avewright/proofwiki-math huggingface dataset:
Theorem: Absolutely Convergent Series is Convergent

Let $V$ be a Banach space with norm $\norm {\, \cdot \,}$.
Let $\ds \sum_{n \mathop = 1}^\infty a_n$ be an absolutely convergent series in $V$.
Then $\ds \sum_{n \mathop = 1}^\infty a_n$ is convergent.

Proof:
That $\ds \sum_{n \mathop = 1}^\infty a_n$ is absolutely convergent means that $\ds \sum_{n \mathop = 1}^\infty \norm {a_n}$ converges in $\R$.
Hence the sequence of partial sums is a Cauchy sequence by Convergent Sequence is Cauchy Sequence.
Now let $\epsilon > 0$.
Let $N \in \N$ such that for all $m, n \in \N$, $m \ge n \ge N$ implies that:
:$\ds \sum_{k \mathop = n + 1}^m \norm {a_k} = \size {\sum_{k \mathop = 1}^m \norm {a_k} - \sum_{k \mathop = 1}^n \norm {a_k} } < \epsilon$
This $N$ exists because the sequence is Cauchy.
Now observe that, for $m \ge n \ge N$, one also has:
{{begin-eqn}}
{{eqn | l = \norm {\sum_{k \mathop = 1}^m a_k - \sum_{k \mathop = 1}^n a_k}
      | r = \norm {\sum_{k \mathop = n + 1}^m a_k}
}}
{{eqn | o = \le
      | r = \sum_{k \mathop = n + 1}^m \norm {a_k}
      | c = Triangle inequality for $\norm {\, \cdot \,}$
}}
{{eqn | o = <
      | r = \epsilon
}}
{{end-eqn}}
It follows that the sequence of partial sums of $\ds \sum_{n \mathop = 1}^\infty a_n$ is Cauchy.
As $V$ is a Banach space, this implies that $\ds \sum_{n \mathop = 1}^\infty a_n$ converges.
{{qed}}
Theorem: Limit Comparison Test

Let $\sequence {a_n}$ and $\sequence {b_n}$ be sequences in $\R$.
Let $\ds \frac {a_n} {b_n} \to l$ as $n \to \infty$ where $l \in \R_{>0}$.
Then the series $\ds \sum_{n \mathop = 1}^\infty a_n$ and $\ds \sum_{n \mathop = 1}^\infty b_n$ are either both convergent or both divergent.

Proof:
Let $\ds \sum_{n \mathop = 1}^\infty b_n$ be convergent.
Then by Terms in Convergent Series Converge to Zero, $\sequence {b_n}$ converges to zero.
A Convergent Sequence is Bounded.
So it follows that:
:$\exists H: \forall n \in \N_{>0}: a_n \le H b_n$
Thus, by {{Corollary|Comparison Test|1}}, $\ds \sum_{n \mathop = 1}^\infty a_n$ is convergent.
Since $l > 0$, from Sequence Converges to Within Half Limit:
:$\exists N: \forall n > N: a_n > \dfrac 1 2 l b_n$
Hence the convergence of $\ds \sum_{n \mathop = 1}^\infty a_n$ implies the convergence of $\ds \sum_{n \mathop = 1}^\infty b_n$.
{{qed}}
Theorem: Area of Parallelogram

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

Proof:
There are three cases to be analysed: the square, the rectangle and the general parallelogram.
Theorem: Area of Triangle in Terms of Inradius and Exradii

Let $\triangle ABC$ be a triangle whose sides are $a$, $b$ and $c$ opposite vertices $A$, $B$ and $C$ respectively.
Then the area $\AA$ of $\triangle ABC$ is given by:
:$\AA = \sqrt {r \rho_a \rho_b \rho_c}$
where:
:$r$ is the inradius
:$\rho_a$, $\rho_b$ and $\rho_c$ are the exradii of $\triangle ABC$ {{WRT}} $a$, $b$ and $c$ respectively.

