nyuuzyou/wikis huggingface dataset:
Inner Limit in Normed Spaces by Open Balls

== Theorem ==
Let $\sequence {C_n}_{n \mathop \in \N}$ be a sequence of sets in a normed vector space $\struct {\XX, \norm {\, \cdot \,} }$.
Then the inner limit of $\sequence {C_n}_{n \mathop \in \N}$ is:
:$\ds \liminf_n C_n = \set {x: \forall \epsilon > 0: \exists N \in \NN_\infty: \forall n \in N: x \in C_n + B_\epsilon}$
where $B_\epsilon$ denotes the open $\epsilon$-ball of the space.
== Proof ==
The proof is an immediate result of Inner Limit in Hausdorff Space by Open Neighborhoods since the arbitrary open sets can be here replaced by open balls.
== Also see ==
* Inner Limit in Hausdorff Space by Open Neighborhoods
* Inner Limit of Sequence of Sets in Normed Space
* Inner Limit in Hausdorff Space by Set Closures
Category:Limits of Sequence of Sets
Inner Limit is Closed Set

== Theorem ==
Let $\struct {S, \tau}$ be a Hausdorff topological space.
Let $\sequence {C_n}_{n \mathop \in \N}$ be a sequence of sets in $S$.
Then the inner limit $\liminf_n C_n$ is a closed set.
== Proof ==
According to Inner Limit in Hausdorff Space by Set Closures, the inner limit is given by an arbitrary intersection of closed sets which is closed in the topology $\tau$.
Category:Limits of Sequence of Sets
Inner Limit of Sequence of Sets in Normed Space

== Theorem ==
Let $\struct {\XX, \norm {\, \cdot \, } }$ be a normed vector space.
Let $\sequence {C_n}_{n \mathop \in \N}$ be a sequence of sets in $\XX$.
The inner limit of $\sequence {C_n}_{n \mathop \in \N}$ is:
:$\ds \liminf_n C_n = \set {x \in X: \lim_n \map d {x, C_n} = 0}$
where $d$ stands for the point-to-set distance mapping.
== Proof ==
$(1): \quad$ We need to show that:
:$\ds \limsup_n \ \map d {x, C_n} = 0$
 that:
:$\ds \limsup_n \ \map d {x, C_n} > 0$
that is, that there exists an increasing sequence of indices $\sequence {n_k}_{k \mathop \in \N}$ so that $\map d {x, C_{n_k} } \to_k a > 0$.
This suggests that there exists a $\epsilon_0 \in \R_{>0}$ such that:
:$\forall k \in \N: \map d {x, C_{n_k} } > \epsilon_0$
However, according to Inner Limit in Hausdorff Space by Set Closures:
:$\ds x \in \map \cl {\bigcup_{k \mathop \in \N} C_{n_k} }$
while:
:$\ds \map d {x, \map \cl {\bigcup_{k \mathop \in \N} C_{n_k} } } \ge \epsilon_0$
which is a contradiction.
Hence:
:$\ds \limsup_n \map d {x, C_n} = 0$
That is:
:$\ds \lim_n \map d {x, C_n} = 0$
and thus we have proven that $x$ is in the set $\ds \set {x \in X: \lim_n \ \map d {x, C_n} = 0}$.
$(2): \quad$ Let $\ds x \in \set {x \in X : \lim_n \ \map d {x, C_n} = 0}$.
This is:
:$\ds \lim_n \map d {x, C_n} = 0$
For any $\epsilon > 0$, we can find $n_0 \in \N$ such that:
:$\forall n \ge n_0: \map d {x, C_n} \le \dfrac \epsilon 2$
By definition:
:$\map d {x, C_n} = \inf \set {\norm {x - y}: y \in C_n}$
Thus we can find a $y_n \in C_n$ such that:
:$\norm {y_n - x} < \map d {x, C_n} + \dfrac \epsilon 2 = \epsilon$
That is:
:$\exists y_n \in C_n: \norm {y_n - x} < \epsilon$
Therefore:
:$x \in C_n + \epsilon \BB$
where $\BB$ is the open unit ball of $\XX$.
That is:
:$\BB := \set {x: \norm x < 1}$
Thus:
:$\epsilon \BB = \set {x: \norm x < \epsilon}$
where $C_n + \epsilon \BB$ denotes the Minkowski sum of $C_n$ and $\epsilon \BB$.
From this observation, it follows from Inner Limit in Normed Spaces by Open Balls that:
:$\ds x \in \liminf_n C_n$
== Also see ==
* Inner Limit in Hausdorff Space by Open Neighborhoods
Category:Normed Spaces
Category:Limits of Sequence of Sets
Inner Product/Examples/Lebesgue 2-Space

