nyuuzyou/wikis huggingface dataset:
Star of Ideal of *-Algebra is Ideal

== Theorem ==
Let $\struct {A, \ast}$ be a $\ast$-algebra over $\C$. 
Let $I$ be an ideal of $A$.
Define:
:$I^\ast = \set {x^\ast : x \in I}$
Then $I^\ast$ is an ideal of $A$.
 
== Proof ==
From Star of Vector Subspace of *-Algebra is Vector Subspace, $I^\ast$ is a vector subspace of $A$.
Let $u \in I^\ast$ and $y \in A$.
Then $u = x^\ast$ for some $x \in I$.
Then we have:
:$u y = x^\ast y = \paren {y^\ast x}^\ast$
from $(\text C^\ast 1)$ and $(\text C^\ast 3)$ in the definition of an involution.
Since $x \in I$, we have $y^\ast x \in I$ since $I$ is an ideal.
Hence $u y \in I$.
Similarly we have:
:$y u = y x^\ast = \paren {x y^\ast}^\ast$
from $(\text C^\ast 1)$ and $(\text C^\ast 3)$ in the definition of an involution.
Hence $y u \in I$.
So $I$ is an ideal of $A$.
Category:*-Algebras
Star of Product of Elements in *-Algebra

== Theorem ==
Let $\tuple {A, \ast}$ be a $\ast$-algebra.
Let $n \ge 2$. 
Let $x_1, \ldots, x_n \in A$.
Then:
:$\ds \paren {\prod_{j \mathop = 1}^n x_j}^\ast = \prod_{j \mathop = 1}^n x_{n - j + 1}^\ast$
== Proof ==
We proceed by induction on $n$. 
For all $n \in \N$, let $\map P n$ be the proposition:
:$\ds \paren {\prod_{j \mathop = 1}^n x_j}^\ast = \prod_{j \mathop = 1}^n x_{n - j + 1}^\ast$ for all $x_1, \ldots, x_n \in A$. 
=== Basis for the Induction ===
We have:
:$\paren {x y}^\ast = y^\ast x^\ast$
from $(\text C^\ast 3)$ in the definition of an involution.
Hence $\map P 2$ is seen to hold.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\ds \paren {\prod_{j \mathop = 1}^k x_j}^\ast = \prod_{j \mathop = 1}^k x_{n - j + 1}^\ast$ for all $x_1, \ldots, x_k \in A$. 
Then we need to show:
:$\ds \paren {\prod_{j \mathop = 1}^{k + 1} x_j}^\ast = \prod_{j \mathop = 1}^{k + 1} x_{k - j + 2}^\ast$ for all $x_1, \ldots, x_{k + 1} \in A$.
=== Induction Step ===
This is our induction step.
Let $x_1, \ldots, x_{k + 1} \in A$.
We have:
Category:*-Algebras
Star of Vector Subspace of *-Algebra is Vector Subspace

== Theorem ==
Let $\struct {A, \ast}$ be a $\ast$-algebra over $\C$. 
Let $B$ be a vector subspace of $A$.
Define:
:$B^\ast = \set {x^\ast : x \in B}$ 
Then $B^\ast$ is a vector subspace of $A$.
== Proof ==
We use the One-Step Vector Subspace Test.
From Zero Vector in *-Algebra is Hermitian, we have:
:${\mathbf 0}_A^\ast = {\mathbf 0}_A$
Hence ${\mathbf 0}_A \in B$.
In particular $B \ne \O$. 
Now let $u, v \in B^\ast$ and $\lambda \in \C$.
We can then write $u = x^\ast$ and $v = y^\ast$ for $x, y \in B$. 
Then we have:
Since $B$ is a vector subspace, we have $x + \overline \lambda y \in B$.
Hence $u + \lambda v \in B^\ast$.
So by One-Step Vector Subspace Test, $B^\ast$ is a vector subspace of $A$.
Category:*-Algebras
State Space of C*-Algebra is Weak-* Compact