Proof:
{{begin-eqn}}
{{eqn | l = \AA
      | r = \rho_a \paren {s - a}
      | c = Area of Triangle in Terms of Exradius
}}
{{eqn | r = \rho_b \paren {s - b}
      | c = Area of Triangle in Terms of Exradius
}}
{{eqn | r = \rho_c \paren {s - c}
      | c = Area of Triangle in Terms of Exradius
}}
{{eqn | r = r s
      | c = Area of Triangle in Terms of Inradius
}}
{{eqn | ll= \leadsto
      | l = \AA^4
      | r = \rho_a \paren {s - a} \rho_b \paren {s - b} \rho_c \paren {s - c} r s
      | c = multiplying them all together
}}
{{eqn | r = s \paren {s - a} \paren {s - b} \paren {s - c} r \rho_a \rho_b \rho_c
      | c = rearranging
}}
{{eqn | r = \AA^2 r \rho_a \rho_b \rho_c
      | c = Heron's Formula
}}
{{eqn | ll= \leadsto
      | l = \AA^2
      | r = r \rho_a \rho_b \rho_c
}}
{{eqn | l = \AA
      | r = \sqrt {r \rho_a \rho_b \rho_c}
}}
{{end-eqn}}
{{Qed}}
Theorem: Stewart's Theorem

Let $\triangle ABC$ be a triangle with sides $a, b, c$.
Let $CP$ be a cevian from $C$ to $P$.
:400px
Then:
:$a^2 \cdot AP + b^2 \cdot PB = c \paren {CP^2 + AP \cdot PB}$

Proof:
{{begin-eqn}}
{{eqn | n = 1
      | l = b^2
      | r = AP^2 + CP^2 - 2 AP \cdot CP \cdot \map \cos {\angle APC}
      | c = Law of Cosines
}}
{{eqn | n = 2
      | l = a^2
      | r = PB^2 + CP^2 - 2 CP \cdot PB \cdot \map \cos {\angle BPC}
      | c = Law of Cosines
}}
{{eqn | r = PB^2 + CP^2 + 2 CP \cdot PB \cdot \map \cos {\angle APC}
      | c = Cosine of Supplementary Angle
}}
{{eqn | n = 3
      | ll= \leadsto \quad
      | l = b^2 \cdot PB
      | r = AP^2 \cdot PB + CP^2 \cdot PB - 2 PB \cdot AP \cdot CP \cdot \map \cos {\angle APC}
      | c = $(1) \times PB$
}}
{{eqn | n = 4
      | l = a^2 \cdot AP
      | r = PB^2 \cdot AP + CP^2 \cdot AP + 2 AP \cdot CP \cdot PB \cdot \map \cos {\angle APC}
      | c = $(2) \times AP$
}}
{{eqn | ll= \leadsto \quad
      | l = a^2 \cdot AP + b^2 \cdot PB
      | r = AP^2 \cdot PB + PB^2 \cdot AP + CP^2 \cdot PB + CP^2 \cdot AP
      | c = $(3) + (4)$
}}
{{eqn | r = CP^2 \paren {PB + AP} + AP \cdot PB \paren {PB + AP}
}}
{{eqn | r = c \paren {CP^2 + AP \cdot PB}
      | c = as $PB + AP = c$
}}
{{end-eqn}}
{{qed}}
Theorem: Median Formula

Let $\triangle ABC$ be a triangle.
Let $CD$ be the median of $\triangle ABC$ which bisects $AB$.
:400px
The length $m_c$ of $CD$ is given by:
:${m_c}^2 = \dfrac {a^2 + b^2} 2 - \dfrac {c^2} 4$

Proof:
{{begin-eqn}}     
{{eqn | l = a^2 \cdot AD + b^2 \cdot DB
      | r = CD^2 \cdot c + AD \cdot DB \cdot c
      | c = Stewart's Theorem
}}
{{eqn | ll= \leadsto
      | l = a^2 \frac c 2 + b^2 \frac c 2
      | r = {m_c}^2 \cdot c + \paren {\frac c 2}^2 c
      | c = substituting $AD = DB = \dfrac c 2$ and $CD = m_c$
}}
{{eqn | ll= \leadsto
      | l = \frac c 2 \paren {a^2 + b^2}
      | r = m_c^2 \cdot c + \frac {c^2} 4 \cdot c
}}
{{eqn | ll= \leadsto
      | l = \frac {a^2 + b^2} 2 
      | r = m_c^2 + \frac {c^2} 4
}}
{{eqn | ll= \leadsto
      | l = {m_c}^2 
      | r = \frac {a^2 + b^2} 2 - \frac {c^2} 4
      | c = after algebra
}}
{{end-eqn}}
{{qed}}
Theorem: Median Formula

