nyuuzyou/wikis huggingface dataset:
Distance-Preserving Mapping is Injection of Metric Spaces

== Theorem ==
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $\phi: M_1 \to M_2$ be a distance-preserving mapping.
Then $\phi$ is an injection.
== Proof ==
Let $a, b \in A_1$ and suppose that $\map \phi a = \map \phi b$.
Then by the definition of a metric space:
:$\map {d_2} {\map \phi a, \map \phi b} = 0$
By the definition of a distance-preserving mapping then:
:$\map {d_1} {a, b} = 0$
Thus by the definition of a metric space:
:$a = b$
Hence $\phi$ is injective.
Category:Metric Spaces
Distance-Preserving Surjection is Isometry of Metric Spaces

== Theorem ==
Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be metric spaces.
Let $\phi: M_1 \to M_2$ be a surjective distance-preserving mapping.
That is:
:$\forall a, b \in M_1: d_1 \tuple {a, b} = d_2 \tuple {\map \phi a, \map \phi b}$
Then $\phi$ is an isometry.
== Proof ==
The premises satisfy all elements of the definition of isometry except for bijectivity.
As we presume $\phi$ to be surjective we need only show that it is injective.
By Distance-Preserving Mapping is Injection of Metric Spaces then $\phi$ is injective.
Category:Isometries (Metric Spaces)
Distance Between Points of Same Latitude along Parallel of Latitude

== Theorem ==
Let $J$ and $K$ be points on Earth's surface that have the same latitude.
Let $JK$ be the length of the arc joining $JK$ measured along the parallel of latitude on which they both lie.
Let $R$ denote the center of the parallel of latitude holding $J$ and $K$.
Let $\operatorname {Long}_J$ and $\operatorname {Long}_K$ denote the longitude of $J$ and $K$ respectively, measured in degrees.
Let $\operatorname {Lat}_J$ denote the latitude of $J$ (and $K$).
Then: 
:$JK \approx 60 \times \size {\operatorname {Long}_J - \operatorname {Long}_K} \cos \operatorname {Lat}_J$ nautical miles
== Proof ==
We have that $\size {\operatorname {Long}_J - \operatorname {Long}_K}$ is the magnitude of the spherical angle between the meridians on which $J$ and $K$ lie.
Thus from Length of Arc of Small Circle:
:$JK \approx \size {\operatorname {Long}_J - \operatorname {Long}_K} \cos \operatorname {Lat}_J$ degrees of arc
Then by definition of nautical mile:
:$1$ nautical mile $= \dfrac 1 {60}$ degree of arc of latitude along a meridian measured along Earth's surface.
Thus, approximately, $1$ degree of arc along the equator equals $60$ nautical miles.
The result follows.
== Sources ==
* : Chapter $\text I$. Spherical Trigonometry: $4$. Terrestrial latitude and longitude.
Category:Geodesy
Distance Formula

== Theorem ==
The distance $d$ between two points $A = \tuple {x_1, y_1}$ and $B = \tuple {x_2, y_2}$ on a Cartesian plane is:
:$d = \sqrt {\paren {x_1 - x_2}^2 + \paren {y_1 - y_2}^2}$
=== 3 Dimensions ===
== Proof ==
The distance in the horizontal direction between $A$ and $B$ is given by $\size {x_1 - x_2}$.
The distance in the vertical direction between $A$ and $B$ is given by $\size {y_1 - y_2}$.
By definition, the angle between a horizontal and a vertical line is a right angle.
So when we place a point $C = \tuple {x_1, y_2}$, $\triangle ABC$ is a right triangle.
Thus, by Pythagoras's Theorem:
:$d^2 = \size {x_1 - x_2}^2 + \size {y_1 - y_2}^2$
and the result follows.
== Sources ==
* : Chapter $\text I$. Coordinates: $4$. Distance between two points
* : Chapter $\text I$: Preliminary Remarks on Analytical Geometry and Vector Analysis: $1$. Rectangular Co-ordinates and Vectors: $1$. Coordinate Axes
* : $\text {III}$. Analytical Geometry: The Straight Line
* : Chapter $1$: Introduction to Coordinate Geometry: $(1.1)$
* : $\S 10$: Formulas from Plane Analytic Geometry: Distance $d$ between Two Points $\map {P_1} {x_1, y_1}$ and $\map {P_2} {x_2, y_2}$: $10.1$
* : $\S 8$: Formulas from Plane Analytic Geometry: Distance $d$ between Two Points $\map {P_1} {x_1, y_1}$ and $\map {P_2} {x_2, y_2}$: $8.1.$
* 
Category:Euclidean Geometry
Category:Analytic Geometry
Category:Distance Formula
Distance Formula/3 Dimensions

