nyuuzyou/wikis huggingface dataset:
Short Exact Sequence Condition of Noetherian Modules

== Theorem ==
Let $A$ be a commutative ring with unity.
Let:
:$0 \longrightarrow M' \stackrel {\alpha} {\longrightarrow} M \stackrel {\beta} {\longrightarrow} M \longrightarrow 0$
be a short exact sequence of $A$-modules.
Then:
:$M$ is Noetherian
:
:$M'$ and $M$ are Noetherian
== Proof ==
=== Necessary condition ===
Assume $M$ is Noetherian.
==== $M'$ is Noetherian ====
Let $N'\subseteq M'$ be a submodule.
Then, $\alpha \sqbrk {N'}\subseteq M$ is a submodule by Image of Linear Transformation is Submodule.
As $M$ is Noetherian, $\alpha \sqbrk {N'}$ has a finite generator:
:$\set {\map\alpha {n'_1}, \ldots , \map\alpha {n' _n}}$
We claim $\set {n' _1,\ldots n'_n}$ is a generator of $N'$.
Indeed, let $x'\in M'$.
Then, there are $a_1,\ldots, a_n\in A$ such that:
As $\map\ker\alpha = 0$:
:$x' = a_1 n'_1 + \cdots + a_n n' _n$
==== $M$ is Noetherian ====
Let $N\subseteq M$ be a submodule.
Then, $\beta ^{-1} \sqbrk {N}\subseteq M$ is a submodule by Preimage of Submodule under Linear Transformation is Submodule.
As $M$ is Noetherian, $\beta ^{-1} \sqbrk {N}$ has a finite generator:
:$\set {m_1, \ldots , m_k}$
We claim $\set {\map\beta {m_1},\ldots , \map\beta {m_k}}$ is a generator of $N$.
Indeed, let $x\in N$.
As $\image\beta = M$, thee is a $x\in M$ such that:
:$\map\beta x = x$
Since $x\in\beta ^{-1} \sqbrk {N}$, there are $b_1,\ldots, b_k\in A$ such that:
:$x = b_1 m_1 + \cdots + b_k m_k$
Therefore:
=== Sufficient condition ===
Let $M'$ and $M$ be Noetherian.
Let $N\subseteq M$ be a submodule.
Then, $\alpha ^{-1} \sqbrk N \subseteq M'$ is a submodule by Preimage of Submodule under Linear Transformation is Submodule.
As $M'$ is Noetherian, $\alpha ^{-1} \sqbrk N$ has a finite generator:
:$\set {m'_1, \ldots , m'_k}$
Furthermore, $\beta \sqbrk N \subseteq M$ is a submodule by Image of Linear Transformation is Submodule.
As $M$ is Noetherian, $\beta \sqbrk N$ has a finite generator:
:$\set {\map\beta {m_1}, \ldots , \map\beta {m_l}}$
We claim $\set {\map\alpha {m'_1},\ldots \map\beta {m'_k}, m_1, \ldots , m_l}$ is a generator of $N$.
Indeed, let $x\in N$.
As $\map\beta x \in \beta \sqbrk N$, there exist $b_1,\ldots, b_l\in A$ such that:
:$\map\beta x = b_1\map\beta {m_1} + \cdots + b_l \map\beta {m_l}$
Therefore:
that implies:
:$x - \paren { b_1 m_1 + \cdots + b_l m_l } \in \map\ker\beta = \image\alpha$
Thus, there is a $x'\in\alpha ^{-1} \sqbrk N$ such that:
:$\map\alpha {x'} = x - \paren { b_1 m_1 + \cdots + b_l m_l }$
On the other hand, there are $a_1,\ldots,a_k\in A$ such that:
:$x' = a_1 m'_1 + \cdots + a_k m'_k$
Therefore:
Category:Noetherian Modules
Shortest Distance between Two Points is Straight Line

