nyuuzyou/wikis huggingface dataset:
Brahmagupta's Formula

== Theorem ==
The area of a cyclic quadrilateral with sides of lengths $a, b, c, d$ is:
:$\sqrt {\paren {s - a} \paren {s - b} \paren {s - c} \paren {s - d} }$
where $s$ is the semiperimeter:
:$s = \dfrac {a + b + c + d} 2$
=== Corollary ===
== Proof ==
Let $ABCD$ be a cyclic quadrilateral with sides $a, b, c, d$.
:
Area of $ABCD$ = Area of $\triangle ABC$ + Area of $\triangle ADC$
From Area of Triangle in Terms of Two Sides and Angle:
From Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles, $\angle ABC + \angle ADC$ equals two right angles, that is, are supplementary.
Hence we have:
This leads to:
Applying the Law of Cosines for $\triangle ABC$ and $\triangle ADC$ and equating the expressions for side $AC$:
:$a^2 + b^2 - 2 a b \cos \angle ABC = c^2 + d^2 - 2 c d \cos \angle ADC$
From the above:
:$\cos \angle ABC = -\cos \angle ADC$
Hence:
:$2 \cos \angle ABC \paren {a b + c d} = a^2 + b^2 - c^2 - d^2$
Substituting this in the above equation for the area:
This is of the form $x^2 - y^2$.
Hence, by Difference of Two Squares, it can be written in the form $\paren {x + y} \paren {x - y}$ as:
When we introduce the expression for the semiperimeter:
:$s = \dfrac {a + b + c + d} 2$
the above converts to:
:$16 \paren {\Area}^2 = 16 \paren {s - a} \paren {s - b} \paren {s - c} \paren {s - d}$
Taking the square root:
:$\Area = \sqrt {\paren {s - a} \paren {s - b} \paren {s - c} \paren {s - d} }$
== Also known as ==
== Also see ==
* This formula is a generalization of Heron's Formula for the area of a triangle, which can be obtained from this by setting $d = 0$.
* The relationship between the general and extended form of Brahmagupta's Formula is similar to how the Law of Cosines extends Pythagoras's Theorem.
* Bretschneider's Formula, which extends this result to the general quadrilateral.
* Area of Quadrilateral with Given Sides is Greatest when Quadrilateral is Cyclic
== Historical Note ==
== Sources ==
* 
* : Chapter $\text {B}.1$: The Pythagorean Theorem: Appendix: The Formulas of Heron and Brahmagupta
* : The Area Enclosed Against The Seashore: $31$
* 
* 
Category:Brahmagupta's Formula
Category:Areas of Quadrilaterals
Brahmagupta's Formula/Also known as

== Brahmagupta's Formula: Also known as == 
Some sources refer to Brahmagupta's Formula as the Archimedes-Heron-Brahmagupta Formula, for and as well as .
The connection comes from the application of this to the triangle to obtain Heron's Formula.
The reference to comes from the possibility that (despite being the one to publish) he may have been the one to first come up with Heron's Formula.
== Sources ==
* 
Category:Brahmagupta's Formula
Brahmagupta's Formula/Corollary

== Corollary to Brahmagupta's Formula ==
The area of a cyclic quadrilateral with sides of lengths $a, b, c, d$ is:
:$\dfrac {\sqrt {\paren {a^2 + b^2 + c^2 + d^2}^2 + 8 a b c d - 2 \paren {a^4 + b^4 + c^4 + d^4} } } 4$
== Proof ==
Brahmagupta's Formula:
:$\AA = \sqrt {\paren {s - a} \paren {s - b} \paren {s - c} \paren {s - d} }$
where $s$ is the semiperimeter:
:$s = \dfrac {a + b + c + d} 2$
Making the substitutions:
results in:
:$\AA = \dfrac {\sqrt {\paren {a^2 + b^2 + c^2 + d^2}^2 + 8 a b c d - 2 \paren {a^4 + b^4 + c^4 + d^4} } } 4$
Category:Brahmagupta's Formula
Category:Areas of Quadrilaterals
Brahmagupta-Fibonacci Identity