== Theorem ==
The distance $d$ between two points $A = \tuple {x_1, y_1, z_1}$ and $B = \tuple {x_2, y_2, z_2}$ in a Cartesian space of 3 dimensions is:
:$d = \sqrt {\paren {x_1 - x_2}^2 + \paren {y_1 - y_2}^2 + \paren {z_1 - z_2}^2}$
Hence $d$ is the length of the straight line segment $AB$.
== Proof ==
:
Let $d$ be the distance to be found between $A = \tuple {x_1, y_1, z_1}$ and $B = \tuple {x_2, y_2, z_2}$.
Let the points $C$ and $D$ be defined as:
:$C = \tuple {x_2, y_1, z_1}$
:$D = \tuple {x_2, y_2, z_1}$
Let $d'$ be the distance between $A$ and $D$.
From Distance Formula, it can be seen that:
:$d' = \sqrt {\paren {x_1 - x_2}^2 + \paren {y_1 - y_2}^2}$
We note that $\triangle ADB$ is a right triangle.
Thus by Pythagoras's Theorem:
:$AB^2 = AD^2 + DB^2$
Thus:
and so:
:$d = \sqrt {\paren {x_1 - x_2}^2 + \paren {y_1 - y_2}^2 + \paren {z_1 - z_2}^2}$
as it was to be proved.
== Sources ==
* : Chapter $\text I$: Preliminary Remarks on Analytical Geometry and Vector Analysis: $1$. Rectangular Co-ordinates and Vectors: $1$. Coordinate Axes
* : Chapter $\text {I}$: Coordinate System: Directions: $2$. Cartesian Coordinates
* : $\S 12$: Formulas from Solid Analytic Geometry: Distance $d$ between Two Points $\map {P_1} {x_1, y_1, z_1}$ and $\map {P_2} {x_2, y_2, z_2}$: $12.1$
* : $\S 10$: Formulas from Solid Analytic Geometry: Distance $d$ between Two Points $\map {P_1} {x_1, y_1, z_1}$ and $\map {P_2} {x_2, y_2, z_2}$: $10.1$
Category:Euclidean Geometry
Category:Solid Analytic Geometry
Category:Distance Formula
Distance Function for Distinct Elements in Metric Space is Strictly Positive

== Theorem ==
Let $A$ be a set.
Let $d: A \times A \to \R$ be a real-valued function on $A$ with the following properties:
which can be considered as being an alternative formulation of the metric space axioms.
Then:
:$\forall x, y \in A: x \ne y \implies \map d {x, y} > 0$
which is .
Thus $d$ is a distance function, so making $M := \struct {A, d}$ a metric space.
== Proof ==
Category:Metric Spaces
Distance Function of Metric Space is Continuous