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra.
Let $S_A$ be the set of states on $A$.
Let $w^\ast$ be the weak-$\ast$ topology. 
Then $\struct {S_A, w^\ast}$ is compact.
== Proof ==
By definition, we have:
:$S_A \subseteq B_{A^\ast}$
where $B_{A^\ast}$ is the closed unit ball of $A^\ast$.
From Banach-Alaoglu Theorem, $\struct {B_{A^\ast}, w^\ast}$ is compact.
From Closed Subspace of Compact Space is Compact, it is enough to show that $S_A$ is closed in $\struct {B_{A^\ast}, w^\ast}$.
Let $\family {f_\lambda}_{\lambda \mathop \in \Lambda}$ be a net in $S_A$ converging to $f$ in $B_{A^\ast}$.
From Characterization of Convergent Net in Weak-* Topology, we have:
:$\family {\map {f_\lambda} a}_{\lambda \mathop \in \Lambda}$ converges to $f$ for each $x \in A$. 
Let $a \in A$ be positive.
Then:
:$\map {f_\lambda} a \ge 0$ for each $\lambda \in \Lambda$. 
Hence from Inequality Rule for Real Convergent Nets, we have:
:$\map f a \ge 0$ 
So $f$ is a positive linear functional. 
Further, we have:
:$\norm {f_\lambda}_{A^\ast} = 1$ for each $\lambda \in \Lambda$.
From Norm is Continuous and Characterization of Continuity in terms of Nets, we have:
:$\norm f_{A^\ast} = 1$ 
So $f \in S_A$.
Hence whenever $\family {f_\lambda}_{\lambda \mathop \in \Lambda}$ be a net in $S_A$ converging to $f$ in $B_{A^\ast}$, we have $f \in S_A$.
Hence from Characterization of Closedness in terms of Nets, $S_A$ is closed in $\struct {B_{A^\ast}, w^\ast}$. 
Hence from Closed Subspace of Compact Space is Compact, $\struct {S_A, w^\ast}$ is compact.
== Sources ==
* $\text {VIII}.5.15$
Category:State Spaces of C*-Algebras
Category:Weak-* Topologies
State Space of Unital C*-Algebra is Convex

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra.
Let $S_A$ be the set of states on $A$.
Then $S_A$ is convex. 
== Proof ==
Let $t \in \closedint 0 1$ and $f, g \in S_A$.
Then $t f + \paren {1 - t} g$ is a linear functional.
Further, for each $a \in A$ positive, we have:
:$\map f a \ge 0$
and:
:$\map g a \ge 0$
Hence:
:$t \map f a + \paren {1 - t} \map g a \ge 0$
Hence $t f + \paren {1 - t} g$ is a positive linear functional.
From Norm of Positive Linear Functional on Unital C*-Algebra, we have:
:$\norm {t f + \paren {1 - t} g}_{A^\ast} = \map {\paren {t f + \paren {1 - t} g} } { {\mathbf 1}_A}$
We have:
:$\map f { {\mathbf 1}_A} = 1$
and:
:$\map g { {\mathbf 1}_A} = 1$
Hence:
:$t \map f { {\mathbf 1}_A} + \paren {1 - t} \map g { {\mathbf 1}_A} = t + \paren {1 - t} = 1$
So:
:$\norm {t f + \paren {1 - t} g}_{A^\ast} = 1$
Hence $S_A$ is convex.
== Sources ==
* $\text {VIII}.5.15$
Category:State Spaces of C*-Algebras
State on Unital C*-Subalgebra extends to whole C*-Algebra