Let $\triangle ABC$ be a triangle.
Let $CD$ be the median of $\triangle ABC$ which bisects $AB$.
:400px
The length $m_c$ of $CD$ is given by:
:${m_c}^2 = \dfrac {a^2 + b^2} 2 - \dfrac {c^2} 4$

Proof:
Let $\triangle ABC$ be embedded in the complex plane.
:300px
Let $\mathbf a = \overrightarrow {AC}$ and $\mathbf b = \overrightarrow {BC}$.
Then:
{{begin-eqn}}
{{eqn | l = \overrightarrow {AB}
      | r = \mathbf a - \mathbf b
      | c = 
}}
{{eqn | l = \overrightarrow {AD}
      | r = \dfrac {\overrightarrow {AB} } 2
      | c = 
}}
{{eqn | r = \dfrac {\mathbf a - \mathbf b} 2
      | c = 
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \overrightarrow {AC} + \overrightarrow {CD}
      | r = \overrightarrow {AD}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \overrightarrow {CD}
      | r = \overrightarrow {AD} - \overrightarrow {AC}
      | c = 
}}
{{eqn | r = \frac {\mathbf a - \mathbf b} 2 - \mathbf a
      | c = 
}}
{{eqn | r = -\frac {\mathbf a + \mathbf b} 2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = {m_c}^2
      | r = \size {\overrightarrow {CD} }^2
}}
{{eqn | r = \size {-\frac {\mathbf a + \mathbf b} 2}^2
}}
{{eqn | r = \frac 1 4 \paren {\paren {\mathbf a + \mathbf b} \cdot \paren {\mathbf a + \mathbf b} }
      | c = Dot Product of Vector with Itself
}}
{{eqn | r = \frac 1 4 \paren {\mathbf a \cdot \mathbf a + \mathbf b \cdot \mathbf b + 2 \mathbf a \cdot \mathbf b}
      | c = Square of Sum of Vectors
}}
{{eqn | r = \frac 1 4 \paren {\mathbf a \cdot \mathbf a + \mathbf b \cdot \mathbf b - \paren {\mathbf a - \mathbf b} \cdot \paren {\mathbf a - \mathbf b} + \mathbf a \cdot \mathbf a + \mathbf b \cdot \mathbf b}
      | c = Square of Sum of Vectors
}}
{{eqn | r = \frac 1 4 \paren {2 \size {\mathbf a}^2 + 2 \size {\mathbf b}^2 - \size {\mathbf a - \mathbf b}^2}
      | c = Dot Product of Vector with Itself
}}
{{eqn | r = \frac 1 4 \paren {2 a^2 + 2 b^2 - c^2}
}}
{{eqn | r = \frac {a^2 + b^2} 2 - \frac {c^2} 4
}}
{{end-eqn}}
{{qed}}
Theorem: Median Formula

Let $\triangle ABC$ be a triangle.
Let $CD$ be the median of $\triangle ABC$ which bisects $AB$.
:400px
The length $m_c$ of $CD$ is given by:
:${m_c}^2 = \dfrac {a^2 + b^2} 2 - \dfrac {c^2} 4$

Proof:
{{begin-eqn}}     
{{eqn | l = {m_c}^2
      | r = b^2 + \paren {\frac c 2}^2 - 2 b \paren {\frac c 2} \cos A
      | c = Law of Cosines
}}
{{eqn | r = b^2 + \frac {c^2} 4 - 2 b \paren {\frac c 2} \paren {\frac {b^2 + c^2 - a^2} {2 b c} }
      | c = Law of Cosines
}}
{{eqn | r = b^2 + \frac {c^2} 4 - \frac {b^2 + c^2 - a^2} 2
      | c = 
}}
{{eqn | r = \frac {a^2} 2 + \frac {b^2} 2 - \frac {c^2} 4
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Length of Angle Bisector

Let $\triangle ABC$ be a triangle.
Let $AD$ be the angle bisector of $\angle BAC$ in $\triangle ABC$.
:300px
Let $d$ be the length of $AD$.
Then $d$ is given by:
:$d^2 = \dfrac {b c} {\paren {b + c}^2} \paren {\paren {b + c}^2 - a^2}$
where $a$, $b$, and $c$ are the sides opposite $A$, $B$ and $C$ respectively.