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $\tau_A$ be the topology on $A$ induced by $d$.
Let $\struct {A \times A, \tau}$ be the product space of $\struct {A, \tau_A}$ and itself.
Then the distance function $d: A \times A \to \R$ is a continuous mapping.
== Proof ==
Let $d_\infty: \paren {A \times A} \times \paren {A \times A} \to \R$ be the metric on $A \times A$ defined by:
:$\map {d_\infty} {\tuple {x, y}, \tuple {x', y'} } = \max \set {\map d {x, x'}, \map d {y, y'} }$
By P-Product Metric Induces Product Topology, $\tau$ is the topology on $A \times A$ induced by $d_\infty$.
Let $\epsilon \in \R_{>0}$ be a strictly positive real number.
Let $\tuple {x_0, y_0} \in A \times A$.
Suppose that $\tuple {x, y} \in A \times A$ and $\map {d_\infty} {\tuple{x, y}, \tuple{x_0, y_0} } < \dfrac 1 2 \epsilon$.
Then:
The result follows from the definition of a continuous mapping.
Category:Continuous Mappings
Category:Metric Spaces
Category:Product Spaces
Distance Moved by Body from Rest under Constant Acceleration

== Theorem ==
Let a body $B$ be stationary.
Let $B$ be subject to a constant acceleration.
Then the distance travelled by $B$ is proportional to the square of the length of time $B$ is under the acceleration.
== Proof ==
From Body under Constant Acceleration: Distance after Time:
:$\mathbf s = \mathbf u t + \dfrac {\mathbf a t^2} 2$
where:
:$\mathbf s$ is the displacement of $B$ at time $t$ from its initial position at time $t$
:$\mathbf u$ is the velocity at time $t = 0$
:$\mathbf a$ is the constant acceleration $t$
In this scenario, $\mathbf u = \mathbf 0$.
Thus:
:$\mathbf s = \dfrac {\mathbf a} 2 t^2$
and so by taking the magnitudes of the vector quantities:
:$s = \dfrac a 2 t^2$
Hence the result, by definition of proportional.
== Historical Note ==
== Sources ==
* : Chapter $8$: The System of the World: Galileo
Category:Mechanics
Distance between Closed Sets in Euclidean Space

== Theorem ==
Let $S, T \subseteq \R^n$ be closed, non-empty subsets of the real Euclidean space $R^n$.
Suppose that $S$ is bounded, and $S$ and $T$ are disjoint.
Then there exists $x \in S$ and $y \in T$ such that:
:$\map d {x, y} = \map d {S, T} > 0$
where:
:$d$ denotes the Euclidean metric
:$\map d {S, T}$ is the distance between $S$ and $T$.
== Proof ==
By definition of distance from subset, we can for all $n \in \N$ find $x_n \in S, y_n \in T$ such that:
:$\map d {S, T} \le \map d {x_n, y_n} < \map d {S, T} + \dfrac 1 n$
so:
:$\ds \lim_{n \mathop \to \infty} \map d {x_n, y_n} = \map d {S, T}$
By definition of bounded space, there exists $a \in S$ and $K \in \R$ such that for all $x \in S$, we have $\map d {x, a} \le K$.
It follows that $\sequence {x_n}$ is a bounded sequence.
Then $\sequence {y_n}$ is also a bounded sequence, as:
The sequence $\sequence {\tuple {x_n, y_n} }$ in $\R^{2 n}$ is also bounded, as:
From Bounded Sequence in Euclidean Space has Convergent Subsequence, it follows that $\sequence {\tuple {x_n, y_n} }$ has a subsequence $\sequence {\tuple {x_{n_r}, y_{n_r} } }_{r \mathop \in N}$ that converges to a limit $\tuple {x, y} \in \R^{2n}$.
Then $\ds \lim_{r \mathop \to \infty} x_{n_r} = x$, and $\ds \lim_{r \mathop \to \infty} y_{n_r} = y$.
From Closed Set iff Contains all its Limit Points, it follows that $x \in S$, and $y \in T$.
Then from Distance Function of Metric Space is Continuous:
:$\ds \lim_{r \mathop \to \infty} \map d {x_{n_r}, y_{n_r} } = \map d {x, y}$
As a Convergent Sequence in Metric Space has Unique Limit, we have:
:$\ds \map d {x, y} = \lim_{r \mathop \to \infty} \map d {x_{n_r}, y_{n_r} } = \lim_{n \mathop \to \infty} \map d {x_n, y_n} = \map d {S, T}$
As $S$ and $T$ are disjoint, it follows that $x \ne y$.
Hence:
:$0 < \map d {x, y} = \map d {S, T}$
== Sources ==
* : $\S A$
Category:Euclidean Space
Distance between Disjoint Compact Set and Closed Set in Metric Space is Positive

