avewright/proofwiki-math huggingface dataset:
Theorem: Space of Piecewise Linear Functions on Closed Interval is Dense in Space of Continuous Functions on Closed Interval

Let $I = \closedint a b$. 
Let $\map \CC I$ be the set of continuous functions on $I$. 
Let $\map {\mathrm {PL} } I$ be the set of piecewise linear functions on $I$. 
Let $d$ be the metric induced by the supremum norm.
Then $\map {\mathrm {PL} } I$ is dense in $\struct {\map \CC I, d}$.

Proof:
Let $f \in \map \CC I$. 
Let $\epsilon \in \R_{>0}$ be a real number.
From Open Ball Characterization of Denseness:
:it suffices to find a $p \in \map {\mathrm {PL} } I$ such that $p$ is contained in the open ball $\map {B_\epsilon} f$.
From Continuous Function on Closed Real Interval is Uniformly Continuous:
:$f$ is uniformly continuous on $I$.
That is: 
:there exists a $\delta > 0$ such that for all $x, y \in I$ with $\size {x - y} < \delta$ we have $\size {\map f x - \map f y} < \epsilon/3$.
Let:
:$P = \{a_0 = a, a_1, a_2, \ldots, a_n = b\}$
be a finite subdivision of $I$, with: 
:$\size {a_{i + 1} - a_i} < \delta$
for each $i$.
Let $p \in \map {\mathrm {PL} } I$ be such that: 
:$\map p {a_i} = \map f {a_i}$
for each $i$, with $p$ continuous. 
We can explicitly construct such a $p$ by connecting $\tuple {a_i, \map f {a_i} }$ to $\tuple {a_{i + 1}, \map f {a_{i + 1} } }$ with a straight line segment for each $i$.
Fix $x \in I$.
Note that there exists precisely one $i$ such that $a_i \le x \le a_{i + 1}$, fix this $i$.
We then have: 
{{begin-eqn}}
{{eqn	| l = \size {\map p x - \map f {a_i} }
	| r = \size {\map p x - \map p {a_i} }
}}
{{eqn	| o = \le
	| r = \size {\map p {a_{i + 1} } - \map p {a_i} }
}}
{{eqn	| r = \size {\map f {a_{i + 1} } - \map f {a_i} }
}}
{{eqn	| o = <
	| r = \epsilon/3
}}
{{end-eqn}}
since $\size {a_{i + 1} - a_i} < \delta$.
Since $\size {x - a_i} < \size {a_{i + 1} - a_i} < \delta$, we also have: 
:$\size {\map f x - \map f {a_i} } < \epsilon/3$
So:
{{begin-eqn}}
{{eqn	| l = \size {\map p x - \map f x}
	| o = \le
	| r = \size {\map p x - \map p {a_i} } + \size {\map f x - \map f {a_i} }
	| c = Triangle Inequality
}}
{{eqn	| o = <
	| r = 2 \epsilon/3
}}
{{end-eqn}}
Note that $x \in I$ was arbitrary, so:
{{begin-eqn}}
{{eqn	| l = \map d {f, p}
	| r = \norm {f - p}_\infty
	| c = {{Defof|Metric Induced by Norm}}
}}
{{eqn	| r = \sup_{x \mathop \in I} \size {\map f x - \map p x}
}}
{{eqn	| o = \le
	| r = 2 \epsilon/3
}}
{{eqn	| o = <
	| r = \epsilon
}}
{{end-eqn}}
so $p \in \map {B_\epsilon} f$.
{{qed}}
Category:Functional Analysis
p0z4b6yktt2olzz5r5o6dt2lvh1m62u
Theorem: Legendre Symbol is Multiplicative

Let $p$ be a odd prime.
Let $a, b \in \Z$.
Then:
:$\paren {\dfrac {a b} p} = \paren {\dfrac a p} \paren {\dfrac b p}$
where $\paren {\dfrac a p}$ is the Legendre symbol.