Proof:
{{begin-eqn}}
{{eqn | l = \frac {BD} {DC} 
      | r = \frac c b
      | c = Angle Bisector Theorem
}}
{{eqn | ll= \leadsto
      | l = \frac {BD} {DC} + 1
      | r = \frac c b + 1
}}
{{eqn | ll= \leadsto
      | l = \frac {BD + DC} {DC}
      | r = \frac {b + c} b
}}
{{eqn | ll= \leadsto
      | l = \frac a {DC}
      | r = \frac {b + c} b
}}
{{eqn | ll= \leadsto
      | l = DC
      | r = \frac {a b} {b + c}
}}
{{end-eqn}}
Similarly, or by symmetry, we get:
:$BD = \dfrac {a c} {b + c}$
From Stewart's Theorem, we have:
:$b^2 \cdot BD + c^2 \cdot DC = d^2 \cdot a + BD \cdot DC \cdot a$
Substituting the above expressions for $BD$ and $DC$:
{{begin-eqn}}
{{eqn | l = b^2 \dfrac {a c} {b + c} + c^2 \frac {a b} {b + c}
      | r = d^2 \cdot a + \dfrac {a c} {b + c} \cdot \frac {a b} {b + c} \cdot a
}}
{{eqn | ll= \leadsto
      | l = a b c \frac {b + c} {b + c}
      | r = d^2 \cdot a + \frac{a^2 b c} {\paren {b + c}^2} \cdot a
}}
{{eqn | ll= \leadsto
      | l = b c
      | r = d^2 + \frac {a^2 b c} {\paren {b + c}^2}
}}
{{eqn | ll= \leadsto
      | l = d^2
      | r = b c \paren {1 - \frac {a^2} {\paren {b + c}^2} }
}}
{{eqn | ll= \leadsto
      | l = d^2
      | r = \frac {b c} {\paren {b + c}^2} \paren {\paren {b + c}^2 - a^2}
}}
{{end-eqn}}
{{qed}}
Theorem: Length of Angle Bisector

Let $\triangle ABC$ be a triangle.
Let $AD$ be the angle bisector of $\angle BAC$ in $\triangle ABC$.
:300px
Let $d$ be the length of $AD$.
Then $d$ is given by:
:$d^2 = \dfrac {b c} {\paren {b + c}^2} \paren {\paren {b + c}^2 - a^2}$
where $a$, $b$, and $c$ are the sides opposite $A$, $B$ and $C$ respectively.

Proof:
:400px
From Length of Angle Bisector: Proof 1, we have:
:$BD = \dfrac {a c} {b + c}$
:$DC = \dfrac {a b} {b + c}$
Then we have:
{{begin-eqn}}
{{eqn | l = \angle BAD
      | o = \cong
      | r = \angle FAC
      | c = {{Defof|Angle Bisector}}
}}
{{eqn | l = \angle ABD
      | o = \cong
      | r = \angle AFC
      | c = Angles in Same Segment of Circle are Equal
}}
{{end-eqn}}
Then from Triangles with Two Equal Angles are Similar we have:
:$\triangle ABD \sim \triangle AFC$
So:
{{begin-eqn}}
{{eqn | l = \frac c d
      | r = \frac {AF} b
      | c = as $\triangle ABD$ and $\triangle AFC$ are similar
}}
{{eqn | ll= \leadsto
      | l = \frac c d
      | r = \frac {d + DF} b
}}
{{end-eqn}}
Now we use the Intersecting Chords Theorem, which gives us $BD \cdot DC = d \cdot DF$.
{{begin-eqn}}
{{eqn | l = \frac c d
      | r = \frac {d + \frac {BD \cdot DC} d} b
}}
{{eqn | ll= \leadsto
      | l = b c
      | r = d^2 + BD \cdot DC
}}
{{eqn | ll= \leadsto
      | l = d^2
      | r = b c - BD \cdot DC
}}
{{eqn | ll= \leadsto
      | l = d^2
      | r = b c - \frac {a c} {b + c} \cdot \frac {a b} {b + c}
}}
{{eqn | ll= \leadsto
      | l = d^2
      | r = \frac {b c} {\paren {b + c}^2} \paren {\paren {b + c}^2 - a^2}
}}
{{end-eqn}}
{{qed}}
Theorem: Supremum of Subset