== Theorem ==
Let $\struct {X, d}$ be a metric space.
Let $A$ and $B$ be disjoint subsets of $X$ such that $A$ is compact and $B$ is closed.
Then $\map d {A, B} > 0$, where $\map d {A, B}$ is the distance between $A$ and $B$. 
== Proof ==
Define $f : X \to \R$ by: 
:$\map f x = \map d {x, B}$
for each $x \in X$.
From the definition of the distance between $A$ and $B$, we have: 
:$\ds \map d {A, B} = \inf_{a \mathop \in A} \map f a$ 
From Compact Subspace of Hausdorff Space is Closed and Metric Space is Hausdorff, $A$ is closed and hence contains all its limit points.
From Point at Zero Distance from Subset of Metric Space is Limit Point or Element we then have:
:$\map f a > 0$
for each $a \in A$.
From Distance from Point to Subset is Continuous Function, $f$ is continuous.
From Restriction of Continuous Mapping is Continuous: Metric Spaces, the restriction $f \restriction_A$ is continuous.
From Continuous Image of Compact Space is Compact: Corollary 3, we have that $f$ attains its bounds. 
That is, there exists $\alpha \in A$ such that: 
:$\ds \map d {\alpha, B} = \inf_{a \mathop \in A} \map f a$
giving:
:$\ds \inf_{a \in A} \map f a > 0$
and so:
:$\map d {A, B} > 0$ 
Category:Compact Spaces
Category:Distance Function
Distance between Element and Subset is Nonnegative

== Theorem ==
Let $\struct {M, d}$ be a metric space.
Let $x \in M$ and $S \subseteq M$.
Then:
:$\map d {x, S} \ge 0$
where $\map d {x, S}$ is the distance between $x$ and $S$.
== Proof ==
By definition of the distance between $x$ and $S$:
:$\map d {x, S} = \ds \inf_{s \mathop \in S} \map d {x, s}$
From the metric space axioms:
:$\forall s \in M: \map d {x, s} \ge 0$
Hence by the nature of the infimum:
:$\map d {x, S} \ge 0$
as desired.
Category:Distance Function
Distance between Element and Subset of Real Numbers/Examples/Example 1

== Example of Distance between Element and Subset of Real Numbers ==
Let $S \subseteq \R$ be the subset of the set of real numbers $\R$ defined as:
:$S := \set {0, 1, 2}$
Then:
:$\map d {3, S} = 1$
where $d$ denotes the distance function.
== Sources ==
* : $\S 2$: Continuum Property: Exercise $\S 2.13 \ (4) \ \text{(i)}$
Category:Examples of Distance Function
Distance between Element and Subset of Real Numbers/Examples/Example 2

== Example of Distance between Element and Subset of Real Numbers ==
Let $S \subseteq \R$ be the subset of the set of real numbers $\R$ defined as:
:$S := \openint 0 1$
Then:
:$\map d {3, S} = 1$
where $d$ denotes the distance function.
== Sources ==
* : $\S 2$: Continuum Property: Exercise $\S 2.13 \ (4) \ \text{(ii)}$
Category:Examples of Distance Function
Distance between Element and Subset of Real Numbers/Examples/Example 3

== Example of Distance between Element and Subset of Real Numbers ==
Let $S \subseteq \R$ be the subset of the set of real numbers $\R$ defined as:
:$S := \closedint 1 2$
Then:
:$\map d {3, S} = 1$
where $d$ denotes the distance function.
== Sources ==
* : $\S 2$: Continuum Property: Exercise $\S 2.13 \ (4) \ \text{(iii)}$
Category:Examples of Distance Function
Distance between Element and Subset of Real Numbers/Examples/Example 4