Proof:
We have: 
{{begin-eqn}}
{{eqn	| l = \paren {\frac {a b} p}
	| r = \paren {a b}^{\frac {p - 1} 2} \bmod p
	| c = {{Defof|Legendre Symbol|index = 2}}
}}
{{eqn	| r = a^{\frac {p - 1} 2} b^{\frac {p - 1} 2} \bmod p
        | c = Power of Product
}}
{{eqn	| r = \paren {\frac a p} \paren {\frac b p} 
	| c = {{Defof|Legendre Symbol|index = 2}}, Congruence of Product
}}
{{end-eqn}}
{{qed}}
Category:Legendre Symbol
dini0s4q7m5l177a6ec2bexy9ibia9n
Theorem: Legendre Symbol of Congruent Integers

Let $p$ be a odd prime.
Let $a, b \in \Z$ be such that $a \equiv b \pmod p$.
Then:
:$\paren {\dfrac a p} = \paren {\dfrac b p}$
where $\paren {\dfrac a p}$ is the Legendre symbol.

Proof:
{{begin-eqn}}
{{eqn	| l = \paren {\frac a p}
	| r = a^{\frac {p - 1} 2} \bmod p
	| c = {{Defof|Legendre Symbol|index = 2}}
}}
{{eqn	| r = b^{\frac {p - 1} 2} \bmod p
	| c = Congruence of Powers
}}
{{eqn	| r = \paren {\frac b p}
	| c = {{Defof|Legendre Symbol|index = 2}}
}}
{{end-eqn}}
{{qed}}
Category:Legendre Symbol
i8i4cwjfhwclfvd86kdogwtcurm3xcl
Theorem: Derivative of Reciprocal

Let $f: \R_{\ne 0} \to \R$ be the reciprocal function defined as:
:$\map f x = \dfrac 1 x$
Then its derivative is given by:
:$\map {f'} x = -\dfrac 1 {x^2}$

Proof:
We have:
{{begin-eqn}}
{{eqn | l = \dfrac 1 x
      | r = x^{-1}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map {\dfrac \d {\d x} } {\dfrac 1 x}
      | r = \paren {-1} x^{-2}
      | c = Power Rule for Derivatives
}}
{{eqn | r = -\dfrac 1 {x^2}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Distance between Two Points in Plane in Polar Coordinates

Let $A = \polar {r_1, \theta_1}$ and $B = \polar {r_2, \theta_2}$ be two points in a polar coordinate plane
The distance $d$ between $A$ and $B$ is given by:
:$d = \sqrt {r_1^2 + r_2^2 + 2 r_1 r_2 \map \cos {\theta_1 - \theta_2} }$

Proof:
Let $A$ and $B$ be embedded as suggested in a polar coordinate plane whose pole is at $O$.
:320px
The distance $d$ is the side $AB$ of the triangle $AOB$.
We have that:
:$OA = r_1$
:$OB = r_2$
and:
:$\theta_2 - \theta_1$ is the opposite angle to $AB$.
Hence we can use the Cosine Rule:
:$AB^2 = r_1^2 + r_2^2 - 2 r_1 r_2 \map \cos {\theta_2 - \theta_1}$
From Cosine Function is Even we have that:
:$\map \cos {\theta_2 - \theta_1} = \map \cos {\theta_1 - \theta_2}$
and the result follows.
{{qed}}
Theorem: Distance between Two Points in Plane in Polar Coordinates

Let $A = \polar {r_1, \theta_1}$ and $B = \polar {r_2, \theta_2}$ be two points in a polar coordinate plane
The distance $d$ between $A$ and $B$ is given by:
:$d = \sqrt {r_1^2 + r_2^2 + 2 r_1 r_2 \map \cos {\theta_1 - \theta_2} }$