== Theorem ==
Let $a, b, c, d$ be numbers.
Then:
:$\paren {a^2 + b^2} \paren {c^2 + d^2} = \paren {a c + b d}^2 + \paren {a d - b c}^2$
=== Corollary ===
=== General result ===
=== Extension ===
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Also see ==
Lagrange's Identity, of which this the case where $n = 2$.
== Also known as ==
This identity is also known as Diophantus's Identity, for .
Some sources also give it as as Fibonacci's Identity.
== Historical Note ==
== Sources ==
* : $13$
* : $13$
Category:Algebra
Category:Sums of Squares
Category:Brahmagupta-Fibonacci Identity
Brahmagupta-Fibonacci Identity/Corollary

== Theorem ==
Let $a, b, c, d$ be numbers.
Then:
:$\paren {a^2 + b^2} \paren {c^2 + d^2} = \paren {a c - b d}^2 + \paren {a d + b c}^2$
== Proof ==
== Historical Note ==
Category:Brahmagupta-Fibonacci Identity
Brahmagupta-Fibonacci Identity/Extension

== Extension to Brahmagupta-Fibonacci Identity ==
Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n$ be integers.
Then:
:$\ds \prod_{j \mathop = 1}^n \paren { {a_j}^2 + {b_j}^2} = c^2 + d^2$
where $c, d \in \Z$.
That is: the product of any number of sums of two squares is also a sum of two squares.
=== General Version ===
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
Category:Brahmagupta-Fibonacci Identity
Brahmagupta-Fibonacci Identity/Extension/General

== Extension to Brahmagupta-Fibonacci Identity ==
Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n, m$ be integers.
Then:
:$\ds \prod_{j \mathop = 1}^n \paren { {a_j}^2 + m {b_j}^2} = c^2 + m d^2$
for some $c, d \in \Z$.
That is: the set of all integers of the form $a^2 + m b^2$ is closed under multiplication.
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
:$\ds \prod_{j \mathop = 1}^n \paren { {a_j}^2 + m {b_j}^2} = c^2 + m d^2$
for some $c, d \in \Z$.
$\map P 1$ is the trivial case:
Thus $\map P 1$ is seen to hold.
=== Basis for the Induction ===
$\map P 2$ is the case:
Thus $\map P 2$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$\ds \prod_{j \mathop = 1}^k \paren { {a_j}^2 + m {b_j}^2} = c^2 + m d^2$
for some $c, d \in \Z$.
from which it is to be shown that:
:$\ds \prod_{j \mathop = 1}^{k + 1} \paren { {a_j}^2 + m {b_j}^2} = c^2 + m d^2$
for some $c, d \in \Z$.
=== Induction Step ===
This is the induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore, for all $n \in \Z_{> 0}$:
:$\ds \prod_{j \mathop = 1}^n \paren { {a_j}^2 + m {b_j}^2} = c^2 + m d^2$
for some $c, d \in \Z$.
Category:Brahmagupta-Fibonacci Identity
Brahmagupta-Fibonacci Identity/Extension/Proof 1

== Extension to Brahmagupta-Fibonacci Identity ==
== Proof ==
From the extension to the general Brahmagupta-Fibonacci Identity:
:$\ds \prod_{j \mathop = 1}^n \paren { {a_j}^2 + m {b_j}^2} = c^2 + m d^2$
for some $c, d \in \Z$, for all $m \in \Z$.
The result follows by setting $m = 1$.
Category:Brahmagupta-Fibonacci Identity
Brahmagupta-Fibonacci Identity/Extension/Proof 2