== Example of Distance between Element and Subset of Real Numbers ==
Let $S \subseteq \R$ be the subset of the set of real numbers $\R$ defined as:
:$S := \openint 2 3$
Then:
:$\map d {3, S} = 0$
where $d$ denotes the distance function.
== Sources ==
* : $\S 2$: Continuum Property: Exercise $\S 2.13 \ (4) \ \text{(iv)}$
Category:Examples of Distance Function
Distance between Excenters of Triangle in Terms of Circumradius

== Theorem ==
Let $\triangle ABC$ be a triangle whose sides are $a$, $b$ and $c$ opposite vertices $A$, $B$ and $C$ respectively.
Let $I_b$ and $I_c$ be the excenters of $\triangle ABC$ $b$ and $c$ respectively.
Let $R$ be the circumradius of $\triangle ABC$.
Then:
:$I_b I_c = 4 R \cos \dfrac A 2$
== Proof ==
== Sources ==
* : $\text V$. Trigonometry: Formulae $(57)$
Category:Distance between Excenters of Triangle in Terms of Circumradius
Category:Circumcircles of Triangles
Category:Excircles of Triangles
Distance between Excenters of Triangle in Terms of Circumradius/Proof

== Theorem ==
== Proof ==
:
From Triangle is Orthic Triangle of Triangle formed from Excenters, we establish that $\triangle ABC$ is the orthic triangle of $\triangle I_a I_b I_c$.
Hence $I_b B$ is an altitude of $\triangle I_a I_b I_c$.
Thus $\angle I_b B I_a$ is a right angle.
From Altitudes of Triangle Bisect Angles of Orthic Triangle:
:$\angle CBI = \dfrac B 2$
So:
:$\angle I_a B C = 90 \degrees - \dfrac B 2$
By a similar argument, :
:$\angle I_a C B = 90 \degrees - \dfrac C 2$
Hence:
== Sources ==
* : $\text V$. Trigonometry: The ex-circles
Category:Distance between Excenters of Triangle in Terms of Circumradius
Distance between Images of Points under Central Dilatation Mapping

== Theorem ==
Let $f$ be a central dilatation mapping.
Let the scale factor of $f$ be $k$.
Let $P$ and $Q$ be points in the domain of $f$.
Let $d$ be the distance between $P$ and $Q$.
Then the distance between the image of $P$ and $Q$ is $\size k d$.
== Proof ==
== Sources ==
* 
* 
Category:Image of Point under Central Dilatation Mapping with Fixed Point at Origin
Distance between Incenter and Excenter of Triangle in Terms of Circumradius

== Theorem ==
Let $\triangle ABC$ be a triangle whose sides are $a$, $b$ and $c$ opposite vertices $A$, $B$ and $C$ respectively.
Let $I$ be the incenters of $\triangle ABC$.
Let $I_a$ be the excenters of $\triangle ABC$ $a$.
Let $R$ be the circumradius of $\triangle ABC$.
Then:
:$I I_a = 4 R \sin \dfrac A 2$
== Proof ==
== Sources ==
* : $\text V$. Trigonometry: Formulae $(58)$
Category:Distance between Incenter and Excenter of Triangle in Terms of Circumradius
Category:Incircles of Triangles
Category:Excircles of Triangles
Category:Circumcircles of Triangles
Distance between Incenter and Excenter of Triangle in Terms of Circumradius/Proof

== Theorem ==
== Proof ==
:
From Triangle is Orthic Triangle of Triangle formed from Excenters, we establish that $\triangle ABC$ is the orthic triangle of $\triangle I_a I_b I_c$.
By the Nine Point Circle Theorem, the Feuerbach circle of $\triangle I_a I_b I_c$ passes through each of $A$, $B$ and $C$.
Therefore the Feuerbach circle of $\triangle I_a I_b I_c$ is the circumcircle of $\triangle ABC$.
Hence the radius of the Feuerbach circle is $R$.
From Radius of Circumcircle is Twice Radius of Feuerbach Circle, the radius of the circumcircle of $\triangle I_a I_b I_c$ is $2 R$.
Hence:
== Sources ==
* : $\text V$. Trigonometry: The ex-circles
Category:Distance between Incenter and Excenter of Triangle in Terms of Circumradius
Distance between Points in Complex Plane