Proof:
From Conversion between Cartesian and Polar Coordinates in Plane, $A$ and $B$ can be expressed in Cartesian coordinates as follows:
{{begin-eqn}}
{{eqn | l = x_1
      | r = r_1 \cos \theta_1
}}
{{eqn | l = y_1
      | r = r_1 \sin \theta_1
}}
{{eqn | l = x_2
      | r = r_2 \cos \theta_2
}}
{{eqn | l = y_2
      | r = r_2 \sin \theta_2
}}
{{end-eqn}}
Thus the distance $d$ between $A = \tuple {x_1, y_1}$ and $B = \tuple {x_2, y_2}$ is given by:
{{begin-eqn}}
{{eqn | l = d^2
      | r = \paren {x_1 - x_2}^2 + \paren {y_1 - y_2}^2
      | c = Distance Formula
}}
{{eqn | r = \paren {r_1 \cos \theta_1 - r_2 \cos \theta_2}^2 + \paren {r_1 \sin \theta_1 - r_2 \sin \theta_2}^2
      | c = 
}}
{{eqn | r = r_1^2 \paren {\cos^2 \theta_1 + \sin^2 \theta_1} + r_2^2 \paren {\cos^2 \theta_2 + \sin^2 \theta_2} + 2 r_1 r_2 \paren {\cos \theta_1 \cos \theta_2 + \sin \theta_1 \sin \theta_2}
      | c = 
}}
{{eqn | r = r_1^2 + r_2^2 + 2 r_1 r_2 \paren {\cos \theta_1 \cos \theta_2 + \sin \theta_1 \sin \theta_2}
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = r_1^2 + r_2^2 + 2 r_1 r_2 \map \cos {\theta_1 - \theta_2}
      | c = Cosine of Difference
}}
{{end-eqn}}
The result follows.
{{qed}}
Theorem: Linear Mappings between Vector Spaces form Vector Space

Let $\struct {F, +, \times}$ be a field whose unity is $1_F$.
Let $X, Y$ be vector spaces over the same field $\struct {F, +, \times}$.
Let $\map \LL {X, Y}$ be the set of linear mappings.
Let $x \in X$.
Define pointwise addition $T + S \in \map \LL {X, Y}$ such that:
:$\forall x \in X: \map {\paren {T + S} } x := \map T x + \map S x$
For $\alpha \in F$, define pointwise scalar multiplication $\alpha \cdot T$ such that:
:$\forall x \in X: \map {\paren {\alpha \cdot T} } x := \alpha \cdot \map T x$
Let $\mathbf 0: X \to Y$ be the zero mapping.
Then $\struct {\map \LL {X, Y}, +, \, \cdot \,}$ is a vector space over $\struct {F, +, \times}$.

Proof:
Let $T, S, P \in \map \LL {X, Y}$ such that:
:$T, S, P: X \to Y$
Let $\lambda, \mu \in F$.
Theorem: Exponential of a x by Cosine of b x and Exponential of a x by Sine of b x are Linearly Independent

Let $x, a, b \in \R$.
Then $e^{a x} \map \cos {b x}$ and $e^{b x} \map \sin {b x}$ are linearly independent.

Proof:
Let $\alpha_1, \alpha_2 \in \R$ be such that:
:$\alpha_1 e^{a x} \map \cos {b x} + \alpha_2 e^{a x} \map \sin {b x} = 0$
Suppose $x = 0$.
Then $\alpha_1 = 0$.
Suppose $\ds x = \frac \pi {2 b}$.
Then:
:$\ds \alpha_2 \map \exp {\frac {\pi a} {2 b}} = 0$
Hence, $\alpha_2 = 0$.
The conclusion follows from the definition of linearly independent real functions.
{{qed}}
Theorem: Difference of Squares of Sines