== Theorem ==
The shortest distance between $2$ points is a straight line.
== Proof ==
Let $s$ be the length of a curve between $2$ points $A$ and $B$.
The problem becomes one of finding the curve for which $\ds \int_a^B \rd s$ is a minimum.
Hence such a curve has the equation:
:$y = m x + c$
which defines a straight line.
== Sources ==
* : Chapter $\text I$. The First Variation: $1.1$. Introduction
Category:Euclidean Geometry
Shortest Possible Distance between Lattice Points on Straight Line in Cartesian Plane

== Theorem ==
Let $\LL$ be the straight line defined by the equation:
:$a x - b y = c$
Let $p_1$ and $p_2$ be lattice points on $\LL$.
Then the shortest possible distance $d$ between $p_1$ and $p_2$ is:
:$d = \dfrac {\sqrt {a^2 + b^2} } {\gcd \set {a, b} }$
where $\gcd \set {a, b}$ denotes the greatest common divisor of $a$ and $b$.
== Proof ==
Let $p_1 = \tuple {x_1, y_1}$ and $p_2 = \tuple {x_2, y_2}$ be on $\LL$.
Thus:
From Solution of Linear Diophantine Equation, it is necessary and sufficient that:
:$\gcd \set {a, b} \divides c$
for such lattice points to exist.
Also from Solution of Linear Diophantine Equation, all lattice points on $\LL$ are solutions to the equation:
:$\forall k \in \Z: x = x_1 + \dfrac b m k, y = y_1 - \dfrac a m k$
where $m = \gcd \set {a, b}$.
So we have:
for some $k \in \Z$ such that $k \ne 0$.
The distance $d$ between $p_1$ and $p_2$ is given by the Distance Formula:
This is a minimum when $k = 1$.
Hence the result.
== Sources ==
* : $\text {2-3}$ The Linear Diophantine Equation: Exercise $8$
Category:Straight Lines
Category:Linear Diophantine Equations
Side of Area Contained by Rational Straight Line and Fifth Apotome

== Theorem ==
== Proof ==
:
Let the area $AB$ be contained by the rational straight line $AC$ and the fifth apotome $AD$.
It is to be proved that the "side" of $AB$ is a straight line which produces with a rational area a medial whole.
Let $DG$ be the annex of the fifth apotome $AD$.
Then, by definition:
:$AG$ and $GD$ are rational straight lines which are commensurable in square only
:the annex $GD$ is commensurable with the rational straight line $AC$
:the square on the whole $AG$ is greater than the square on the annex $GD$ by the square on a straight line which is incommensurable in length with $AG$.
Let there be applied to $AG$ a parallelogram equal to the fourth part of the square on $GD$ and deficient by a square figure.
From :
:that parallelogram divides $AG$ into incommensurable parts.
Let $DG$ be bisected at $E$.
Let the rectangle contained by $AF$ and $FG$ be applied to $AG$ which is equal to the square on $EG$ and deficient by a square figure.
Therefore $AF$ is incommensurable with $FG$.
We have that $AG$ commensurable in length with $AC$, and both are rational.
Therefore from :
:the rectangle $AK$ is medial.
We have that $DG$ isincommensurable in length with $AC$, while both are rational.
Therefore from : 
:the rectangles $DK$ is medial.
Let the square $LM$ be constructed equal to $AI$.
Let the square $NO$ be subtracted from $LM$ having the common angle $\angle LPM$ equal to $FK$.
Therefore from :
:the squares $LM$ and $NO$ are about the same diameter.
Let $PR$ be the diameter of $LM$ and $NO$.
It is to be shown that $LN$ is the "side" of the area $AB$.
We have that $AK$ is medial and equals $LP^2 + PN^2$.
Therefore $LP^2 + PN^2$ is medial.
We have that the rectangle $DK$ is rational.
But $DK$ equals twice the rectangle contained by $LP$ and $PN$.
Therefore twice the rectangle contained by $LP$ and $PN$ is rational.
We have that $AI$ is incommensurable with $FK$.
Therefore $LP^2$ is incommensurable with $PN^2$.
Therefore $LP$ and $PN$ are straight lines which are incommensurable in square such that $LP^2 + PN^2$ is medial and such that $2 \cdot LP \cdot PN$ is rational.
Therefore by definition $LN$ is a straight line which produces with a rational area a medial whole.
But $LN$ is the "side" of the area $AB$.
Hence the result.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Side of Area Contained by Rational Straight Line and Sixth Apotome