== Theorem ==
Let $A$ and $B$ be points in the complex plane such that:
:$A = \tuple {x_1, y_1}$
:$B = \tuple {x_2, y_2}$
Then the distance between $A$ and $B$ is given by:
where $z_1$ and $z_2$ are represented by the complex numbers $z_1$ and $z_2$ respectively.
== Proof ==
We have:
== Sources ==
* : $1$: Solved Problems: Graphical Representations of Complex Numbers. Vectors: $8 \ \text {(b)}$
* : $1$ The complex plane: Complex numbers $\S 1.5$ Subsets of the complex plane
Category:Complex Addition
Category:Geometry of Complex Plane
Distance between Points in Regular Hexagon

== Theorem ==
Let $H$ be a regular hexagon embedded in the Euclidean plane $\R^2$.
Let $s \in \R_{>0}$ be the side length of $H$.
Let $\mathbf x, \mathbf y \in \R^2$ such that $\mathbf x$ and $\mathbf y$ lie in the interior of $H$, or on the circumference of $H$.
Then:
:$\map d {\mathbf x, \mathbf y} \le 2s$
where $\map d {\mathbf x, \mathbf y}$ denotes the Euclidean distance between $\mathbf x$ and $\mathbf y$.
== Proof ==
From Regular Polygon is Cyclic, it follows that $H$ can be inscribed in a circle with center $\mathbf c$.
The circumcircle intersects all vertices of $H$.
From Regular Hexagon is composed of Equilateral Triangles, it follows that the side length $s$ is equal to the distance from $\mathbf c$ to any vertex of $H$.
It follows that the radius of the circumcircle is equal to $s$.
Hence:
Category:Hexagons
Distance between Two Parallel Straight Lines is Everywhere the Same

== Theorem ==
Let $AB$ and $CD$ be parallel straight lines.
Let perpendiculars $EF$ and $GH$ be drawn from $AB$ to $CD$, where $E, G$ are on $AB$ and $F, H$ are on $CD$.
Then $EF = GH$.
That is, the distance between $AB$ and $CD$ is the same everywhere along their length.
== Proof ==
$\angle EFH, \angle FEG, \angle EGH, \angle FHG$ are all right angles.
Then $EF$ and $GH$ are parallel.
Thus $\Box EFHG$ is by definition a parallelogram.
From Opposite Sides and Angles of Parallelogram are Equal it follows that $EF = GH$.
== Sources ==
* : Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.22$: Corollary $1$
Category:Parallel Lines
Distance between Two Points in Plane in Polar Coordinates

== Theorem ==
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 1 ==
== Proof 2 ==
== Also presented as ==
This result can also be seen presented as:
:$d = \sqrt {r_1^2 + r_2^2 + 2 r_1 r_2 \map \cos {\theta_2 - \theta_1} }$
The equivalence of the results follows from the fact that $\map \cos {\theta_2 - \theta_1} = \map \cos {\theta_1 - \theta_2}$.
== Also see ==
* Distance Formula
* Complex Modulus of Difference of Complex Numbers
Category:Distance between Two Points in Plane in Polar Coordinates
Category:Distance Formula
Category:Analytic Geometry
Distance between Two Points in Plane in Polar Coordinates/Proof 1

== Theorem ==
== Proof ==
Let $A$ and $B$ be embedded as suggested in a polar coordinate plane whose pole is at $O$.
:
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.
== Sources ==
* : Chapter $\text I$. Coordinates: $5$. Distance between two points in polar coordinates
Category:Distance between Two Points in Plane in Polar Coordinates
Distance between Two Points in Plane in Polar Coordinates/Proof 2