Let $\left({U, \preceq}\right)$ be an ordered set.
Let $S \subseteq U$.
Let $T \subseteq S$.
Let $S$ admit a supremum (in $U$).
If $T$ also admits a supremum (in $U$), then $\sup \left({T}\right) \preceq\sup \left({S}\right)$.

Proof:
The number $\sup S$ is an upper bound for $S$.
Therefore, $\sup S$ is an upper bound for $T$ as $T$ is a non-empty subset of $S$.
Accordingly, $T$ has a supremum by the Continuum Property.
The number $\sup S$ is an upper bound for $T$.
Therefore, $\sup S$ is greater than or equal to $\sup T$ as $\sup T$ is the least upper bound of $T$.
{{qed}}
Theorem: Supremum of Subset

Let $\left({U, \preceq}\right)$ be an ordered set.
Let $S \subseteq U$.
Let $T \subseteq S$.
Let $S$ admit a supremum (in $U$).
If $T$ also admits a supremum (in $U$), then $\sup \left({T}\right) \preceq\sup \left({S}\right)$.

Proof:
By the Continuum Property, $T$ admits a supremum.
It follows from Supremum of Subset that $\sup T \le \sup S$.
{{qed}}
Theorem: Supremum of Subset

Let $\left({U, \preceq}\right)$ be an ordered set.
Let $S \subseteq U$.
Let $T \subseteq S$.
Let $S$ admit a supremum (in $U$).
If $T$ also admits a supremum (in $U$), then $\sup \left({T}\right) \preceq\sup \left({S}\right)$.

Proof:
$S$ is bounded above as $S$ has a supremum.
Therefore, $T$ is bounded above as $T$ is a subset of $S$.
Accordingly, $T$ admits a supremum by the Continuum Property as $T$ is non-empty.
We know that $\sup T$ and $\sup S$ exist.
Therefore by Suprema of two Real Sets:
:$\forall \epsilon \in \R_{>0}: \forall t \in T: \exists s \in S: t < s + \epsilon \iff \sup T \le \sup S$
We have:
{{begin-eqn}}
{{eqn | q = \forall \epsilon \in \R_{>0}
      | l = 0
      | o = <
      | r = \epsilon
}}
{{eqn | ll= \leadsto
      | q = \forall \epsilon \in \R_{>0}: \forall t \in T
      | l = t
      | o = < 
      | r = t + \epsilon
}}
{{eqn | ll= \leadsto
      | q = \forall \epsilon \in \R_{>0}: \forall t \in T
      | l = t
      | o = <
      | r = s + \epsilon
}}
{{eqn | lo= \land
      | l = s
      | r = t
}}
{{eqn | ll= \leadsto
      | q = \forall \epsilon \in \R_{>0}: \forall t \in T: \exists s \in S
      | l = t
      | o = <
      | r = s + \epsilon
      | c = as $T \subseteq S$
}}
{{eqn | ll= \leadsto
      | l = \sup T
      | o = \le
      | r = \sup S
}}
{{end-eqn}}
{{qed}}
Theorem: Supremum of Subset

Let $\left({U, \preceq}\right)$ be an ordered set.
Let $S \subseteq U$.
Let $T \subseteq S$.
Let $S$ admit a supremum (in $U$).
If $T$ also admits a supremum (in $U$), then $\sup \left({T}\right) \preceq\sup \left({S}\right)$.