== Theorem ==
== Proof ==
:
Let the area $AB$ be contained by the rational straight line $AC$ and the sixth apotome $AD$.
It is to be proved that the "side" of $AB$ is a straight line which produces with a medial area a medial whole.
Let $DG$ be the annex of the sixth apotome $AD$.
Then, by definition:
:$AG$ and $GD$ are rational straight lines which are commensurable in square only
:neither the whole $AG$ nor the annex $GD$ is commensurable with the rational straight line $AC$
:the square on the whole $AG$ is greater than the square on the annex $GD$ by the square on a straight line which is incommensurable in length with $AG$.
Let there be applied to $AG$ a parallelogram equal to the fourth part of the square on $GD$ and deficient by a square figure.
From :
:that parallelogram divides $AG$ into incommensurable parts.
Let $DG$ be bisected at $E$.
Let the rectangle contained by $AF$ and $FG$ be applied to $AG$ which is equal to the square on $EG$ and deficient by a square figure.
Therefore $AF$ is incommensurable with $FG$.
We have that $AF$ is incommensurable in length with $FG$.
But from :
:$AF : FG = AI : FK$
Therefore from :
:$AI$ is incommensurable with $FK$.
We have that $AG$ commensurable in length with $AC$, and both are rational.
Therefore from :
:the rectangle $AK$ is medial.
We have that $DG$ is incommensurable in length with $AC$, while both are rational.
Therefore from : 
:the rectangles $DK$ is medial.
But $AG$ and $GD$ are rational straight lines which are commensurable in square only.
Therefore $AG$ is incommensurable in length with $GD$.
But:
:$AG : GD = AK : KD$
Therefore from:
:
and:
::
:$AK$ is incommensurable with $KD$.
Let the square $LM$ be constructed equal to $AI$.
Let the square $NO$ be subtracted from $LM$ having the common angle $\angle LPM$ equal to $FK$.
Therefore from :
:the squares $LM$ and $NO$ are about the same diameter.
Let $PR$ be the diameter of $LM$ and $NO$.
It is to be shown that $LN$ is the "side" of the area $AB$.
We have that $AK$ is medial and equals $LP^2 + PN^2$.
Therefore $LP^2 + PN^2$ is medial.
We have that the rectangle $DK$ is medial.
But $DK$ equals twice the rectangle contained by $LP$ and $PN$.
Therefore twice the rectangle contained by $LP$ and $PN$ is medial.
We have that $AK$ is incommensurable with $DK$.
Therefore $LP^2 + PN^2$ is incommensurable with $2 \cdot LP \cdot PN$.
We have that $AI$ is incommensurable with $FK$.
Therefore $LP^2$ is incommensurable with $PN^2$.
Therefore $LP$ and $PN$ are straight lines which are incommensurable in square such that $LP^2 + PN^2$ is medial and such that $2 \cdot LP \cdot PN$ is also medial.
Further, $LP^2 + PN^2$ is incommensurable with $2 \cdot LP \cdot PN$.
Therefore by definition $LN$ is a straight line which produces with a medial area a medial whole.
But $LN$ is the "side" of the area $AB$.
Hence the result.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Side of Rational Plus Medial Area is Divisible Uniquely

