \sqbrk {\theta + \frac 1 2 \map \sin {2 \theta} }_{-\frac {\pi} 2}^{\frac {\pi} 2}       | c = Integration of Constant and Primitive of Cosine Function }} {{eqn | r = a b \paren {\frac {\pi} 2 + \frac 1 2 \map \sin {2 \cdot \frac {-\pi} 2} - \frac {-\pi} 2 - \frac 1 2 \map \sin {2 \cdot \frac {\pi} 2} } }} {{eqn | r = a b \paren {2 \cdot \frac {\pi} 2 + 2 \cdot \frac 1 2 \cdot 0} }} {{eqn | r = \pi a b }} {{end-eqn}} {{qed}}