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra. 
Let $B$ be a unital $\text C^\ast$-subalgebra of $A$.
Let $f_1 : B \to \C$ be a state.
Then there exists a state $f : A \to \C$ extending $f_1$. 
== Proof ==
Let $\le_A$ be the canonical preordering of $A$.
Let $B_{\mathbf {SA} }$ and $A_{\mathbf {SA} }$ be the set of Hermitian elements of $B$ and $A$ respectively.
From Hermitian Elements of *-Algebra form Real Vector Subspace, these are vector spaces over $\R$. 
From Bounds on Hermitian Element of Unital C*-Algebra in terms of Bounds on Spectrum, for each $a \in A_{\mathbf {SA} }$ we have:
:$a \le_A \norm a {\mathbf 1}_A$
where $\norm a {\mathbf 1}_A \in B_{\mathbf {SA} }$. 
Hence $B_{\mathbf {SA} }$ is cofinal in the preordered vector space $\struct {A_{\mathbf {SA} }, \le_A}$.
From Positive Linear Functional on C*-Algebra preserves Star and Complex Number equals Conjugate iff Wholly Real, $f_1$ is real-valued on $B_{\mathbf {SA} }$. 
Hence from Extension Theorem for Positive Linear Functional defined on Cofinal Linear Subspace, there exists a positive linear functional $f_0 : A_{\mathbf {SA} } \to \R$ extending $f_1 \restriction {B_{\mathbf {SA} } }$.
For general $a \in A$, define:
:$\map f a = \map {f_0} {\map \Re a} + i \map {f_0} {\map \Im a}$
For $b \in B$, we have: 
Hence $f$ extends $f_1$. 
We show that $f$ is as desired.
We first show that $f$ is linear.
Firstly, for $a \in A$ we have:
Now let $a \in A$ and $\lambda \in \R$, we have that:
We also have, for $a, b \in A$:
Hence for $a, b \in A$ and $\lambda \in \C$ we have:
For $a \in A$ positive, we have:
:$\map f a = \map {f_0} a \ge 0$
Hence $f$ is a positive linear functional.
From Norm of Positive Linear Functional on Unital C*-Algebra, we have:
Hence $f$ is a state that extends $f_0$ as required.
== Sources ==
* $\text {VIII}.5.16$
Category:State Spaces of C*-Algebras
Statement Form/Examples

== Examples of Statement Forms ==
=== Napoleon ===
=== Shape of Eggs ===
=== Barometer ===
=== Arbitrary Example 1 ===
=== Arbitrary Example 2 ===
=== Arbitrary Example 3 ===
=== Arbitrary Example 4 ===
=== Arbitrary Example 5 ===
=== Arbitrary Example 6 ===
=== Arbitrary Example 7 ===
=== Arbitrary Example 8 ===
Of the above statements:
:Arbitrary Example 1 and Arbitrary Example 4 have the same statement form
:Shape of Eggs and Arbitrary Example 2 have the same statement form.
Of the above statements:
:Arbitrary Example 5 and Arbitrary Example 6 have the same meaning
and:
:Arbitrary Example 4 and Arbitrary Example 8 have the same meaning.
== Sources ==
* : $\S 1$: Informal statement calculus: $\S 1.1$: Statements and connectives: Exercises $2 \ \text {(a)}, \text {(b)}$
Category:Examples of Statements
Statement Form/Examples/Arbitrary Example 1

== Example of Statement Form ==
:If demand has remained constant and prices have been increased, then turnover must have decreased
has the statement form
:$\paren {A \land B} \implies C$
where:
:$A$ stands for Demand has remained constant
:$B$ stands for Prices have been increased
:$C$ stands for Turnover must have decreased.
== Sources ==
* : $\S 1$: Informal statement calculus: $\S 1.1$: Statements and connectives: Exercises $1 \ \text {(a)}$
Category:Examples of Statements
Statement Form/Examples/Arbitrary Example 2

== Example of Statement Form ==
:We shall win the election, provided that Jones is elected leader of the party
has the statement form
:$A \implies B$
where:
:$A$ stands for Jones is elected leader of the party
:$B$ stands for We shall win the election.
== Sources ==
* : $\S 1$: Informal statement calculus: $\S 1.1$: Statements and connectives: Exercises $1 \ \text {(b)}$
Category:Examples of Statements
Statement Form/Examples/Arbitrary Example 3

== Example of Statement Form ==
:If Jones is not elected leader of the party, then either Smith or Robinson will leave the cabinet, and we shall lose the election
has the statement form
:$\neg A \implies \paren {\paren {B \lor C} \land D}$
where:
:$A$ stands for Jones is elected leader of the party
:$B$ stands for Smith will leave the cabinet
:$C$ stands for Robinson will leave the cabinet
:$D$ stands for We shall lose the election.
== Sources ==
* : $\S 1$: Informal statement calculus: $\S 1.1$: Statements and connectives: Exercises $1 \ \text {(c)}$
Category:Examples of Statements
Statement Form/Examples/Arbitrary Example 4