== Theorem ==
== Proof ==
Let $AB$ be a side of a rational plus a medial area.
Let $AB$ be divided at $C$ to create $AC$ and $CB$ such that:
:$AC$ and $CB$ are incommensurable in square
:$AC^2 + CB^2$ is medial
:$AC$ and $CB$ contain a rational rectangle.
From :
:$2 \cdot AC \cdot CB$ is a rational rectangle.
Let $AB$ be divided at $D$ such that $AD$ and $DB$ have the same properties as $AB$ and $CB$.
From :
:$AB^2 = \paren {AC + CB}^2 = AC^2 + CB^2 + 2 \cdot AC \cdot CB$
and:
:$AB^2 = \paren {AD + DB}^2 = AD^2 + DB^2 + 2 \cdot AD \cdot DB$
and so:
:$\paren {AC^2 + CB^2} - \paren {AD^2 + DB^2} = 2 \cdot AD \cdot DB - 2 \cdot AC \cdot CB$
Since $2 \cdot AD \cdot DB - 2 \cdot AC \cdot CB$ is rational:
:$\paren {AC^2 + CB^2} - \paren {AD^2 + DB^2}$ is rational.
But $\paren {AC^2 + CB^2}$ and $\paren {AD^2 + DB^2}$ are medial.
From this cannot be the case.
So there can be no such $D$.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Side of Rational plus Medial Area is Irrational

== Theorem ==
== Proof ==
From :
:Let $AB$ and $BC$ be straight lines which are incommensurable in square which make the sum of the squares on them medial but the rectangle contained by them rational.
We have that:
:$AB^2 + BC^2$ is medial
and from :
:$2 \cdot AB \cdot BC$ is rational.
So:
:$AB^2 + BC^2$ is incommensurable with $2 \cdot AB \cdot BC$.
So from :
:$AC^2 = \left({AB + BC}\right)^2 = AB^2 + BC^2 + 2 \cdot AB \cdot BC$ is also incommensurable with $2 \cdot AB \cdot BC$.
But $2 \cdot AB \cdot BC$ is rational.
Therefore $AC^2$ is irrational.
Hence $AC$ is irrational.
Such a straight line is called the side of a rational plus a medial area.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Side of Remaining Area from Rational Area from which Medial Area Subtracted

== Theorem ==
== Proof ==
:
Let $BC$ be a rational area.
Let the medial area $BD$ be subtracted from $BC$.
It needs to be demonstrated that the "side" of the remainder $EC$ is either an apotome or a minor straight line.
Let $FG$ be a rational straight line.
Let the rectangle $GH$ be applied to $FG$ equal to $BC$ producing $FH$ as breadth.
Let the area $GK$ equal to $BC$ be subtracted from $GH$.
Then the remainder $EC$ is equal to $LH$.
We have that $BC$ is rational and $BD$ is medial.
We also have:
:$BC = GH$
and:
:$BD = GK$
Therefore $GH$ is rational and $GK$ is medial.
But $GH$ and $GK$ are applied to a rational straight line $FG$.
Therefore from :
:$FH$ is rational and commensurable in length with $FG$
while from :
:$FK$ is rational and incommensurable in length with $FG$.
Therefore by : 
:$FH$ is incommensurable in length with $FK$.
Therefore $FH$ and $FK$ are rational straight lines which are commensurable in square only.
Therefore $KH$ is an apotome and $KF$ is the annex to $KH$.
We have that:
: $HF^2 = FK^2 + \lambda^2$
where either:
:$\lambda$ is commensurable in length with $HF$
or:
:$\lambda$ is incommensurable in length with $HF$.
First suppose $\lambda$ is commensurable in length with $HF$.
Then $HF$ is commensurable in length with the rational straight line $FG$.
Therefore $KH$ is a first apotome.
But from : 
:the "side" of $LH$ is an apotome.
Next suppose $\lambda$ is incommensurable in length with $HF$.
Then $HF$ is incommensurable in length with the rational straight line $FG$.
Therefore $KH$ is a fourth apotome.
But from : 
:the "side" of $LH$ is minor.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Side of Spherical Triangle is Less than 2 Right Angles

== Theorem ==
Let $ABC$ be a spherical triangle on a sphere $S$.
Let $AB$ be a side of $ABC$.
The length of $AB$ is less than $2$ right angles.
== Proof ==
$A$ and $B$ are two points on a great circle $E$ of $S$ which are not both on the same diameter.
So $AB$ is not equal to $2$ right angles.
Then it is noted that both $A$ and $B$ are in the same hemisphere, from Three Points on Sphere in Same Hemisphere.
That means the distance along $E$ is less than one semicircle of $E$.
The result follows by definition of spherical angle and length of side of $AB$.
== Sources ==
* : Chapter $\text I$. Spherical Trigonometry: $2$. The spherical triangle.
Category:Spherical Triangles
Side of Spherical Triangle is Supplement of Angle of Polar Triangle

