Theorem: Area of Circle

The area $A$ of a circle is given by:
:$A = \pi r^2$
where $r$ is the radius of the circle.

Proof:
By the method of exhaustion:
:800px
== Construction ==
{{tidy}}
{{MissingLinks}}
For step $1$ of the construction, construct a circle $C$ and a diameter of $C$, thereby dividing the circumference of $C$ into $2$ arcs.
Let $A$ be the area of $C$ and $r$ be the radius of $C$.
For $n > 1$, step $n$ of the construction consists of constructing a regular $2^n$-gon $P_n$ inscribed in $C$.
For each arc $EB$ of $C$ which has not been subdivided, construct the midpoint $G$ of arc $EB$ and join segments $\overline{EG}$ and $\overline{GB}$.
=== Lemma: The area of $P_n$ converges to $A$ as $n \to \infty$ ===
Let $E_n = A - P_n$ be the area outside of $P_n$ and inside of $C$.
As this area is divided into $2^n$ disjoint, congruent sectors, let $e_n = \dfrac {E_n} {2^n}$ be the area of one such sector.
Consider a sector with endpoints $E$ and $B$, with $G$ the midpoint of the bounding arc.
Construct a line $\ell$ through $G$ parallel to $\overline{EB}$.
Construct perpendiculars to $\ell$ through $E$ and $B$ and let them meet $\ell$ at $H$ and $I$ respectively.
Then quadrilateral $EHIB$ is a rectangle, as it has four right angles by construction.
Therefore, its area is twice that of $\triangle EGB$. 
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We have:
{{explain|What is $e_n$?}}
{{begin-eqn}}
{{eqn | l = e_n
      | o = \le
      | r = \Box EHIB
      | c = as arc $EB$ is contained in $EHIB$
}}
{{eqn | r = 2 \triangle EGB
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac {e_n} 2
      | o = \le
      | r = \triangle EGB
      | c = 
}}
{{eqn | ll= \leadsto
      | l = e_n
      | o = \le
      | r = \triangle EGB + \dfrac {e_n} 2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = e_n - \triangle EGB
      | o = \le
      | r = \dfrac {e_n} 2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 2 e_{n + 1}
      | o = \le
      | r = \dfrac {e_n} 2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = e_{n + 1}
      | o = \le
      | r = \dfrac {e_n} 4
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac {E_{n+1} } {2^{n + 1} }
      | o = \le
      | r = \dfrac 1 4 \cdot \dfrac {E_n} {2^n}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = E_{n+1}
      | o = \le
      | r = \dfrac {E_n}2
      | c = 
}}
{{end-eqn}}
It can be shown by induction that $E_n$ is bounded above by the geometric sequence $\paren {\dfrac 1 2}^n E_1$ which converges to $0$.
{{finish|Show that inductive proof}}
Since each term of $E_n$ is an area, it is bounded below by the constant sequence $0$, which converges to $0$.
By the Squeeze Theorem, $E_n$ converges to $0$.
By Combination Theorem for Sequences, $P_n$ converges to $A$. 
{{qed|lemma}}
Observe that $P_n$ is the union of $2^n$ congruent isosceles triangles whose bases are the edges of $P_n$, whose legs are radii of $C$, and which have a common vertex at the center of $C$.
Thus if $T$ is the area of one of these triangles, the area of $P_n$ is $2^n T$. 
For each such triangle, the angle at the vertex is $\dfrac {2 \pi} {2^n}$.
By Area of Isosceles Triangle, the area of each triangle is $\dfrac 1 2 r^2 \map \sin {\dfrac {2 \pi} {2^n} }$.
Thus, the area of $P_n$ is $2^{n - 1} r^2 \map \sin {\dfrac {2 \pi} {2^n} }$.
By the Lemma, the area of $C$ is then:
:$\ds \lim_{n \mathop \to \infty} 2^{n - 1} r^2 \map \sin {\dfrac \pi {2^{n - 1} } }$
Let $m = \dfrac \pi {2^{n-1} }$.
Because $m \to 0$ as $n \to \infty$, this leads to:
{{begin-eqn}}
{{eqn | l = A
      | r = r^2 \lim_{m \mathop \to 0} \dfrac \pi m \map \sin m
      | c = 
}}
{{eqn | r = \pi r^2 \lim_{m \mathop \to 0} \dfrac {\map \sin m} m
      | c = 
}}
{{eqn | r = \pi r^2
      | c = Limit of Sine of X over X at Zero
}}
{{end-eqn}}
{{qed}}