== Example of Statement Form ==
:If $x$ is a rational number and $y$ is an integer, then $z$ is not real
has the statement form
:$\paren {A \land B} \implies \neg C$
where:
:$A$ stands for $x$ is a rational number
:$B$ stands for $y$ is an integer
:$C$ stands for $z$ is real.
== Sources ==
* : $\S 1$: Informal statement calculus: $\S 1.1$: Statements and connectives: Exercises $1 \ \text {(d)}$
Category:Examples of Statements
Statement Form/Examples/Arbitrary Example 5

== Example of Statement Form ==
:Either the murderer has left the country or somebody is harbouring him
has the statement form
:$A \lor B$
where:
:$A$ stands for The murderer has left the country
:$B$ stands for Somebody is harbouring him.
== Sources ==
* : $\S 1$: Informal statement calculus: $\S 1.1$: Statements and connectives: Exercises $1 \ \text {(e)}$
Category:Examples of Statements
Statement Form/Examples/Arbitrary Example 6

== Example of Statement Form ==
:If the murderer has not left the country, then somebody is harbouring him
has the statement form
:$\neg A \implies B$
where:
:$A$ stands for The murderer has left the country
:$B$ stands for Somebody is harbouring him.
== Sources ==
* : $\S 1$: Informal statement calculus: $\S 1.1$: Statements and connectives: Exercises $1 \ \text {(f)}$
Category:Examples of Statements
Statement Form/Examples/Arbitrary Example 7

== Example of Statement Form ==
:The sum of two numbers is even if and only if either both numbers are even or both numbers are odd
has the statement form
:$A \iff \paren {B \lor C}$
where:
:$A$ stands for The sum of two numbers is even
:$B$ stands for Both numbers are even.
:$C$ stands for Both numbers are odd.
== Sources ==
* : $\S 1$: Informal statement calculus: $\S 1.1$: Statements and connectives: Exercises $1 \ \text {(g)}$
Category:Examples of Statements
Statement Form/Examples/Arbitrary Example 8

== Example of Statement Form ==
:If $y$ is an integer then $z$ is not real, provided that $x$ is a rational number
has the statement form
:$A \implies \paren {B \implies \neg C}$
where:
:$A$ stands for $x$ is a rational number
:$B$ stands for $y$ is an integer.
:$C$ stands for $z$ is real.
== Sources ==
* : $\S 1$: Informal statement calculus: $\S 1.1$: Statements and connectives: Exercises $1 \ \text {(h)}$
Category:Examples of Statements
Statement Form/Examples/Barometer

== Example of Statement Form ==
:If the barometer falls then either it will rain or it will snow
has the statement form
:$A \implies \paren {B \lor C}$
where:
:$A$ stands for The barometer falls
:$B$ stands for It will rain
:$C$ stands for It will snow
== Sources ==
* : $\S 1$: Informal statement calculus: $\S 1.1$: Statements and connectives
Category:Examples of Statements
Statement Form/Examples/Napoleon

== Example of Statement Form ==
:Napoleon is dead and the world is rejoicing
has the statement form:
:$A \land B$
where:
:$A$ stands for Napoleon is dead
:$B$ stands for The world is rejoicing
== Sources ==
* : $\S 1$: Informal statement calculus: $\S 1.1$: Statements and connectives
Category:Examples of Statements
Statement Form/Examples/Shape of Eggs

== Example of Statement Form ==
:If all eggs are not square then all eggs are round
has the statement form
:$A \implies B$
where:
:$A$ stands for All eggs are not square
:$B$ stands for All eggs are round
== Sources ==
* : $\S 1$: Informal statement calculus: $\S 1.1$: Statements and connectives
Category:Examples of Statements
Stationary Point is Saddle Point iff Hessian is Negative

== Theorem ==
Let $\SS$ be a surface defined by the Cartesian equation $z = \map f {x, y}$.
Let $P$ be a stationary point on $\SS$.
Then $P$ is a saddle point of $\SS$ the Hessian of $\SS$ at $P$ is negative.
== Proof ==
== Sources ==
* 
* 
Category:Saddle Points (Geometry)
Category:Stationary Points
Statistical Time Series/Examples/Batch Process

== Example of Statistical Time Series ==
Consider a manufacturing plant which outputs a product at some variable rate.
Let the output generate a discrete time series, obtained by accumulation.
It is not possible to know exactly what the output will be at a particular timestamp.
It is necessary instead to identify what particular probability distribution is obeyed by the data.
== Sources ==
* :
::Part $\text {I}$: Stochastic Models and their Forecasting:
:::$2$: Autocorrelation Function and Spectrum of Stationary Processes:
::::$2.1$ Autocorrelation Properties of Stationary Models:
:::::$2.1.1$ Time Series and Stochastic Processes: Deterministic and statistical time series
Category:Statistical Time Series
Statistical Transformation/Examples