== Example of Inner Product ==
Let $\tuple {X, \Sigma, \mu}$ be a measure space.
Let $\map {L^2} \mu$ be the Lebesgue $2$-space of $\mu$.
Let $\innerprod \cdot \cdot: \map {L^2} \mu \times \map {L^2} \mu \to \C$ be the mapping defined by:
:$\ds \innerprod f g = \int f \, \overline g \rd \mu$
Then $\innerprod \cdot \cdot$ is an inner product on $\map {L^2} \mu$.
== Proof ==
First of all, by Hölder's Inequality for Integrals with $p = q = 2$, it follows that:
:$\ds \int f \, \overline g \rd \mu$
is defined.
Now checking the axioms for an inner product in turn:
=== $(1)$ Conjugate Symmetry ===
=== $(2)$ Sesquilinearity ===
=== $(3)$ Non-Negative Definiteness ===
=== $(4)$ Positivity ===
Suppose that $\innerprod f f = 0$.
That is:
:$\ds \int \cmod f^2 \rd \mu = 0$
Hence:
:$\ds \int \cmod {f - 0}^2 \rd \mu = 0$
which is to say that $f = 0$ in $\map {L^2} \mu$ by definition of Lebesgue space.
Having verified all the axioms, we conclude $\innerprod \cdot \cdot$ is an inner product.
== Sources ==
* : $\text{I}$ Hilbert Spaces: $\S 1.$ Elementary Properties and Examples: Example $1.3$
Category:Examples of Inner Products
Category:Lebesgue Spaces
Category:Functional Analysis
Inner Product/Examples/Sequences with Finite Support

== Example of Inner Product ==
Let $\GF$ be either $\R$ or $\C$.
Let $V$ be the vector space of sequences with finite support over $\GF$.
Let $f: \N \to \R_{>0}$ be a mapping.
Let $\innerprod \cdot \cdot: V \times V \to \GF$ be the mapping defined by:
:$\ds \innerprod {\sequence {a_n} } {\sequence {b_n} } = \sum_{n \mathop = 1}^\infty \map f n a_n \overline{ b_n }$
Then $\innerprod \cdot \cdot$ is an inner product on $V$.
== Proof ==
First of all, note that $V$ contains only the sequences with finite support.
Therefore, for each $\sequence {a_n}, \sequence{b_n}$ there exists $N \in \N$ such that:
:$\forall n \ge N: a_n = b_n = 0$
and hence:
:$\ds \innerprod {\sequence {a_n} } {\sequence {b_n} } = \sum_{n \mathop = 1}^\infty \map f n a_n \overline {b_n} = \sum_{n \mathop = 1}^{N-1} \map f n a_n \overline {b_n }$
so that $\innerprod \cdot \cdot: V \times V \to \GF$ is indeed defined.
Now checking the axioms for an inner product in turn:
=== $(1)$ Conjugate Symmetry ===
=== $(2)$ Sesquilinearity ===
=== $(3)$ Non-Negative Definiteness ===
=== $(4)$ Positivity ===
Suppose that $\innerprod { \sequence{a_n} } { \sequence{a_n} } = 0$.
That is:
:$\ds \sum_{n \mathop = 1}^\infty \map f n \cmod{ a_n }^2 = 0$
For each $n \in \N$, we have $\map f n \cmod{ a_n }^2 \ge 0$.
Hence, for each $n \in \N$:
:$\ds \sum_{n \mathop = 1}^\infty \map f n \cmod{ a_n }^2 \ge \map f n \cmod{ a_n }^2$
Thus for each $n \in \N$:
:$\map f n \cmod{ a_n }^2 = 0$
Since $\map f n > 0$ it follows that $\cmod{ a_n }^2 = 0$.
Therefore $a_n = 0$ for all $n \in \N$.
Having verified all the axioms, we conclude $\innerprod \cdot \cdot$ is an inner product.
== Sources ==
* : $\text{I}$ Hilbert Spaces: $\S 1.$ Elementary Properties and Examples: Example $1.2$
Category:Examples of Inner Products
Inner Product Norm is Norm