:$\sin^2 A - \sin^2 B = \map \sin {A + B} \map \sin {A - B}$

Proof:
{{begin-eqn}}
{{eqn | l = \sin^2 A - \sin^2 B 
      | r = \sin^2 A - \sin^2 B + 0
      | c =
}}
{{eqn | r = \sin^2 A - \sin^2 B + \paren {\sin^2 A \sin^2 B - \sin^2 A \sin^2 B}
      | c =
}}
{{eqn | r = \sin^2 A \paren {1 - \sin^2 B} - \sin^2 B \paren {1 - \sin^2 A}
      | c =
}}
{{eqn | r = \sin^2 A \paren {\cos^2 B} - \sin^2 B \paren {\cos^2 A}
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = \sin^2 A \cos^2 B - \sin^2 B \cos^2 A + 0
      | c =
}}
{{eqn | r = \sin^2 A \cos^2 B - \sin^2 B \cos^2 A + \paren {\sin A \sin B \cos A \cos B - \sin A \sin B \cos A \cos B}
      | c =
}}
{{eqn | r = \paren {\sin A \cos B + \cos A \sin B} \paren { \sin A \cos B - \cos A \sin B}  
      | c =
}}
{{eqn | r = \map \sin {A + B} \map \sin {A - B}
      | c =
}}
{{end-eqn}}
{{qed}}
Category:Sine Function
Category:Difference of Two Powers
1ywl6jjgfukom9yxg7cwj1obpa1fzdr
Theorem: Reversal of Order of Vertices of Triangle causes Reversal of Sign of Area

Let $\triangle ABC$ be a triangle embedded in the plane.
Let $\Area \triangle ABC = \AA$.
Then:
:$\Area \triangle CBA = -\AA$.

Proof:
$\triangle CBA$ is the same as $\triangle ABC$ but with its vertices in the reverse order.
We have that:
:if $\triangle ABC$ is traversed anticlockwise going $AB \to BC \to CA$, then $\triangle CBA$ is traversed clockwise going $CB \to BA \to AC$
:if $\triangle ABC$ is traversed clockwise going $AB \to BC \to CA$, then $\triangle CBA$ is traversed anticlockwise going $CB \to BA \to AC$.
The result follows by definition of sign of $\Area \triangle ABC$
{{qed}}
Theorem: Condition for 3 Points in Plane to be Collinear

Let $A = \tuple {x_1, y_1}, B = \tuple {x_2, y_2}, C = \tuple {x_3, y_3}$ be points in the Cartesian plane.
Then:
:$A$, $B$ and $C$ are collinear
{{iff}} the determinant:
:<nowiki>$\begin {vmatrix}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1 \\
\end {vmatrix}$</nowiki>
equals zero.

Proof:
We have that:
:$A$, $B$ and $C$ are collinear
{{iff}}:
:the area of $\triangle ABC = 0$
{{iff}}:
:$\dfrac 1 2 \size {\paren {\begin {vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \\ \end {vmatrix} } } = 0$ (from Area of Triangle in Determinant Form)
{{iff}}:
:$\begin {vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \\ \end {vmatrix} = 0$
{{qed}}
Theorem: Joachimsthal's Section-Formulae

Let $P = \tuple {x_1, y_1}$ and $Q = \tuple {x_2, y_2}$ be points in the Cartesian plane.
Let $R = \tuple {x, y}$ be a point on $PQ$ dividing $PQ$ in the ratio:
:$PR : RQ = l : m$
Then:
{{begin-eqn}}
{{eqn | l = x
      | r = \dfrac {l x_2 + m x_1} {l + m}
}}
{{eqn | l = y
      | r = \dfrac {l y_2 + m y_1} {l + m}
}}
{{end-eqn}}

Proof:
:600px
Let the ordinates $PL$, $QM$ and $RN$ be constructed for $P$, $Q$ and $R$ respectively.
Then we have:
{{begin-eqn}}
{{eqn | l = OL
      | r = x_1
}}
{{eqn | l = OM
      | r = x_2
}}
{{eqn | l = ON
      | r = x
}}
{{eqn | l = LP
      | r = y_1
}}
{{eqn | l = MQ
      | r = y_2
}}
{{eqn | l = NR
      | r = y
}}
{{eqn | l = LN : NM = PR : RQ
      | r = l : m
}}
{{eqn | ll= \leadsto
      | l = \dfrac {x - x_1} {x_2 - x}
      | r = \dfrac l m
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x
      | r = \dfrac {l x_2 + m x_1} {l + m}
}}
{{eqn | l = y
      | r = \dfrac {l y_2 + m y_1} {l + m}
}}
{{end-eqn}}
{{qed}}
{{Namedfor|Ferdinand Joachimsthal|cat = Joachimsthal}}
Theorem: Value of Position-Ratio