== Extension to Brahmagupta-Fibonacci Identity ==
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
:$\ds \prod_{j \mathop = 1}^n \paren { {a_j}^2 + {b_j}^2} = c^2 + d^2$
for some $c, d \in \Z$.
$\map P 1$ is the trivial case:
Thus $\map P 1$ is seen to hold.
=== Basis for the Induction ===
$\map P 2$ is the case:
Thus $\map P 2$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$\ds \prod_{j \mathop = 1}^k \paren { {a_j}^2 + {b_j}^2} = c^2 + d^2$
for some $c, d \in \Z$.
from which it is to be shown that:
:$\ds \prod_{j \mathop = 1}^{k + 1} \paren { {a_j}^2 + {b_j}^2} = c^2 + d^2$
for some $c, d \in \Z$.
=== Induction Step ===
This is the induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore, for all $n \in \Z_{>0}$:
:$\ds \prod_{j \mathop = 1}^n \paren { {a_j}^2 + {b_j}^2} = c^2 + d^2$
for some $c, d \in \Z$.
Category:Brahmagupta-Fibonacci Identity
Brahmagupta-Fibonacci Identity/Extension/Proof 3

== Extension to Brahmagupta-Fibonacci Identity ==
== Proof ==
Let $z_j = a_j + i b_j$ for each $j = 1, 2, \ldots, n$.
Let $\ds c + i d = \prod_{j \mathop = 1}^n z_j$.
Then:
As $a_1, a_2, \ldots, a_n$ and $b_1, b_2, \ldots, b_n$ are integers, so are $c$ and $d$.
Thus:
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Miscellaneous Problems: $142$
Category:Brahmagupta-Fibonacci Identity
Brahmagupta-Fibonacci Identity/General

== General version of Brahmagupta-Fibonacci Identity ==
Let $a, b, c, d, n$ be numbers.
:$\paren {a^2 + n b^2} \paren {c^2 + n d^2} = \paren {a c + n b d}^2 + n \paren {a d - b c}^2$
=== Corollary ===
=== Extension ===
== Proof ==
Category:Brahmagupta-Fibonacci Identity
Brahmagupta-Fibonacci Identity/General/Corollary

== Corollary to General Version of Brahmagupta-Fibonacci Identity ==
Let $a, b, c, d, n$ be numbers.
Then:
:$\paren {a^2 + n b^2} \paren {c^2 + n d^2} = \paren {a c - n b d}^2 + n \paren {a d + b c}^2$
== Proof ==
Category:Brahmagupta-Fibonacci Identity
Brahmagupta-Fibonacci Identity/Historical Note

== Historical Note on Brahmagupta-Fibonacci Identity ==
Both and described what is now known as the Brahmagupta-Fibonacci Identity in their writings:
* 628: : Brahmasphutasiddhanta (The Opening of the Universe)
* 1225: : Liber quadratorum (The Book of Squares)
However, it appeared earlier than either of those in 's of the third century C.E.
Category:Historical Notes
Brahmagupta-Fibonacci Identity/Proof 2

== Theorem ==
== Proof ==
From the more general version of Brahmagupta-Fibonacci Identity:
:$\paren {a^2 + n b^2} \paren {c^2 + n d^2} = \paren {a c + n b d}^2 + n \paren {a d - b c}^2$
The result follows by setting $n = 1$.
== Historical Note ==
Category:Brahmagupta-Fibonacci Identity
Brahmagupta-Fibonacci Identity/Proof 3

== Theorem ==
== Proof ==
Lagrange's Identity gives:
:$\ds \paren {\sum_{k \mathop = 1}^n {a_k}^2} \paren {\sum_{k \mathop = 1}^n {b_k}^2} - \paren {\sum_{k \mathop = 1}^n a_k b_k}^2 = \sum_{i \mathop = 1}^{n - 1} \sum_{j \mathop = i + 1}^n \paren {a_i b_j - a_j b_i}^2$
Setting $n = 2$:
== Historical Note ==
Category:Brahmagupta-Fibonacci Identity
Brahmagupta-Fibonacci Identity/Proof 4

== Theorem ==
== Proof ==
Let $z_1 = a - b i, z_2 = c + d i$ be complex numbers.
Let $\cmod z$ denote the complex modulus of a given complex number $z \in \C$.
By definition of complex multiplication:
:$(1): \quad z_1 z_2 = \paren {a c + b d} + \paren {a d - b c} i$
Then:
== Historical Note ==
== Sources ==
* : $\S 1.2$. The Algebraic Theory
Category:Brahmagupta-Fibonacci Identity
Brahmagupta Theorem