Proof:
By definition $\sup S$ is an upper bound for $S$.
Thus:
:$\forall x \in S: x \le \sup S$
As $T \subseteq S$ we have by definition of subset that:
:$\forall x \in T: x \in S$
Hence:
:$\forall x \in T: x \le \sup S$
So by definition $\sup S$ is an upper bound for $T$.
So $\sup S$ is at least as big as the smallest upper bound for $T$
Thus by definition of supremum:
:$\sup T \le \sup S$
{{qed}}
Theorem: Supremum of Subset

Let $\left({U, \preceq}\right)$ be an ordered set.
Let $S \subseteq U$.
Let $T \subseteq S$.
Let $S$ admit a supremum (in $U$).
If $T$ also admits a supremum (in $U$), then $\sup \left({T}\right) \preceq\sup \left({S}\right)$.

Proof:
Let $B = \sup \left({S}\right)$.
Then $B$ is an upper bound for $S$.
As $T \subseteq S$, it follows by the definition of a subset that $x \in T \implies x \in S$.
Because $x \in S \implies x \preceq B$ (as $B$ is an upper bound for $S$) it follows that $x \in T \implies x \preceq B$.
So $B$ is an upper bound for $T$.
Therefore $B$ succeeds the supremum of $T$ in $S$.
Hence the result.
{{qed}}
Category:Order Theory
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Theorem: Area of Triangle

This page gathers a variety of formulas for the area of a triangle.

Proof:
The point $\tuple {x = 0, y = 0}$ satisfies $y = x^2$.
Since $x^2 \ge 0$, $0$ is the minimum value for $y$.
Thus, the vertex of $T$ lies at the origin.
The coordinates of $C$ are given as the average those for $A$ and $B$.
Given:
:$A = \tuple {u, u^2}$
:$B = \tuple {v, v^2}$, with $u > v$
By the definition of average: 
:$C = \tuple {\dfrac 1 2 \paren {u + v}, \dfrac 1 4 \paren {u + v}^2 } $
Let $P$ be the midpoint of $AB$, such that $AP = BP$.
Let the coordinates of $P$ be $\tuple {P_x, P_y}$.
By the Midline Theorem:
:$P_x = \dfrac 1 2 \paren {u + v}$
Also by the Midline Theorem:
:$P_y = \dfrac 1 2 \paren {u^2 + v^2}$
Let $h$ be one-half the distance between the $x$-coordinates of $A$ and $B$:
:$h = \dfrac 1 2 \paren {u - v}$
By construction:
:the height of both $\triangle APC$ and $\triangle BPC$ is equal to $h$.
Let $b$ be the length of $PC$, the shared base of $\triangle APC$ and $\triangle BPC$.
{{begin-eqn}} 
{{eqn | l = b
      | r = P_y - C_y
      | c = Definition above
}}
{{eqn | l = b
      | r = \dfrac 1 2 \paren {u^2 + v^2} - \dfrac 1 4 \paren {u + v}^2
      | c = Substitution
}}
{{eqn | l = b
      | r = \dfrac 1 4 \paren {2 u^2 + 2 v^2 - \paren {u + v}^2}
      | c = Factor out $\dfrac 1 2$
}}
{{eqn | l = b
      | r = \dfrac 1 4 \paren {u^2 - 2 u v + v^2}
      | c = Expansion and subtraction
}}
{{eqn | l = b
      | r = \dfrac 1 4 \paren {u - v}^2
      | c = Factor
}}
{{end-eqn}}
{{:Euclid:Proposition/I/38}}
By Area of Triangle:
:$\map \AA {\triangle APC} = \dfrac 1 2 h b$
Substituting from above:
:$\map \AA {\triangle APC} = \dfrac 1 {16} \paren {u - v}^3$
Since they have the same height and a shared base:
:$\map \AA {\triangle APC} = \map \AA {\triangle BPC}$
But:
:$\map \AA {\triangle ABC} = \map \AA {\triangle APC} + \map \AA {\triangle BPC}$
By addition:
:$\map \AA {\triangle ABC} = h b = \dfrac 1 8 \paren {u - v}^3$
{{qed}}
Theorem: Area of Triangle

This page gathers a variety of formulas for the area of a triangle.

Proof:
{{begin-eqn}}
{{eqn | l = \map \Area T
      | r = \dfrac 1 2 \size {\paren {\begin {vmatrix} 0 & 0 & 1 \\ b & a & 1 \\ x & y & 1 \end {vmatrix} } }
      | c = Area of Triangle in Determinant Form
}}
{{eqn | r = \dfrac 1 2 \size {b y - a x}
      | c = Determinant of Order 3
}}
{{end-eqn}}
{{qed}}
Theorem: Area of Triangle

This page gathers a variety of formulas for the area of a triangle.