Let $k$ denote the position-ratio of $R$.
Then:
:$k = \dfrac {PQ} {RQ} - 1$

Proof:
{{begin-eqn}}
{{eqn | l = k
      | r = \dfrac {PR} {RQ}
      | c = 
}}
{{eqn | r = \dfrac {PQ + QR} {RQ}
      | c = 
}}
{{eqn | r = \dfrac {PQ} {RQ} - 1
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Span of Exponential of a x by Cosine of b x and Exponential of a x by Sine of b x is preserved under Differentiation wrt x

Let $a, b, x, \alpha_1, \alpha_2 \in \R$ be real numbers.
Let $f_1$ and $f_2$ be the real functions defined as:
{{begin-eqn}}
{{eqn | q = \forall x \in \R
      | l = \map {f_1} x
      | r = \map \exp {a x} \map \cos {b x}
}}
{{eqn | l = \map {f_2} x
      | r = \map \exp {a x} \map \sin {b x}
}}
{{end-eqn}}
Let $\map \CC \R$ be the space of continuous real-valued functions.
Let $\tuple {\map {\CC^1} \R, +, \, \cdot \,}_\R$ be the vector space of continuously differentiable real-valued functions.
Let $S = \set {f_1, f_2} \subset \map {\CC^1} \R$ be a vector space.
Let $\map \span S$ be the span of $S$.
Then differentiation {{WRT|Differentiation}} $x$ preserves $\map \span S$.

Proof:
{{begin-eqn}}
{{eqn | l = \map {\dfrac \d {\d x} } {\alpha_1 f_1 + \alpha_2 f_2}
      | r = \map {\dfrac \d {\d x} } {\alpha_1 \map \exp {a x} \map \cos {b x} + \alpha_2 \map \exp {a x} \map \sin {b x} }
}}
{{eqn | r = \alpha_1 a \map \exp {a x} \map \cos {b x} - \alpha_1 b \map \exp {a x} \map \sin {b x} + \alpha_2 a \map \exp {a x} \map \sin {b x} + \alpha_2 b \map \exp {a x} \map \cos {b x}
}}
{{eqn | r = \paren {\alpha_1 a + \alpha_2 b} \map \exp {a x} \map \cos {b x} + \paren {\alpha_2 a - \alpha_1 b} \map \exp {a x} \map \sin {b x}
}}
{{eqn | r = \beta_1 f_1 + \beta_2 f_2
}}
{{end-eqn}}
{{qed}}
{{explain|Why this proves the result <br> Because the basis does not reduce to a single function nor it introduces a new function, while the coefficients are real.<br/>That will need to be explained then.}}
Theorem: Position of Centroid of Triangle on Median

Let $\triangle ABC$ be a triangle.
Let $AL$, $BM$ and $CN$ be the medians of $\triangle ABC$ meeting at the centroid $G$ of $\triangle ABC$.
Then $G$ is $\dfrac 1 3$ of the way along $AL$ from $L$, and similarly for the other medians.