== Theorem ==
== Proof ==
From Conversion between Cartesian and Polar Coordinates in Plane, $A$ and $B$ can be expressed in Cartesian coordinates as follows:
Thus the distance $d$ between $A = \tuple {x_1, y_1}$ and $B = \tuple {x_2, y_2}$ is given by:
The result follows.
Category:Distance between Two Points in Plane in Polar Coordinates
Distance from Foot of Altitude of Triangle to Orthocenter

== Theorem ==
Let $\triangle ABC$ be a triangle with sides $a$, $b$ and $c$ opposite vertices $A$, $B$ and $C$ respectively.
Let $D$ be the foot of the altitude of $\angle A$
Let $H$ be the orthocenter of $\triangle ABC$.
Then:
:$HD = 2 R \cos B \cos C$
where $R$ denotes the circumradius of $\triangle ABC$.
== Proof ==
== Sources ==
* : $\text V$. Trigonometry: Formulae $(47)$
Category:Distance from Foot of Altitude of Triangle to Orthocenter
Category:Orthocenters of Triangles
Distance from Foot of Altitude of Triangle to Orthocenter/Proof

== Theorem ==
== Proof ==
:
We construct the circumcircle of $\triangle ABC$, whose circumcenter is $K$ and whose circumradius is $R$.
We construct the orthocenter $H$ of $\triangle ABC$ as the intersection of the altitudes $AD$ and $BE$.
First we note the following:
$\angle CDH$ and $\angle CEH$ are both right angles.
Hence from Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles, $\Box EHCD$ is a cyclic quadrilateral.
Thus:
Hence:
Then we have:
The result follows.
== Sources ==
* : $\text V$. Trigonometry: The orthocentre
Category:Distance from Foot of Altitude of Triangle to Orthocenter
Distance from Point to Subset is Continuous Function

== Theorem ==
Let $M = \struct {X, d}$ be a metric space.
Let $A \subseteq X$ be a non-empty subset of $X$.
Let $f: X \to \R$ be the function defined as:
:$\forall x \in X: \map f x = \map d {x, A}$
where $\map d {x, A}$ denotes the distance from $x$ to $A$.
Then $f$ is continuous.
== Proof ==
$\forall x, y \in X, \forall z \in A$, by the definition of the distance from $x$ to $A$, we have:
:$\map d {x, A} \le \map d {x, z} \le \map d {x, y} + \map d {y, z}$
From Triangle Inequality on Distance from Point to Subset:
:$\map d {x, A} \le \map d {x, y} + \map d {y, A}$
and:
:$\map d {y, A} \le \map d {x, y} + \map d {x, A}$
Therefore:
:$\norm {\map d {x, A} - \map d {y, A} } \le \map d {x, y}$
We can conclude that $f$ is continuous.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 5$: Limits: Exercise $7$
* : $4$: The Hausdorff condition: Exercise $4.3: 5 \ \text {(c)}$
Category:Continuous Mappings
Category:Distance Function
Distance from Subset of Real Numbers

== Theorem ==
Let $S$ be a subset of the set of real numbers $\R$.
Let $x \in \R$ be a real number.
Let $\map d {x, S}$ be the distance between $x$ and $S$.
Then:
=== Distance from Subset of Real Numbers to Element ===
=== Distance from Subset of Real Numbers to Supremum ===
=== Distance from Subset of Real Numbers to Infimum ===
=== Real Number at Distance Zero from Closed Real Interval is In Interval ===
=== Existence of Real Number at Distance Zero from Open Real Interval not in Interval ===
== Sources ==
* : $\S 2$: Continuum Property: Exercise $\S 2.13 \ (5)$
Category:Distance Function
Distance from Subset of Real Numbers to Element

== Theorem ==
Let $S$ be a subset of the set of real numbers $\R$.
Let $x \in \R$ be a real number.
Let $\map d {x, S}$ be the distance between $x$ and $S$.
Then:
:$x \in S \implies \map d {x, S} = 0$
== Proof 1 ==
== Proof 2 ==
== Examples ==
Category:Distance Function
Category:Distance from Subset of Real Numbers to Element