Proof:
Let the polar coordinates of $B$ and $C$ be:
{{begin-eqn}}
{{eqn | l = B
      | r = \polar {r_1, \theta_1}
}}
{{eqn | l = C
      | r = \polar {r_2, \theta_2}
}}
{{end-eqn}}
Let $\theta$ be the angle between $AB$ and $AC$.
Then we have:
{{begin-eqn}}
{{eqn | l = \map \Area {\triangle ABC}
      | r = \dfrac 1 2 AB \cdot AC \sin \theta
      | c = Area of Triangle in Terms of Two Sides and Angle
}}
{{eqn | r = \dfrac 1 2 r_1 r_2 \map \sin {\theta_2 - \theta_1}
      | c = Definition of $\theta$
}}
{{eqn | r = \dfrac 1 2 r_1 r_2 \paren {\sin \theta_2 \cos \theta_1 - \cos \theta_2 \sin \theta_1}
      | c = Sine of Difference
}}
{{eqn | r = \dfrac 1 2 \paren {r_1 \cos \theta_1 r_2 \sin \theta_2 - r_1 \sin \theta_1 r_2 \cos \theta_2}
      | c = rearranging
}}
{{eqn | r = \dfrac 1 2 \paren {b y - a x}
      | c = rearranging
}}
{{end-eqn}}
We can define the area of $\triangle ABC$ as being positive or negative according to the sign of $\dfrac 1 2 \paren {b y - a x}$.
However, if we are interested only in the absolute value of $\triangle ABC$, as in this context, we can report:
:$\map \Area {\triangle ABC} = \dfrac {\size {b y - a x} } 2$
{{qed}}
Theorem: Area of Triangle

This page gathers a variety of formulas for the area of a triangle.

Proof:
:400px
Let $A$, $B$ and $C$ be defined as complex numbers in the complex plane.
The vectors from $C$ to $A$ and from $C$ to $B$ are given by:
:$z_1 = \paren {x_1 - x_3} + i \paren {y_1 - y_3}$
:$z_2 = \paren {x_2 - x_3} + i \paren {y_2 - y_3}$
From Area of Triangle in Terms of Side and Altitude, $\AA$ is half that of a parallelogram contained by $z_1$ and $z_2$.
Thus:
{{begin-eqn}}
{{eqn | l = \AA
      | r = \frac 1 2 z_1 \times z_2
      | c = Area of Parallelogram in Complex Plane
}}
{{eqn | r = \frac 1 2 \size {\paren {\map \Im {\paren {x_1 - x_3} - i \paren {y_1 - y_3} } \paren {\paren {x_2 - x_3} - i \paren {y_2 - y_3} } } }
      | c = {{Defof|Vector Cross Product|subdef = Complex|index = 3}}
}}
{{eqn | r = \frac 1 2 \size {\paren {x_1 - x_3} \paren {y_2 - y_3} - \paren {y_1 - y_3} \paren {x_2 - x_3} }
      | c = {{Defof|Complex Multiplication}}
}}
{{eqn | r = \frac 1 2 \size {x_1 y_2 - y_1 x_2 + x_2 y_3 - y_2 x_3 + x_3 y_1 - y_3 x_1}
      | c = multiplying out
}}
{{eqn | r = \frac 1 2 \size {\paren {\begin {vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \\ \end {vmatrix} } }
      | c = Determinant of Order 3
}}
{{end-eqn}}
{{qed}}
Theorem: Area of Triangle

This page gathers a variety of formulas for the area of a triangle.