Proof:
:520px
Let $\triangle ABC$ be embedded in a Cartesian plane such that $A = \tuple {x_1, y_1}$, $B = \tuple {x_2, y_2}$ and $C = \tuple {x_3, y_3}$.
The coordinates of $L$ are $\tuple {\dfrac {x_2 + x_3} 2, \dfrac {y_2 + y_3} 2}$.
Let $G$ be the point dividing $AL$ in the ratio $2 : 1$.
The coordinates of $G$ are $\tuple {\dfrac {x_1 + \paren {x_2 + x_3} } {1 + 2}, \dfrac {y_1 + \paren {y_2 + y_3} } {1 + 2} }$.
By similarly calculating the coordinates of $M$ and $N$, we get:
{{begin-eqn}}
{{eqn | l = M
      | r = \tuple {\dfrac {x_1 + x_3} 2, \dfrac {y_1 + y_3} 2}
}}
{{eqn | l = N
      | r = \tuple {\dfrac {x_1 + x_2} 2, \dfrac {y_1 + y_2} 2}
}}
{{end-eqn}}
Similarly:
:calculating the position of the point $G'$ dividing $BM$ in the ratio $2 : 1$
:calculating the position of the point $G''$ dividing $CN$ in the ratio $2 : 1$
we find that:
:$G = G' = G'' = \tuple {\dfrac {x_1 + x_2 + x_3} 3, \dfrac {y_1 + y_2 + y_3} 3}$
and the result follows.
{{qed}}
Theorem: Position of Centroid of Triangle on Median

Let $\triangle ABC$ be a triangle.
Let $AL$, $BM$ and $CN$ be the medians of $\triangle ABC$ meeting at the centroid $G$ of $\triangle ABC$.
Then $G$ is $\dfrac 1 3$ of the way along $AL$ from $L$, and similarly for the other medians.

Proof:
:520px
Let $\triangle ABC$ be a triangle.
Let $AL$, $BM$ and $CN$ be the medians of $\triangle ABC$.
Let the medians be concurrent at the centroid, $G$.
By the definition of median, the sides of $\triangle ABC$ are bisected.
{{begin-eqn}}
{{eqn | l = BL
      | r = LC
      | c = {{hypothesis}}
}}
{{eqn | l = BC
      | r = BL + LC
      | c = addition
}}
{{eqn | ll = \leadsto
      | n = 1
      | l = \frac {BC} {LC}
      | r = 2
      | c =
}}
{{eqn | l = AN
      | r = NB
      | c = {{hypothesis}}
}}
{{eqn | ll = \leadsto
      | n = 2
      | l = \frac {AN} {NB}
      | r = 1
      | c =
}}
{{eqn | l = \dfrac {LG} {GA} \cdot \dfrac {AN} {NB} \cdot \dfrac {BC} {CL}
      | r = -1
      | c = Menelaus's Theorem
}}
{{eqn | l = CL
      | r = -LC
      | c = {{Defof|Directed Line Segment}}
}}
{{eqn | ll = \leadsto
      | l = \dfrac {LG} {GA} \cdot \dfrac {AN} {NB} \cdot \dfrac {BC} {LC}
      | r = 1
      | c =
}}
{{eqn | l = \dfrac {LG} {GA} \cdot 2
      | r = 1
      | c = substitute from $(1)$ and $(2)$
}}
{{eqn | l = \dfrac {GA} {LG}
      | r = 2
      | c = rearranging
}}
{{end-eqn}}
Hence:
:the distance from the vertex to the centroid is $2/3$ of the whole length of the median.
and so:
:the distance from the centroid to the side is $1/3$ of the whole length of the median.
The result follows.
{{qed}}
Theorem: Position of Centroid of Triangle on Median

Let $\triangle ABC$ be a triangle.
Let $AL$, $BM$ and $CN$ be the medians of $\triangle ABC$ meeting at the centroid $G$ of $\triangle ABC$.
Then $G$ is $\dfrac 1 3$ of the way along $AL$ from $L$, and similarly for the other medians.