== Theorem ==
Let $V$ be an inner product space over a subfield $\Bbb F$ of $\C$.
Let $\norm {\, \cdot \,}$ denote the inner product norm on $V$.
Then $\norm {\, \cdot \,}$ is a norm on $V$.
== Proof ==
Let us verify the norm axioms in turn.
=== Axiom $(\text N 1)$ ===
=== Axiom $(\text N 2)$ ===
Part $(2)$ of Properties of Semi-Inner Product.
=== Axiom $(\text N 3)$ ===
Part $(3)$ of Properties of Semi-Inner Product.
Hence all the properties of a norm have been shown to hold.
== Sources ==
* : $I.1.5$
Category:Examples of Norms
Category:Inner Product Spaces
Inner Product Space is Uniformly Convex

== Theorem ==
Let $\struct {V, \innerprod \cdot \cdot}$ be an inner product space.
Then $V$ is uniformly convex.
== Proof ==
Let $\norm \cdot$ be the inner product norm for $\struct {V, \innerprod \cdot \cdot}$.
Let $\epsilon > 0$. 
Let $x, y \in V$ be such that: 
:$\norm x = \norm y = 1$
and:
:$\norm {x - y} > \epsilon$
Then from the Parallelogram Law (Inner Product Space), we have:
:$\norm {x + y}^2 + \norm {x - y}^2 = 2 \paren {\norm x^2 + \norm y^2} = 4$
So that: 
:$\norm {x + y}^2 = 4 - \norm {x - y}^2$
Since:
:$\norm {x - y} > \epsilon$
we have:
:$\norm {x + y}^2 < 4 - \epsilon^2$
From positive homogeneity, we have: 
:$\ds \norm {\frac {x + y} 2}^2 < 1 - \frac {\epsilon^2} 4$
We now pick $\delta$ such that: 
:$\ds \paren {1 - \delta}^2 = 1 - \frac {\epsilon^2} 4$
Note that since:
:$\norm {x - y} \le \norm x + \norm y$
we have: 
:$\norm {x - y} \le 2$ 
So if there exists $x, y \in V$ with $\norm x = \norm y = 1$ such that: 
:$\norm {x - y} > \epsilon$
we have $0 < \epsilon < 2$. 
Hence, if $\epsilon \ge 2$, the demand is vacuously true, and we can pick any positive real $\delta > 0$.
Now take $0 < \epsilon < 2$.
We then have: 
:$\ds 0 < 4 - \epsilon^2 < 4$
so that: 
:$\ds 0 < 1 - \frac {\epsilon^2} 4 < 1$
So, we may pick: 
:$\ds \delta = 1 - \sqrt {1 - \frac {\epsilon^2} 4} > 0$
Then, for each $x, y \in V$ such that $\norm x = \norm y = 1$ and $\norm {x - y} > \epsilon$, we have: 
:$\ds \norm {\frac {x + y} 2}^2 < \paren {1 - \delta}^2$
Since $0 < 1 - \delta < 1$, we have: 
:$\ds \norm {\frac {x + y} 2} < 1 - \delta$
Since $\epsilon > 0$ was arbitrary, for each $\epsilon > 0$ we have found $\delta > 0$ such that:
:whenever $x, y \in V$ have $\norm x = \norm y = 1$ and $\norm {x - y} > \epsilon$, we have: 
::$\ds \norm {\frac {x + y} 2} < 1 - \delta$
as required.
So $V$ is uniformly convex.
== Sources ==
* $8$: Exercises
Category:Inner Product Spaces
Category:Uniformly Convex Normed Vector Spaces
Inner Product is Continuous