== Examples of Statistical Transformations ==
Examples of statistical transformations may include:
:taking logarithms of the sample
:taking square roots of the sample
:taking arcsines of the sample (particularly when considering proportions)
in order for the data to be closer to a normal distribution.
Such a statistical transformation may also be to allow fitting of a linear model.
== Sources ==
* 
* 
Category:Examples of Statistical Transformations
Steady-State Solution to Cart attached to Wall by Spring under Forced Vibration

== Theorem ==
=== Problem Definition ===
Then the steady state component of the horizontal position of $C$ at time $t$ can be expressed as:
:$\mathbf x_s = \dfrac {\mathbf F_0} {\sqrt {\paren {k^2 - \omega^2 m^2}^2 + \omega^2 c^2} } \map \cos {\omega t - \phi}$
where:
:$\phi = \map \arctan {\dfrac {\omega c} {k - \omega^2 m} }$
== Proof ==
From Forced Vibration of Cart attached to Wall by Spring, the equation of motion of $C$ is:
:$(1): \quad m \dfrac {\d^2 \mathbf x} {\d t^2} + c \dfrac {\d \mathbf x} {\d t} + k \mathbf x = \mathbf F_0 \cos \omega t$
From Position of Underdamped Cart attached to Wall by Spring under Forced Vibration, the general solution to $(1)$ is:
:$\mathbf x = e^{-b t} \paren {C_1 \cos \alpha t + C_2 \sin \alpha t} + \dfrac {\mathbf F_0} {\sqrt {\paren {k^2 - \omega^2 m^2}^2 + \omega^2 c^2} } \map \cos {\omega t - \phi}$
where:
:$C_1$ and $C_2$ depend upon the conditions of $C$ at time $t = 0$
:$\alpha = \sqrt {\dfrac k m - \dfrac {c^2} {4 m^2} }$
:$\phi = \map \arctan {\dfrac {\omega c} {k - \omega^2 m} }$
We have that $(1)$ is in the form:
:$(2): \quad \dfrac {\d^2 y} {\d x^2} + 2 b \dfrac {\d y} {\d x} + a^2 x = K \cos \omega x$
where:
:$K \in \R: k > 0$
:$a, b \in \R_{>0}: b < a$
whose general solution from Linear Second Order ODE: $y + 2 b y' + a^2 y = K \cos \omega x$ where $b < a$ is:
:$(1): \quad y = e^{-b x} \paren {C_1 \cos \alpha x + C_2 \sin \alpha x} + \dfrac K {\sqrt {4 b^2 \omega^2 + \paren {a^2 - \omega^2}^2} } \map \cos {\omega x - \phi}$
where:
:$\alpha = \sqrt {a^2 - b^2}$
:$\phi = \map \arctan {\dfrac {2 b \omega} {a^2 - \omega^2} }$
By definition, the steady-state component of the general solution to $(2)$ is:
:$\dfrac K {\sqrt {4 b^2 \omega^2 + \paren {a^2 - \omega^2}^2} } \map \cos {\omega x - \phi}$
Thus by definition and comparison, the steady-state component of the general solution to $(1)$ is:
:$\mathbf x_s = \dfrac {\mathbf F_0} {\sqrt {\paren {k^2 - \omega^2 m^2}^2 + \omega^2 c^2} } \map \cos {\omega t - \phi}$
where:
:$\phi = \map \arctan {\dfrac {\omega c} {k - \omega^2 m} }$
== Sources ==
* : $\S 3.20$: Vibrations in Mechanical Systems: $(29)$
Category:Mechanics
Steiner's Calculus Problem