== Theorem ==
If a cyclic quadrilateral has diagonals which are perpendicular, then the perpendicular to a side from the point of intersection of the diagonals always bisects the opposite side.
Specifically:
Let $ABCD$ be a cyclic quadrilateral whose diagonals $AC$ and $BD$ are perpendicular, crossing at $M$.
Let $EF$ be a line passing through $M$ and crossing opposite sides $BC$ and $AD$ of $ABCD$.
Then $EF$ is perpendicular to $BC$ $F$ is the midpoint of $AD$.
== Proof ==
:
=== Sufficient Condition ===
Suppose that $EF$ is perpendicular to $BC$.
We need to prove that $AF = FD$.
Thus:
Then by Triangle with Two Equal Angles is Isosceles, it follows that $AF = FM$.
Similarly:
Then by Triangle with Two Equal Angles is Isosceles, it follows that $FD = FM$.
So $AF = FD$, as we needed to show.
=== Necessary Condition ===
Now suppose that $AF = FD$.
We now need to show that $EF$ is perpendicular to $BC$.
From Thales' Theorem (indirectly) we have that $AF = FM = FD$.
So:
We note the result Sum of Angles of Triangle equals Two Right Angles.
We have that $\angle EBM$ and $\angle ECM$ are complementary, as both are angles in $\triangle CBM$, which is a right triangle.
So $\angle EMC$ and $\angle ECM$ are complementary, which means that $\angle CEM$ must be a right angle.
Hence by definition $EF$ is perpendicular to $BC$, as we were to show.
Category:Cyclic Quadrilaterals
Branch Point/Examples/Cube Root of z-a

== Example of Branch Point ==
Let $f: \C \to \C$ be the complex function defined as:
:$\forall z \in \C: \map f z = \paren {z - a}^{1/3}$
for some $a \in \C$.
Then $a$ is a branch point of $f$.
== Proof ==
We need to show that:
:$f$ has exactly one value at $a$ itself.
:$f$ has more than one value at one or more points in every neighborhood of $a$
We have that:
:$f$ has exactly one value at $a$ itself: 
:$\map f a = \paren {a - a}^{1/3} = 0$
To demonstrate the second part, we recall that the $\epsilon$-neighborhood of $a$ is defined as:
:$\map {N_\epsilon} a := \set {z \in \C: \cmod {z - a} < \epsilon}$
Therefore, we need to show that $\forall \epsilon: \map f {\epsilon + a}$ will have more than one value.
From Cube Roots of Unity we have:
We can rewrite $\epsilon$ as $e^{\map \ln \epsilon}$ thus:
:$\map f {\epsilon + a} = \paren {\paren {\epsilon + a} - a}^{1/3} = e^{\tfrac 1 3 \map \ln \epsilon} \times \paren {1}$
:$\map f {\epsilon + a} = \paren {\paren {\epsilon + a} - a}^{1/3} = e^{\tfrac 1 3 \map \ln \epsilon} \times \paren {-\dfrac 1 2 + \dfrac {i \sqrt 3} 2 }$
:$\map f {\epsilon + a} = \paren {\paren {\epsilon + a} - a}^{1/3} = e^{\tfrac 1 3 \map \ln \epsilon} \times \paren {-\dfrac 1 2 - \dfrac {i \sqrt 3} 2 }$
Hence $f$ has more than one value (three in this example) at one or more points in every neighborhood of $a$.
== Sources ==
* 
* 
Category:Examples of Branch Points
Branch of Finite Propositional Tableau is Finite

== Theorem ==
Let $T$ be a finite propositional tableau.
Let $\Gamma$ be a branch of $T$.
Then $\Gamma$ is a finite branch.
== Proof ==
By definition, a finite propositional tableau $T$ is formed by applying the tableau extension rules a finite number of times.
Each tableau extension rule extends $T$ finitely.
Therefore $T$ is a finite tree.
The result follows from Branch of Finite Tree is Finite.
Category:Propositional Tableaux
Branch of Finite Tree is Finite