Proof:
:400px
By Medians of Triangle Form Six Triangles of Equal Area:
:these six triangles formed by the medians of $\triangle ABC$ have equal area:
* $\triangle AGN$
* $\triangle BGN$
* $\triangle BGL$
* $\triangle CGL$
* $\triangle CGM$
* $\triangle AGM$
{{WLOG}} consider one of the three medians of $\triangle ABC$, $AGL$.
The following triangles both have their base on $AGL$ and share vertex $C$:
* $\triangle CGA$
* $\triangle CGL$
Since $\triangle CGA$ contains two of the small triangles:
{{begin-eqn}}
{{eqn | ll = \leadsto
      | l = \AA \paren { \triangle CGA }
      | r = \AA \paren { \triangle AGM } + \AA \paren { \triangle CGM }
      | c = 
}}
{{eqn | l = \dfrac { \AA \paren { \triangle CGA } } { \AA \paren { \triangle CGL } }
      | r = \dfrac { \AA \paren { \triangle AGM } + \AA \paren { \triangle CGM } } { \AA \paren { \triangle CGL } }
      | c = 
}}
{{end-eqn}}
Since 
:$\AA \paren { \triangle AGM } = \AA \paren { \triangle CGM } = \AA \paren { \triangle CGL }$
{{begin-eqn}}
{{eqn | ll = \leadsto
      | l = \dfrac 2 1
      | r = \dfrac { \AA \paren { \triangle AGM } + \AA \paren { \triangle CGM } } { \AA \paren { \triangle CGL } }
      | c = 
}}
{{eqn | l = \dfrac { \AA \paren { \triangle CGA } } { \AA \paren { \triangle CGL } }
      | r = \dfrac 2 1
      | c = {{EuclidCommonNotionLink|1}}
}}
{{eqn | l = \dfrac { \AA \paren { \triangle CGA } } { \AA \paren { \triangle CGL } }
      | r = \dfrac { AG } { GL }
      | c = Areas of Triangles and Parallelograms Proportional to Base
}}
{{eqn | l = \dfrac { AG } { GL }
      | r = \dfrac 2 1
      | c = {{EuclidCommonNotionLink|1}}
}}
{{eqn | ll = \leadsto
      | l = AG
      | r = 2 \cdot GL
      | c = 
}}
{{end-eqn}}
{{mutatis}}:
:$BG = 2 \cdot GM$
:$CG = 2 \cdot GN$
Hence:
:the distance from the vertex to the centroid is $2/3$ of the whole length of the median.
and so:
:the distance from the centroid to the side is $1/3$ of the whole length of the median.
The result follows.
{{qed}}
Theorem: Equation of Straight Line in Plane/Point-Slope Form/Parametric Form

Let $\LL$ be a straight line embedded in a cartesian plane, given in point-slope form as:
::$y - y_0 = \paren {x - x_0} \tan \psi$
where $\psi$ is the angle between $\LL$ and the $x$-axis.
Then $\LL$ can be expressed by the parametric equations:
:$\begin {cases} x = x_0 + t \cos \psi \\ y = y_0 + t \sin \psi \end {cases}$

Proof:
Let $P_0$ be the point $\tuple {x_0, y_0}$.
Let $P$ be an arbitrary point on $\LL$.
Let $t$ be the distance from $P_0$ to $P$ measured as positive when in the positive $x$ direction.
The equation for $P$ is then:
{{begin-eqn}}
{{eqn | l = y - y_0
      | r = \paren {x - x_0} \tan \psi
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac {x - x_0} {\cos \psi}
      | r = t
      | c = 
}}
{{eqn | l = \dfrac {y - y_0} {\sin \psi}
      | r = t
      | c = 
}}
{{end-eqn}}
The result follows.
{{qed}}
Theorem: Equation of Straight Line in Plane/Two-Point Form/Parametric Form

Let $\LL$ be a straight line embedded in a cartesian plane, given in two-point form as:
::$\dfrac {x - x_1} {x_2 - x_1} = \dfrac {y - y_1} {y_2 - y_1}$
Then $\LL$ can be expressed by the parametric equations:
:$\begin {cases} x = x_1 + t \paren {x_2 - x_1} \\ y = y_1 + t \paren {y_2 - y_1} \end {cases}$