== Theorem ==
Let $\struct {V, \innerprod \cdot \cdot}$ be a inner product space. 
Let $\norm \cdot$ be the inner product norm on $V$. 
Let $x, y \in V$.
Let $\sequence {x_n}_{n \mathop \in \N}$ and $\sequence {y_n}_{n \mathop \in \N}$ be sequences converging in $\struct {V, \norm \cdot}$ to $x$ and $y$ respectively.
Then we have: 
:$\innerprod {x_n} {y_n} \to \innerprod x y$
as $n \to \infty$.
== Proof ==
Let $\sequence {x_n}_{n \mathop \in \N}$ and $\sequence {y_n}_{n \mathop \in \N}$ be sequences converging to $x$ and $y$ respectively.
From Modulus of Limit in Normed Vector Space, we have that: 
:$\norm {x_n} \to \norm x$
and:
:$\norm {y_n} \to \norm y$
We have:
From Convergent Real Sequence is Bounded, we have: 
:$\sequence {\norm {y_n} }$ is bounded.
That is, there exists a positive real number $M$ such that: 
:$\norm {y_n} < M$
Let $\epsilon > 0$. 
From the definition of a convergent sequence in a normed vector space, there exists $N_1 \in \N$ such that: 
:$\norm {x_n - x} < \dfrac \epsilon {2 M}$
for $n > N_1$.
If $x = 0$, we have: 
:$\norm x = 0$
Then:
for $n > N_1$, so we are done in this case. 
Now take $x \ne 0$, then: 
:$\norm x \ne 0$
So by the definition of a convergent sequence in a normed vector space, there exists $N_2 \in \N$ such that: 
:$\norm {y_n - y} < \dfrac \epsilon {2 \norm x}$
Let:
:$N = \max \set {N_1, N_2}$
Then, for $n > N$, we have: 
So we have: 
:$\innerprod {x_n} {y_n} \to \innerprod x y$
as $n \to \infty$, hence the claim.
Category:Inner Product Spaces
Inner Product is Sesquilinear

== Theorem ==
Let $\mathbb F$ be a subfield of $\C$. 
Let $V$ be a inner product space over $V$ with inner product $\innerprod \cdot \cdot$. 
Define the $f : V \times V \to \mathbb F$ by: 
:$\map f {x, y} = \innerprod x y$
for each $x, y \in V$. 
Then $f$ is sesquilinear.
== Proof ==
Let $\alpha \in \mathbb F$. 
Let $x_1, x_2, y \in V$. 
By the definition of the inner product, $f$ is linear in its first argument.
So, we have: 
:$\innerprod {\alpha x_1 + x_2} y = \alpha \innerprod {x_1} y + \innerprod {x_2} y$
From the definition of the inner product, we also have that $f$ is conjugate symmetric, so: 
So we have: 
:$\innerprod {\alpha x_1 + x_2} y = \alpha \innerprod {x_1} y + \innerprod {x_2} y$
and:
:$\innerprod y {\alpha x_1 + x_2} = \overline \alpha \innerprod y {x_1} + \innerprod y {x_2}$
for $\alpha \in \mathbb F$ and $x_1, x_2, y \in V$. 
So $f$ is sesquilinear.
Category:Inner Product Spaces
Inner Product on Space of Square Summable Mappings is Complex Inner Product

== Theorem ==
Let $\GF$ be a subfield of $\C$.
Let $I$ be a set.
Let $\map {\ell^2} I$ be the space of square summable mappings over $I$.
Let $\innerprod \cdot \cdot: \map {\ell^2} I \times \map {\ell^2} I \to \GF$ be the inner product on $\map {\ell^2} I$.
Then $\innerprod \cdot \cdot$ is a complex inner product.
== Proof ==
By Space of Square Summable Mappings is $L^2$ Space, $\map {\ell^2} I$ is equal to $\map {L^2} {I, \powerset I, \mu}$.
Here, $\mu$ is the counting measure on the subsets of $I$.
By Inner Product/Examples/Lebesgue 2-Space, there is a complex inner product on $\map {\ell^2} I = \map {L^2} {I, \powerset I, \mu}$ defined by:
:$\ds \innerprod f g = \int f ~ \overline g \rd \mu$
By Integral over Counting Measure is Sum over Values, it follows that:
:$\ds \innerprod f g = \int f ~ \overline g \rd \mu = \sum_{i \mathop \in I} \map f i \overline{ \map g i }$
which is exactly the definition of the inner product on $\map {\ell^2} I$.
Category:Examples of Inner Products
Inner Product with Vector is Bounded Linear Functional