Proof:
:400px
Let $A$, $B$ and $C$ be as defined..
Let $O$ be the origin of the Cartesian plane in which $\triangle ABC$ is embedded.
Taking into account the signs of the areas of the various triangles involved:
:$\triangle ABC = \triangle OAB + \triangle OBC + \triangle OCA$
as it is seen that $\triangle OBC$ and $\triangle OCA$ are described in clockwise sense.
From proof 2 of Area of Triangle in Determinant Form with Vertex at Origin:
{{begin-eqn}}
{{eqn | l = \triangle OAB
      | r = \dfrac 1 2 \paren {x_1 y_2 - x_2 y_1}
}}
{{eqn | l = \triangle OBC
      | r = \dfrac 1 2 \paren {x_2 y_3 - x_3 y_2}
}}
{{eqn | l = \triangle OBC
      | r = \dfrac 1 2 \paren {x_3 y_1 - x_1 y_3}
}}
{{eqn | ll= \leadsto
      | l = \triangle ABC
      | r = \dfrac 1 2 \paren {\paren {x_1 y_2 - x_2 y_1} + \paren {x_2 y_3 - x_3 y_2} + \paren {x_3 y_1 - x_1 y_3} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \triangle ABC
      | r = \frac 1 2 \paren {\begin {vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \\ \end {vmatrix} }
      | c = Determinant of Order 3
}}
{{end-eqn}}
The result follows.
{{qed}}
Theorem: Area of Triangle

This page gathers a variety of formulas for the area of a triangle.

Proof:
:400px
Let $A$, $B$ and $C$ be defined as $\tuple {x_1, y_1}$, $\tuple {x_2, y_2}$ and $\tuple {x_3, y_3}$ respectively.
From the figure, we see that:
{{begin-eqn}}
{{eqn | l = \map \Area {ABC}
      | r = \map \Area {PACR} + \map \Area {RCBQ} - \map \Area {PABQ}
      | c = 
}}
{{eqn | r = \dfrac {\paren {x_3 - x_1} \paren {y_3 + y_1} } 2 + \dfrac {\paren {x_2 - x_3} \paren {y_2 + y_3} } 2 - \dfrac {\paren {x_2 - x_1} \paren {y_1 + y_2} } 2
      | c = Area of Trapezium
}}
{{eqn | r = \dfrac {x_1 y_2 - x_2 y_1 + x_2 y_3 - x_3 y_2 + x_3 y_1 - x_1 y_3} 2
      | c = simplification
}}
{{eqn | r = \dfrac 1 2 \size {\paren {\begin {vmatrix}  x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \\ \end {vmatrix} } }
      | c = Determinant of Order 3
}}
{{end-eqn}}
{{qed}}
Theorem: Area of Triangle

This page gathers a variety of formulas for the area of a triangle.

Proof:
:500px
Let $C$ be the excircle of $\triangle ABC$ which is tangent to $a$.
By definition:
:$\rho_a$ is the radius of $C$
:$I_a$ is the center of $C$.
Then we have:
{{begin-eqn}}
{{eqn | l = \AA
      | r = \map \Area {\triangle ABI_a} + \map \Area {\triangle ACI_a} - \map \Area {\triangle CBI_a}
      | c = (see figure above)
}}
{{eqn | r = \frac {c \rho_a} 2 + \frac {b \rho_a} 2 - \frac {a \rho_a} 2
      | c = Area of Triangle in Terms of Side and Altitude
}}
{{eqn | r = \rho_a \frac {b + c + a} 2 - \rho_a a
}}
{{eqn | r = \rho_a s - \rho_a a
      | c = {{Defof|Semiperimeter}}
}}
{{eqn | l = (ABC)
      | r = \rho_a \paren {s - a}
}}
{{end-eqn}}
{{qed}}
Theorem: Area of Triangle

This page gathers a variety of formulas for the area of a triangle.

Proof:
:400px
Construct a point $D$ so that $\Box ABDC$ is a parallelogram.
From Opposite Sides and Angles of Parallelogram are Equal:
:$\triangle ABC \cong \triangle DCB$
hence their areas are equal.
The Area of Parallelogram is equal to the product of one of its bases and the associated altitude.
Thus 
{{begin-eqn}}
{{eqn | l = \paren {ABCD}
      | r = c \cdot h_c
}}
{{eqn | ll= \leadsto
      | l = 2 \paren {ABC}
      | r = c \cdot h_c
      | c = because congruent surfaces have equal areas
}}
{{eqn | l = \paren {ABC}
      | r = \frac {c \cdot h_c} 2
}}
{{end-eqn}}
where $\paren {XYZ}$ is the area of the plane figure $XYZ$.
A similar argument can be used to show that the statement holds for the other sides.
{{qed}}