== Theorem ==
Let $\triangle ABC$ be a spherical triangle on the surface of a sphere whose center is $O$.
Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.
Let $\triangle A'B'C'$ be the polar triangle of $\triangle ABC$.
Then $A'$ is the supplement of $a$.
That is:
:$A' = \pi - a$
and it follows by symmetry that:
:$B' = \pi - b$
:$C' = \pi - c$
== Proof ==
:
Let $BC$ be produced to meet $A'B'$ and $A'C'$ at $L$ and $M$ respectively.
Because $A'$ is the pole of the great circle $LBCM$, the spherical angle $A'$ equals the side of the spherical triangle $A'LM$.
That is:
:$(1): \quad \sphericalangle A' = LM$
From Spherical Triangle is Polar Triangle of its Polar Triangle, $\triangle ABC'$ is also the polar triangle of $\triangle A'B'C'$.
That is, $C$ is a pole of the great circle $A'LB'$.
Hence $CL$ is a right angle.
Similarly, $BM$ is also a right angle.
Thus we have:
By definition, we have that:
:$BC = a$
Then:
where $2 \Box$ is $2$ right angles, that is, $\pi$ radians.
That is, $A'$ is the supplement of $a$:
:$A' = \pi - a$
By applying the same analysis to $B'$ and $C'$, it follows similarly that:
:$B' = \pi - b$
:$C' = \pi - c$
== Sources ==
* : Chapter $\text I$. Spherical Trigonometry: $11$. Polar formulae.
* 
* 
Category:Polar Triangles
Category:Supplementary Angles
Side of Sum of Medial Areas is Irrational

== Theorem ==
== Proof ==
:
From :
:Let $AB$ and $BC$ be two straight lines incommensurable in square such that:
:: $AB^2 + BC^2$ is medial
:: $AB \cdot BC$ is medial
:: $AB \cdot BC$ is incommensurable with the $AB^2 + BC^2$.
Let $DE$ be a rational straight line.
Using :
:Let $DF$ be a rectangle set out on $DE$ equal to $AB^2 + BC^2$
:Let $GH$ be a rectangle set out on $DE$ equal to $2 \cdot AB \cdot BC$.
From :
:The rectangle $DH$ equals $AC^2$.
Since $AB^2 + BC^2$ is medial, $DF$ is medial.
We have that $DF$ is applied to the rational straight line $DE$.
Therefore from :
:$DG$ is rational and incommensurable in length with $DE$.
For the same reason $GK$ is rational and incommensurable in length with $GF$, that is, $DE$.
Since $AB^2 + BC^2$ is incommensurable with $2 \cdot AB \cdot BC$:
:$DF$ is incommensurable with $GH$.
So by:
: 
and:
: :
:$DG$ is also incommensurable in length with $GK$.
Also, $DG$ and $GK$ are rational.
Therefore $DG$ and $GK$ are rational straight lines which are commensurable in square only.
Therefore $DK$ is irrational.
From :
:$DK$ is binomial.
But $DE$ is rational.
Therefore from :
:$DH$ is irrational.
Also, the side of the square equal to $DH$ is irrational.
But $AC$ is the side of the square equal to $DH$.
Therefore $AC$ is irrational.
 