== Theorem ==
Let $\GF$ be a subfield of $\C$.
Let $\struct {V, \innerprod \cdot \cdot}$ be an inner product space over $\GF$.
Let $v_0 \in V$.
Then the mapping $L: V \to \GF$ defined by:
:$\map L v := \innerprod v {v_0}$
is a bounded linear functional with norm equal to $\norm {v_0}$.
== Proof ==
By Inner Product with Vector is Linear Functional, $L$ is a linear functional.
To check that $L$ is bounded:
Thus $L$ is bounded by $\norm {v_0}$.
Furthermore, note:
:$\map L {v_0} = \innerprod {v_0} {v_0} = \norm {v_0}^2$
so that indeed $\norm {v_0}$ is the norm of $L$.
== Sources ==
* : $\text{I}$ Hilbert Spaces: $\S 3.$ The Riesz Representation Theorem
Category:Inner Product Spaces
Category:Bounded Linear Functionals
Inner Product with Vector is Linear Functional

== Theorem ==
Let $\GF$ be a subfield of $\C$.
Let $\struct{ V, \innerprod \cdot \cdot }$ be an inner product space over $\GF$.
Let $v_0 \in V$.
Then the mapping $L: V \to \GF$ defined by:
:$\map L v := \innerprod v {v_0}$
is a linear functional.
== Proof ==
Let us directly check the definition of linear functional:
== Sources ==
* : $\text{I}$ Hilbert Spaces: $\S 3.$ The Riesz Representation Theorem
Category:Inner Product Spaces
Category:Linear Functionals
Inner Product with Zero Vector

== Theorem ==
Let $\struct {V, \innerprod \cdot \cdot}$ be an inner product space.
Let $0_V$ be the zero vector of $V$.
Then: 
:$\innerprod {0_V} x = \innerprod x {0_V} = 0$
for all $x \in V$.
== Proof ==
We have: 
so:
:$\innerprod {0_V} x = 0$
From conjugate symmetry, we have: 
:$\innerprod x {0_V} = \overline {\innerprod {0_V} x}$
so:
:$\innerprod x {0_V} = \overline 0 = 0$
Category:Inner Product Spaces
Inradius 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 $r$ denote the inradius of $\triangle ABC$.
Let $R$ denote the circumradius of $\triangle ABC$.
Then:
:$r = 4 R \sin \dfrac A 2 \sin \dfrac B 2 \sin \dfrac C 2$
== Proof ==
== Sources ==
* : $\text V$. Trigonometry: Formulae $(54)$
Category:Inradius in Terms of Circumradius
Category:Incircles of Triangles
Category:Circumcircles of Triangles
Inradius in Terms of Circumradius/Proof

== Theorem ==
== Proof ==
:
Let $D$, $E$ and $F$ be the points where the incircle is tangent to the sides $AC$, $AB$ and $CB$ respectively.
Let $s$ denote the semiperimeter of $\triangle ABC$.
From Tangent Points of Incircle in Terms of Semiperimeter:
Then:
== Sources ==
* : $\text V$. Trigonometry: The incircle
Category:Inradius in Terms of Circumradius
Inradius of Pythagorean Triangle is Integer

== Theorem ==
Let $\triangle ABC$ be a Pythagorean triangle.
Let $r$ be the inradius of $\triangle ABC$.
Then $r$ is an integer.
== Proof 1 ==
== Proof 2 ==
Category:Pythagorean Triangles
Category:Inradius of Pythagorean Triangle is Integer
Inradius of Pythagorean Triangle is Integer/Proof 1

== Theorem ==
== Proof ==
Let $\triangle ABC$ be such that $\angle C$ is a right angle.
Let:
:$A$ be opposite the side $a$
:$B$ be opposite the side $b$
:$C$ be opposite the side $c$
:
From Solutions of Pythagorean Equation, we have that:
:$c = m^2 + n^2$
for some $m, n \in \Z_{>0}$, and that the other sides are $m^2 - n^2$ and $2 m n$
, let the sides $a$ and $b$ be such that:
Hence:
As $m$ and $n$ are both integers, it follows that $r$ is also an integer.
Category:Inradius of Pythagorean Triangle is Integer
Inradius of Pythagorean Triangle is Integer/Proof 2