== Theorem ==
Let $T$ be a finite rooted tree with root node $r_T$.
Let $\Gamma$ be a branch of $T$.
Then $\Gamma$ is a finite branch.
== Proof ==
Let $\Gamma$ be a branch of a finite rooted tree $T$.
 $\Gamma$ were an infinite branch of $T$.
By definition $\Gamma$ contains an infinite number of nodes.
From Subset of Finite Set is Finite, it follows that $\Gamma$ has a finite number of nodes.
From this contradiction it follows that $\Gamma$ can not be an infinite branch of $T$.
Hence the result.
== Sources ==
* : $\S 1.7$: Tableaus
Category:Rooted Trees
Branches of Conchoid of Nicomedes are Asymptotic to Directrix

== Theorem ==
Let $\KK$ be a conchoid of Nicomedes.
Let $\CC$ be the directrix of $\KK$.
Then $\CC$ is an asymptote to both branches of $\KK$ in both directions.
== Proof ==
== Sources ==
* 
* 
* 
* 
Category:Definitions/Conchoid of Nicomedes
Bretschneider's Formula

== Theorem ==
Let $ABCD$ be a general quadrilateral.
Then the area $\AA$ of $ABCD$ is given by:
:$\AA = \sqrt {\paren {s - a} \paren {s - b} \paren {s - c} \paren {s - d} - a b c d \map {\cos^2} {\dfrac {\alpha + \gamma} 2} }$
where:
:$a, b, c, d$ are the lengths of the sides of the quadrilateral
:$s = \dfrac {a + b + c + d} 2$ is the semiperimeter
:$\alpha$ and $\gamma$ are opposite angles.
== Proof ==
:
Let the area of $\triangle DAB$ and $\triangle BCD$ be $\AA_1$ and $\AA_2$.
From Area of Triangle in Terms of Two Sides and Angle:
:$\AA_1 = \dfrac {a b \sin \alpha} 2$ and $\AA_2 = \dfrac {c d \sin \gamma} 2$
From the second axiom of area:
:$\AA = \AA_1 + \AA_2$
so:
The diagonal $p$ can be written in 2 ways using the Law of Cosines:
:$p^2 = a^2 + b^2 - 2 a b \cos \alpha$
:$p^2 = c^2 + d^2 - 2 c d \cos \gamma$
Equality is transitive, so:
Now add this equation to $(1)$.
Then trigonometric identities can be used, as follows:
By expanding the square $\paren {a^2 + b^2 - c^2 - d^2}^2$: 
Adding and subtracting $8 a b c d$ to and from the numerator of the first term of $(2)$:
allows the product $\paren {-a + b + c + d} \paren {a - b + c + d} \paren {a + b - c + d} \paren {a + b + c - d}$ to be formed:
Hence the result.
== Also see ==
* Brahmagupta's Formula is a specific version of Bretschneider's Formula for a cyclic quadrilateral.
In this case, from Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles, $\alpha + \gamma = 180^\circ$ and the formula becomes:
:$\AA = \sqrt {\paren {s - a} \paren {s - b} \paren {s - c} \paren {s - d} }$
* Heron's Formula is Brahmagupta's Formula for triangles, so $d = 0$ and the formula becomes:
:$\AA = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$
== Historical Note ==
== Sources ==
* 
* 
Category:Areas of Quadrilaterals
Brianchon's Theorem

== Theorem ==
Let tangents to $6$ points on a conic section $K$ form a hexagon $H$ to circumscribe the $K$.
Then the main diagonals of $H$ meet at a single point.
:
== Proof ==
== Also see ==
* Pascal's Theorem
== Sources ==
* : $6$
* 
* : $6$
* 
* 
* 
* 
Category:Brianchon's Theorem
Category:Conic Sections
Category:Hexagons
Bridge divides Graph into Two Components