Such a straight line is called the side of the sum of two medial areas.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Side of Sum of Two Medial Areas is Divisible Uniquely

== Theorem ==
== Proof ==
:
Let $AB$ be a side of a sum of two medial areas.
Let $AB$ be divided at $C$ to create $AC$ and $CB$ such that:
:$AC$ and $CB$ are incommensurable in square
:$AC^2 + CB^2$ is medial
:$AC$ and $CB$ contain a medial rectangle
:$AC \cdot CB$ is incommensurable with $AC^2 + CB^2$.
Let $AB$ be divided at $D$ such that $AD$ and $DB$ have the same properties as $AB$ and $CB$.
Let $AC > BD$.
From :
:$AB^2 = \left({AC + CB}\right)^2 = AC^2 + CB^2 + 2 \cdot AC \cdot CB$
and:
:$AB^2 = \left({AD + DB}\right)^2 = AD^2 + DB^2 + 2 \cdot AD \cdot DB$
Let $EF$ be a rational straight line.
Using :
:Let $EG$ be a rectangle set out on $EF$ equal to $AC^2 + CB^2$
and
:Let $HK$ be a rectangle set out on $EF$ equal to $2 \cdot AC \cdot CB$.
Therefore $EK$ is a rectangle equal to $AB^2$.
Using :
:Let $EL$ be a rectangle set out on $EF$ equal to $AD^2 + DB^2$
The remainder $MK$ is a rectangle set out on $EF$ equal to $2 \cdot AD \cdot DB$.
By hypothesis, $AC^2 + CB^2$ is medial.
We have that $AC^2 + CB^2$ is applied to a rational straight line $EF$.
So from :
:$HE$ is rational and incommensurable in length with $EF$.
For the same reason:
:$HN$ is rational and incommensurable in length with $EF$.
Since $AC^2 + CB^2$ is incommensurable with $2 \cdot AC \cdot CB$:
:$EG$ is incommensurable with $GN$.
So by:
: 
and:
: 
it follows that:
:$EH$ is incommensurable with $HN$.
Also, $EH$ and $HN$ are rational.
Therefore $EH$ and $HN$ are rational straight lines commensurable in square only.
Therefore from :
:$EN$ is a binomial straight line divided at $H$.
Using the same argument we can deduce:
:$EN$ is a binomial straight line divided at $M$.
But this contradicts .
So there can be no such $D$.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Sides Appended of Hexagon and Decagon inscribed in same Circle are cut in Extreme and Mean Ratio

== Theorem ==
== Proof ==
:
Let $ABC$ be a circle.
Let $BC$ be the side of a regular decagon which has been inscribed within $ABC$.
Let $BC$ be produced to $D$, where $CD$ is equal to the side of a regular hexagon which can be inscribed within $ABC$.
It is to be demonstrated that the straight line $BCD$ is cut in extreme and mean ratio at the point $C$, and that $CD$ is the greater segment.
Let $E$ be the center of $ABC$.
Let $EB, EC, ED$ be joined.
Let $BE$ be produced to $A$.
We have that $BC$ is the side of a regular decagon.
Therefore the arc $ACB$ is five times the arc $CB$.
Therefore the arc $AC$ is five times the arc $CB$.
But from :
:$\angle AEC = 4 \cdot \angle CEB$
From :
:$\angle EBC = \angle ECB$
Therefore by :
:$\angle AEC = 2 \cdot \angle ECB$
From :
:$EC = CD$
So from :
:$\angle CED = \angle CDE$
Therefore by :
:$\angle ECB = 2 \cdot \angle EDC$
But it has been proved that:
:$\angle AEC = 2 \cdot \angle ECB$
Therefore:
:$\angle AEC = 4 \cdot \angle EDC$
But it has been proved that:
:$\angle AEC = 4 \cdot \angle BEC$
Therefore:
:$\angle EDC = \angle BEC$
But $\angle EBD$ is common to $\triangle BEC$ and $\triangle BED$.
Therefore by :
:$\angle BED = \angle ECB$
Therefore $\triangle EBD$ is equiangular with $\triangle EBC$.
Therefore from :
:$BD : BE = EB : BC$
But $EB = CD$
Therefore:
:$BD : DC = DC : CB$
and:
:$BD > DC$
Therefore:
:$DC > CB$
Therefore $BD$ is cut in extreme and mean ratio at the point $C$, and $CD$ is the greater segment.
== Sources ==
* : Book $\text{XIII}$. Propositions
Category:Hexagons
Category:Decagons
Sides of Equal and Equiangular Parallelograms are Reciprocally Proportional