== Theorem ==
== Proof ==
Let $\triangle ABC$ be such that $\angle C$ is a right angle.
Let:
:$A$ be opposite the side $a$
:$B$ be opposite the side $b$
:$C$ be opposite the side $c$
:
Let $I$ denote the incenter of $\triangle ABC$.
From Solutions of Pythagorean Equation, we have that:
:$c = m^2 + n^2$
for some $m, n \in \Z_{>0}$, and that the other sides are $m^2 - n^2$ and $2 m n$
, let the sides $a$ and $b$ be such that:
We have that:
As $m$ and $n$ are both integers, it follows that $r$ is also an integer.
Category:Inradius of Pythagorean Triangle is Integer
Inradius of Triangle in Terms of Exradii

== Theorem ==
Let $\triangle ABC$ be a triangle whose sides are $a$, $b$ and $c$ opposite vertices $A$, $B$ and $C$ respectively.
Then we have the following relation:
:$\dfrac 1 r = \dfrac 1 {\rho_a} + \dfrac 1 {\rho_b} + \dfrac 1 {\rho_c}$
where:
:$r$ is the inradius
:$\rho_a$, $\rho_b$ and $\rho_c$ are the exradii of $\triangle ABC$ $a$, $b$ and $c$ respectively.
== Proof ==
The area $\AA$ of $\triangle ABC$ is given by:
Similarly:
Hence:
Category:Incircles of Triangles
Category:Excircles of Triangles
Inscribe Square in Circle using Compasses Alone

== Problem ==
:To inscribe a square in a given circle,
:by means of compass alone,
:supposing the center to be known.
=== Corollary ===
== Solution ==
Let $K$ denote the circle.
:
Locate point $B$ on the circumference at which one of the vertices of the square is to be located.
Construct $X$, $C$, $A$ from arcs of radius of $K$ center $B$, $X$ and $C$ respectively.
Construct arcs $CD$ and $XD$ with center $B$ and $A$ respectively of radius $BC$ and $AX$ respectively.
Construct $E$ and $F$ on the circumference of $K$ so as to make $AE$ and $AF$ the same length as $OD$.
$\Box AEBF$ is the square required.
== Sources ==
* 
* : Rational Amusements for Winter Evenings: $158$
Category:Circles
Category:Squares
Inscribe Square in Circle using Compasses Alone/Corollary

== Problem ==
:To inscribe a regular dodecagon in a given circle,
:by means of compass alone,
:supposing the center to be known.
== Solution ==
We use the same construction as for Inscribe Square in Circle using Compasses Alone:
:
Then we note that, for example, $C$ and $E$ are $30 \degrees$ apart.
Hence they are adjacent vertices of a regular dodecagon inscribed in the circle.
It remains to construct the remaining vertices, which is done by, for example, setting the compass to $CE$ and drawing an arc with center $C$ and working round the circle.
== Sources ==
* 
* : Rational Amusements for Winter Evenings: $158$
Category:Circles
Category:Regular Polygons
Inscribed Angle Theorem

== Theorem ==
An inscribed angle is equal to half the angle that is subtended by that arc.
:
Thus, in the figure above:
:$\angle ABC = \frac 1 2 \angle ADC$
:In a circle the angle at the center is double of the angle at the circumference, when the angles have the same circumference as base.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* $(2)$
* $(2)$
Category:Circles
Category:Named Theorems
Category:Inscribed Angle Theorem
Inscribed Angle Theorem/Proof 1

== Theorem ==
== Proof ==
:
Let $ABC$ be a circle, let $\angle BEC$ be an angle at its center, and let $\angle BAC$ be an angle at the circumference.
Let these angles have the same arc $BC$ at their base.
Let $AE$ be joined and drawn through to $F$.
Since $EA = EB$, then from Isosceles Triangle has Two Equal Angles we have that $\angle EBA = \angle EAB$.
So $\angle EBA + \angle EAB = 2 \angle EAB$.
But from Sum of Angles of Triangle equals Two Right Angles we have that $\angle BEF = \angle EBA + \angle EAB$.
That is, $\angle BEF = 2 \angle EAB$.
For the same reason, $\angle FEC = 2 \angle EAC$.
So adding them together, we see that $\angle BEC = 2 \angle BAC$.
Now consider the point $D$, from which we have another angle $\angle BDC$.
Let $DE$ be joined and produced to $G$.
Similarly, we prove that $\angle GEC = 2 \angle EDC$.
Then $\angle GEB = 2 \angle EDB$.
Therefore $\angle BEC$ which remains is equal to $2 \angle BDC$.
Hence the result.
== Sources ==
* : Book $\text{III}$. Propositions
Category:Inscribed Angle Theorem
Inscribed Angle Theorem/Proof 2