== Theorem ==
Let $G$ be a connected graph.
Let $e$ be a bridge of $G$.
Then the edge deletion $G - e$ contains exactly $2$ components.
== Proof ==
Let $G$ be a connected graph and $e = u v$ be a bridge of $G$.
By definition of bridge, $G - e$ has to be of at least $2$ components.
 $G - e$ were of more than $2$ components.
Let $G_1, G_2, G_3$ be $3$ of those components such that $u \in G_1$ and $v \in G_2$.
Note that $u$ and $v$ cannot both be in the same component or $e$ would not be a bridge.
Now let $e$ be replaced so as to connect $G_1$ and $G_2$ again.
Then $G_3$ would still be disconnected from the rest of $G$, and $G$ itself would therefore be a disconnected graph.
From this contradiction it follows that $G - e$ can only be of two components.
== Sources ==
* : $\S 2.4$: Cut-Vertices and Bridges
Category:Graph Theory
Bridges of Königsberg

== Problem ==
Through the city of flowed the . In this river were two large islands, which were part of the city.
Joining the mainland either side of the river and those two islands there stood seven bridges.
:
It was a popular exercise among the citizens to take a pleasure stroll across the bridges.
The question naturally arose: was it possible to cross each bridge once and once only during the course of a single walk?
More to the point: could this be done such that the walkers end up back at their starting point?
== Solution ==
The shape of the land and the positions of the bridges are irrelevant.
What is important is the relationships between the land and the bridges.
We can represent each of the landmasses as the vertices and the bridges as the edges of a multigraph:
:
The question now evolves into: does this graph allow the construction of an Eulerian circuit, or failing that, an Eulerian trail?
It can be seen that this is a graph in which there are four vertices, each of which is odd.
From Characteristics of Eulerian Graph, it is clear that the walkers can not cross all the bridges once each and return to their starting point, as all the vertices would need to be even for that.
From Characteristics of Traversable Graph, it also follows that this graph is not traversable, as it has more than two odd vertices.
The answer to both questions is, therefore: No.
== Also see ==
This problem does not relate, in any way, to graph theoretic bridges.
== Historical Note ==
== Sources ==
* : $\S 3.1$: The Königsberg Bridge Problem: An Introduction to Eulerian Graphs: Theorem $3.1$
* : Chapter $\text {A}.21$: Euler ()
* : The Bridges of Königsberg: $133$
* 
* 
* 
* 
* 
* 
Category:Eulerian Graphs
Category:Graph Theory
Category:Classic Problems
Bridges of Königsberg/Historical Note

== Historical Note on Bridges of Königsberg ==
The solution of the Bridges of Königsberg problem, in a rather different form, was first given by in his $1736$ paper Solutio problematis ad geometriam situs pertinentis.
This is widely considered as the first ever paper in the field of graph theory.
Mathematically-minded tourists are to be aware that the City of Königsberg is currently a part of Russia, and goes by the name of Kaliningrad.
It is very much not the same as it was. There are now only five bridges, and their configuration corresponds to a graph which is now semi-Eulerian.
Like many places that were taken over by Russia in the 20th century, it has been allowed to deteriorate.
== Sources ==
* : $\S 3$: Appendix $\text A$: Euler
* : $\S 3.1$: The Königsberg Bridge Problem: An Introduction to Eulerian Graphs: Theorem $3.1$
* : Chapter $\text {A}.21$: Euler ()
* : The Bridges of Königsberg: $133$
* 
* 
Category:Historical Notes
Brocard's Problem

== Unsolved Problem ==
For which pairs of (strictly) positive integers $\tuple {m, n}$ do the following hold:
:$n! + 1 = m^2$
The only known pairs are:
== Also see ==
* Definition:Brown Numbers
== Sources ==
* : Chapter $1$: Some Preliminary Considerations: $1.1$ Mathematical Induction: Problems $1.1$: $5 \ \text {(a)}$
* : $4$
* : $7$
* : $121$
* : $4$
* : $7$
* : $121$
* 
Category:Factorials
Category:Square Numbers
Category:Unsolved Problems
Brocard Points are Isogonal Conjugates

==Theorem==
The first and second Brocard Points are Isogonal Conjugates.
==Proof==
Let the point $P$ be first Brocard point of $\triangle ABC$.
Then
:$\angle PAB = \angle PBC = \angle PCA=\omega$
Let the point $Q$ be second Brocard point of $\triangle ABC$.
Then
:$\angle QAC = \angle QCB = \angle QBA=\omega'$
To prove $P,Q$ are Isogonal Conjugates, it suffices to prove $\omega=\omega'$.
Category:Brocard Points
Broken Chord Theorem