== Theorem ==
Let $f: \R_{>0} \to \R$ be the real function defined as:
:$\forall x \in \R_{>0}: \map f x = x^{1/x}$
Then $\map f x$ reaches its maximum at $x = e$ where $e$ is Euler's number.
== Proof ==
$\dfrac {x^{1/x} } {x^2}$ is always greater than $0$.
Therefore:
:$\map {f'} x > 0$ for $\ln x < 1$
:$\map {f'} x = 0$ for $\ln x = 1$
:$\map {f'} x < 0$ for $\ln x > 1$
By Derivative at Maximum or Minimum, maximum is obtained when $\ln x = 1$,
that is, when $x = e$.
== Sources ==
* 
* 
* : $1 \cdotp 444 \, 667 \, 861 \ldots$
* : $1 \cdotp 44466 \, 7861 \ldots$
Category:Euler's Number
Category:Steiner's Problem
Steiner-Lehmus Theorem

== Theorem ==
Let $ABC$ be a triangle.
Denote the lengths of the angle bisectors through the vertices $A$ and $B$ by $\omega_\alpha$ and $\omega_\beta$.
Let $\omega_\alpha = \omega_\beta$.
Then $ABC$ is an isosceles triangle.
:
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Converse ==
The converse of the theorem is also true, and is much easier to prove:
Category:Triangles
Category:Steiner-Lehmus Theorem
Steiner-Lehmus Theorem/Converse

== Converse to Steiner-Lehmus Theorem ==
Let $ABC$ be an isosceles triangle with $BA = BC$.
Let $D$ be the point of intersection of the angle bisectors through the vertices $A$ with the side $BC$.
Likewise, let $E$ be the point of intersection of the angle bisectors through the vertices $C$ with the side $BA$.
Then:
:$AD = CE$
== Proof ==
We have:
Category:Steiner-Lehmus Theorem
Steiner-Lehmus Theorem/Lemma 1

== Lemma for Steiner-Lehmus Theorem ==
In the same circle, let one chord $PR$, be larger than another, $PQ$.
Then the inscribed angle subtended by $PR$, with its vertex in the major arc, is larger.
== Proof ==
Let there be a circle on center $O$.
Let $PQ$ and $PR$ be two chords on the circle from the same point $P$.
Let chord $PR > PQ$.
Let arc $PQ$ be the minor arc of the chord $PQ$.
, let arc $PR$ be the arc that contains arc $PQ$.
By :
:arc $PR > $ arc $PQ$
Hence the central angles are such that:
:$\angle POR > \angle POQ$
Then an inscribed angle subtended by $PR$ is greater than an inscribed angle subtended by $PQ$
The result follows.
Category:Steiner-Lehmus Theorem
Steiner-Lehmus Theorem/Lemma 2

== Lemma for Steiner-Lehmus Theorem ==
Let $\triangle ABC$ be a triangle.
Let $\angle ABC$ be bisected by $BM$.
Let $\angle ACB$ be bisected by $CN$.
Let $\angle ABC < \angle ACB$.
Then:
:$CN < BM$
== Proof ==
:
Let $\angle NBM = \angle MBC = \beta$.
Let $\angle NCB = \angle NCM = \gamma$.
We have that:
:$\angle ABC < \angle ACB$
Hence:
:$\beta < \gamma$
Let a circle be drawn through $BCN$.
Let $M'$ be on the circle such that:
:$\angle NCM' = \beta$
Since $\beta < \gamma$:
:$\angle NCM' < \angle NCM$
and so:
:$BM' < BM$
By , since $\angle NBC < \angle M'CB$:
:$CN < BM'$
Hence:
:$CN < BM' < BM$
Thus:
:$CN < BM$
Category:Steiner-Lehmus Theorem
Steiner-Lehmus Theorem/Proof 1

== Theorem ==
== Proof ==
Let $a$, $b$, and $c$ be the sides opposite $A$, $B$ and $C$ respectively.
By Length of Angle Bisector, $\omega_\alpha, \omega_\beta$ are given by:
:$\omega_\alpha^2 = \dfrac {b c} {\paren {b + c}^2} \paren {\paren {b + c}^2 - a^2}$
:$\omega_\beta^2 = \dfrac {a c} {\paren {a + c}^2} \paren {\paren {a + c}^2 - b^2}$
Equating $\omega_\alpha^2$ with $\omega_\beta^2$:
Therefore $ABC$ is an isosceles triangle.
Category:Steiner-Lehmus Theorem
Steiner-Lehmus Theorem/Proof 2