Proof:
Let $P = \tuple {x, y}$ be an arbitrary point on $\LL$.
Let $t = \dfrac {x - x_1} {x_2 - x_1} = \dfrac {y - y_1} {y_2 - y_1}$.
We then have:
{{begin-eqn}}
{{eqn | l = t
      | r = \dfrac {x - x_1} {x_2 - x_1}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x - x_1
      | r = t \paren {x_2 - x_1}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x
      | r = x_1 + t \paren {x_2 - x_1}
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = t
      | r = \dfrac {y - y_1} {y_2 - y_1}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y - y_1
      | r = t \paren {y_2 - y_1}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y
      | r = y_1 + t \paren {y_2 - y_1}
      | c = 
}}
{{end-eqn}}
The result follows.
{{qed}}
Theorem: Equation of Straight Line in Plane/Two-Point Form/Determinant Form

Let $\LL$ be a straight line embedded in a Cartesian plane, given in two-point form as:
:$\dfrac {x - x_1} {x_2 - x_1} = \dfrac {y - y_1} {y_2 - y_1}$
Then $\LL$ can be expressed in the form:
:$\begin {vmatrix} x & y & 1 \\ x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \end {vmatrix} = 0$

Proof:
{{begin-eqn}}
{{eqn | l = \frac {x - x_1} {x_2 - x_1}
      | r = \frac {y - y_1} {y_2 - y_1}
}}
{{eqn | ll= \leadsto
      | l = \paren {x - x_1} \paren {y_2 - y_1}
      | r = \paren {x_2 - x_1} \paren {y - y_1}
}}
{{eqn | ll= \leadsto
      | l = \paren {x - x_1} \paren {y_2 - y_1} - \paren {x_2 - x_1} \paren {y - y_1}
      | r = 0
}}
{{eqn | ll= \leadsto
      | l = \begin {vmatrix} x - x_1 & y - y_1 \\ x_2 - x_1 & y_2 - y_1 \end {vmatrix}
      | r = 0
      | c = Determinant of Order 2
}}
{{eqn | ll= \leadsto
      | l = \begin {vmatrix} x & y & 1 \\ x - x_1 & y - y_1 & 0 \\ x_2 - x_1 & y_2 - y_1 & 0 \end {vmatrix}
      | r = 0
      | c = Determinant with Unit Element in Otherwise Zero Column
}}
{{eqn | ll= \leadsto
      | l = \begin {vmatrix} x & y & 1 \\ -x_1 & -y_1 & -1 \\ x_2 - x_1 & y_2 - y_1 & 0 \end {vmatrix}
      | r = 0
      | c = Multiple of Row Added to Row of Determinant
}}
{{eqn | ll= \leadsto
      | l = \begin {vmatrix} x & y & 1 \\ x_1 & y_1 & 1 \\ x_2 - x_1 & y_2 - y_1 & 0 \end {vmatrix}
      | r = 0
      | c = Determinant with Row Multiplied by Constant
}}
{{eqn | ll= \leadsto
      | l = \begin {vmatrix} x & y & 1 \\ x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \end {vmatrix}
      | r = 0
      | c = Multiple of Row Added to Row of Determinant
}}
{{end-eqn}}
{{qed}}
Theorem: Equation of Straight Line in Plane/Normal Form/Polar Form

Let $\LL$ be defined in normal form:
:$x \cos \alpha + y \sin \alpha = p$
Then $\LL$ can be presented in polar coordinates as:
:$r \map \cos {\theta - \alpha} = p$

Proof:
Let $O$ be the origin of the Cartesian plane and the pole of the corresponding polar frame.
Let $OX$ denote the polar axis, coincident with the $x$-axis.
Let $P$ be an arbitrary point on $\LL$, expressed in polar coordinates as $\polar {r, \theta}$.
Let $N$ be the point on $\LL$ where the normal to $\LL$ intersects $\LL$.
We have that $OP$ is at an angle $\theta$ to $OX$ and is of length $r$.
We have that $ON$ is at an angle $\alpha$ to $OX$ and is of length $p$.
Hence $\angle NOP = \theta = \alpha$.
We also have that $\angle ONP$ is a right angle.
Thus:
:$p = r \map \cos {\theta - \alpha}$
{{qed}}