== Theorem ==
:
Let $A$ and $C$ be arbitrary points on a circle in the plane.
Let $M$ be a point on the circle with arc $AM = $ arc $MC$.
Let $B$ lie on the minor arc of $AM$.
Draw chords $AB$ and $BC$.
Find $D$ such that $MD \perp BC$.
Then:
:$AB + BD = DC$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Proof 5 ==
Category:Broken Chord Theorem
Category:Circles
Category:Named Theorems
Broken Chord Theorem/Historical Note

== Historical Note on Broken Chord Theorem ==
According to , the Broken Chord Theorem is ascribed to .
The work exists in German (by ) as a translation of a survey by , as well as the original in Arabic, showing about two dozen proofs of the theorem.
Three of these are described as ' own proofs.
* Broken Chord Theorem/Proof 1
* Broken Chord Theorem/Proof 2
* Broken Chord Theorem/Proof 3
The books are:
* 
* 
* 
' book is lost.
Category:Historical Notes
Broken Chord Theorem/Proof 1

== Theorem ==
== Proof ==
Given:
:$MD \perp BC$
Find $E$ such that $BD = DE$ and draw $ME$
Then:
:$MD$ is the perpendicular bisector of $BE$.
By Triangle Side-Angle-Side Congruence:
:$\triangle MDB \cong \triangle MDE$
Let $H$ be a point such that arc $MH$ is equal to arc $BM$.
Label three angles for reference:
Let $\alpha = \angle MBC$.
Let $\beta = \angle MCB$.
Let $\gamma = \angle CMH$.
By the definition of congruence:
:$\angle MBC = \angle MEB = \alpha$
By construction:
:arc $AM$ = arc $AB$ + arc $BM$
Also by construction:
:arc $MC = $ arc $MH + $ arc $HC$
Subtracting equals:
:arc $AB = $ arc $HC$
So:
:$AB = HC$
By construction:
:arc $HC$ subtends $\gamma$
By Inscribed Angle Theorem
:$\angle MCH = \beta$
By construction:
:arc $MHC$ subtends $\alpha$
Equating the results:
:$\alpha = \beta + \gamma$
But $\alpha = \angle MEB$.
By External Angle of Triangle equals Sum of other Internal Angles:
:$\alpha = \angle CME + \beta$
Subtracting:
:$\angle CME = \gamma = \angle CMH$.
$MC$ is shared.
From above:
:$\angle MCH = \angle MCB = \beta$
 
Category:Broken Chord Theorem
Broken Chord Theorem/Proof 2

== Theorem ==
== Proof ==
Let point $E$ be such that $BD = DE$.
Extend $BC$ to $G$ such that $GD = DC$.
 
Given:
:arc $AM$ = arc $MC$
By the definition of isosceles triangles:
:$\triangle MGA$ is isosceles.
By Triangle with Two Equal Angles is Isosceles:
:$\triangle BAG$ is isosceles.
Category:Broken Chord Theorem
Broken Chord Theorem/Proof 3

== Theorem ==
== Proof ==
:
Let $E$ be a point such that $BD = DE$.
Given:
:arc $AM = $ arc $MC$
By Equal Arcs of Circles Subtended by Equal Straight Lines:
:$AM = MC$
By Angles on Equal Arcs are Equal:
:$\angle BAM = \angle MCB$
:$BM$ is shared
We have Ambiguous Case for Triangle Side-Side-Angle Congruence for these three triangles:
* $\triangle BAM$
* $\triangle MBC$
* $\triangle MEC$
Given:
:$\angle MDC$ is a right angle
By External Angle of Triangle is Greater than Internal Opposite:
:$\angle MEC$ is obtuse.
Since $AM = MC$ and $AC$ is the rest of the circumference:
:the major arc of $AM$ is more than half the circle.
It follows that $\angle ABM$ is obtuse.
Category:Broken Chord Theorem