== Theorem ==
== Proof ==
Consider the simplest case that occurs when $AC$ is a diameter of the circle:
:
Because all lines radiating from $D$ to the circumference are radii and thus equal:
:$AD = BD = CD$
Hence the triangles $\triangle ADB$ and $\triangle BDC$ are isosceles.
Therefore from Isosceles Triangle has Two Equal Angles:
:$\angle DBC = \angle DCB$.
From Sum of Angles of Triangle equals Two Right Angles:
:$\angle BDC$ is a supplement of $\angle DBC + \angle DCB = 2 \angle DCB$.
From Thales' Theorem, $\angle ABC$ is right.
By similar reasoning $\angle DAB$ is the complement of $\angle DCB$.
If $\angle BDC$ is the supplement of twice the complement of $\angle DAB$, then $\angle BDC = 2 \angle DAB$.
That proves the theorem for this case.
The general case is illustrated below.
:
A diameter is drawn from $A$ through the center $D$ to $E$. 
By the previous logic:
:$\angle BDE = 2 \angle BAE$
:$\angle CDE = 2 \angle CAE$
Subtracting the latter from the former equation obtains the general result. 
Category:Inscribed Angle Theorem
Inscribed Squares in Right-Angled Triangle

== Theorem ==
For any right-angled triangle, two squares can be inscribed inside it.
One square would share a vertex with the right-angled vertex of the right-angled triangle:
:
The other square would have a side lying on the hypotenuse of the right-angled triangle:
::
=== Compass and Straightedge Construction ===
Both inscribed squares can be constructed using compass and straightedge, thus demonstrating their existence.
=== Side Lengths ===
Let $a, b, c$ be the side lengths of the right-angled triangle, where $c$ is the length of the hypotenuse.
Then the side lengths of the inscribed squares are given by:
==== Shared right angle ====
==== Side lies on Hypotenuse ====
== Proof ==
By definition of inscribed polygon, all four vertices of the inscribed square lies on the sides of the right-angled triangle.
By Pigeonhole Principle, at least two of the vertices must lie on the same side of the right-angled triangle.
The case where this side is the hypotenuse would be the second case above.
For the case where this side is not the hypotenuse, if none of the vertices coincide with the right angle of the right-angled triangle, the angle formed by the three vertices not on the hypotenuse would be obtuse, which is not possible since a square contains no obtuse angles.
Therefore a vertex must coincide with the right angle of the right-angled triangle, which is the first case above.
Category:Inscribed Squares in Right-Angled Triangle
Category:Squares
Category:Right Triangles
Inscribed Squares in Right-Angled Triangle/Compass and Straightedge Construction

== Theorem ==
Both inscribed squares can be constructed using compass and straightedge, thus demonstrating their existence:
=== Shared right angle ===
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=== Side lies on Hypotenuse ===
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Category:Inscribed Squares in Right-Angled Triangle
Category:Compass and Straightedge Constructions
Inscribed Squares in Right-Angled Triangle/Compass and Straightedge Construction/Shared Right Angle

== Theorem ==
:
Let $\triangle ABC$ be a right-angled triangle, where $\angle C = 90^\circ$.
Construct the angle bisector of $\angle C$.
Let the point of intersection of this angle bisector and side $AB$ be $E$.
Construct perpedicular lines from $E$ to sides $AC$ and $BC$, and name their intersections $F$ and $D$ respectively.
Then $CDEF$ is an inscribed square.
== Proof ==
Note that $\angle C = 90^\circ$ and $ED \perp BC$, $EF \perp AC$ by construction.
Therefore $CDEF$ is a rectangle.
By definition of an angle bisector, $\angle ECF = 45^\circ$.
Since $\angle CFE = 90^\circ$, by Sum of Angles of Triangle equals Two Right Angles:
:$\angle CEF + \angle CFE + \angle ECF = 180^\circ$
:$\therefore \angle CEF = 180^\circ - 90^\circ - 45^\circ = 45^\circ = \angle ECF$
By Triangle with Two Equal Angles is Isosceles, $CF = FE$.
Since $CDEF$ is a rectangle with two adjacent equal sides, it must also be a square.
Category:Inscribed Squares in Right-Angled Triangle