== Theorem ==
== Proof ==
Let $a$, $b$, and $c$ be the sides opposite $A$, $B$ and $C$ respectively.
By Length of Angle Bisector, $\omega_\alpha, \omega_\beta$ are given by:
Equating $\omega_\alpha^2$ with $\omega_\beta^2$ yields:
Substituting $b$ for $a$ in $(1)$ proves that $b = a$ is a solution for $(1)$.
We still have to show that $b = a$ is the only solution for $(1)$.
 $b \ne a$.
, suppose $b > a$.
Because $c \paren {b - a} \paren {b + c}^2 \paren {a + c}^2 > 0$, the of $(1)$ is positive.
We have that:
:$a$, $b$, and $c$ are positive real numbers
and :
:$a < b$
By Real Number Ordering is Compatible with Multiplication and since $a > 0$:
:$a^2 < a \cdot b$
For the same reason and since $b > 0$:
:$a \cdot b < b^2$
By the Transitive Law:
:$a^2 < b^2$
Then:
Since $c a > 0$, the of $(1)$ will be negative :
:$a \paren {a + c}^2 - b \paren {b + c}^2 < 0$
We rearrange as follows:
From above, each term in parenthesis in $(2)$ is negative.
As $c > 0$ we have:
Hence each term in $(2)$ is negative.
Hence their sum is also negative.
Therefore, the of $(1)$ is negative.
Hence we have reached a contradiction.
Similarly, we also get a contradiction by assuming $b < a$.
This assumption yields a negative and a positive for $(1)$.
Therefore, by Proof by Contradiction, $b = a$ is the only solution for $(1)$.
Therefore $ABC$ is an isosceles triangle.
Category:Steiner-Lehmus Theorem
Steiner-Lehmus Theorem/Proof 3

== Theorem ==
== Proof ==
:
Draw $DF \parallel BE$ and $EF \parallel BD$.
By Quadrilateral is Parallelogram iff Both Pairs of Opposite Sides are Equal or Parallel:
:$\Box BEFD$ is a parallelogram.
:$\leadsto \angle DFE = \beta$
Draw $FA$.
Let $\angle EFA = \gamma$.
Let $\angle EAF = \delta$.
 $\alpha > \beta$.
Compare $\triangle BAD$ with $\triangle ABE$. 
The base is the same, $BA$.
But this is a contradiction.
Hence by Proof by Contradiction it cannot be the case that $\alpha > \beta$.
 $\beta > \alpha$.
Again compare $\triangle BAD$ with $\triangle ABE$.
They have the same base, $BC$.
But this is a contradiction.
Hence by Proof by Contradiction it cannot be the case that $\alpha > \beta$.
Since neither $\alpha > \beta$ nor $\alpha < \beta$, it must be the case that
:$\alpha = \beta$
So by addition:
:$\angle ABC = \angle BAC$
and by Triangle with Two Equal Angles is Isosceles:
:$\triangle ABC$ is isosceles.
The result follows.
Category:Steiner-Lehmus Theorem
Steiner-Lehmus Theorem/Proof 4

== Theorem ==
== Proof ==
=== ===
=== ===
Let $\triangle ABC$ be a triangle.
Let $\angle ABC$ be bisected by $BM$
Let $\angle ACB$ be bisected by $CN$.
Let $BM = CN$.
Suppose $\angle ABC < \angle ACB$.
By :
:$CN < BM$
This is a contradiction.
Suppose $\angle ABC < \angle ACB$.
By :
:$CN > BM$
This is a contradiction.
Since $\angle ABC$ can be neither less than nor greater than $\angle ACB$, $\angle ABC = \angle ACB$.
By Triangle with Two Equal Angles is Isosceles:
:$\triangle ABC$ is an isosceles triangle.
The result follows.
== Sources ==
* : Chapter $1$ : "Points and Lines Connected with a Triangle."
Category:Steiner-Lehmus Theorem
Steiner Inellipse is Unique

== Theorem ==
For every triangle, there is one and only one Steiner inellipse.
== Proof (outline) ==
Take your triangle and apply an affine transformation to map it to an equilateral triangle.
Inscribe an incircle in that equilateral triangle.
Perform the inverse affine transformation to map the equilateral triangle (complete with incircle) back to the triangle you started with.
By properties of affine transformations the circle will be a Steiner inellipse.
Category:Affine Geometry