== Theorem ==
Note: in the above, equal is to be taken to mean of equal area.
== Proof ==
Let $\Box AB$ and $\Box BC$ be two equiangular parallelograms of equal area such that the angles at $B$ are equal.
Let $DB, BE$ be placed in a straight line.
By Two Angles making Two Right Angles make Straight Line it follows that $FB, BG$ also make a straight line.
We need to show that $DB : BE = GB : BF$, that is, the sides about the equal angles are reciprocally proportional.
:
Let the parallelogram $\Box FE$ be completed.
We have that $\Box AB$ is of equal area with $\Box BC$, and $\Box FE$ is another area.
So from Ratios of Equal Magnitudes:
: $\Box AB : \Box FE = \Box BC : \Box FE$
But from Areas of Triangles and Parallelograms Proportional to Base:
: $\Box AB : \Box FE = DB : BE$
Also from Areas of Triangles and Parallelograms Proportional to Base:
: $\Box BC : \Box FE = GB : BF$
So from Equality of Ratios is Transitive:
: $DB : BE = GB : BF$
Next, suppose that $DB : BE = GB : BF$.
From Areas of Triangles and Parallelograms Proportional to Base:
: $DB : BE = \Box AB : \Box FE$
Also from Areas of Triangles and Parallelograms Proportional to Base:
: $GB : BF = \Box BC : \Box FE$
So from Equality of Ratios is Transitive:
: $\Box AB : \Box FE = \Box BC : \Box FE$
So from Magnitudes with Same Ratios are Equal:
: $\Box AB = \Box BC$
== Sources ==
* : Book $\text{VI}$. Propositions
Category:Parallelograms
Sides of Equiangular Triangles are Reciprocally Proportional

== Theorem ==
Note: in the above, equal is to be taken to mean of equal area.
== Proof ==
Let $\triangle ABC, \triangle ADE$ be triangles of equal area which have one angle equal to one angle, namely $\angle BAC = \angle DAE$.
We need to show that $CA : AD = EA : AB$, that is, the sides about the equal angles are reciprocally proportional.
:
Place them so $CA$ is in a straight line with $AD$.
From Two Angles making Two Right Angles make Straight Line $EA$ is also in a straight line with $AB$.
Join $BD$.
It follows from Ratios of Equal Magnitudes that:
: $\triangle CAB : \triangle BAD = \triangle EAD : \triangle BAD$
But from Areas of Triangles and Parallelograms Proportional to Base:
: $\triangle CAB : \triangle BAD = CA : AD$
Also from Areas of Triangles and Parallelograms Proportional to Base:
:$\triangle EAD : \triangle BAD = EA : AB$
So from Equality of Ratios is Transitive:
: $CA : AD = EA : AB$
Now let the sides in $\triangle ABC, \triangle ADE$ be reciprocally proportional.
That is, $CA : AD = EA : AB$.
Join $BD$.
From Areas of Triangles and Parallelograms Proportional to Base:
: $\triangle CAB : \triangle BAD = CA : AD$
Also from Areas of Triangles and Parallelograms Proportional to Base:
: $\triangle EAD : \triangle BAD = EA : AB$
It follows from Equality of Ratios is Transitive that:
: $\triangle CAB : \triangle BAD = \triangle EAD : \triangle BAD$
So from Magnitudes with Same Ratios are Equal:
: $\triangle ABC = \triangle ADE$
== Sources ==
* : Book $\text{VI}$. Propositions
Category:Triangles
Sides of Orthic Triangle of Acute Triangle

== Theorem ==
Let $\triangle ABC$ be an acute triangle with sides $a$, $b$ and $c$ opposite vertices $A$, $B$ and $C$ respectively.
Let $\triangle DEF$ be the orthic triangle of $\triangle ABC$.
Then the sides of $\triangle DEF$ are $a \cos A$, $b \cos B$ and $c \cos C$.
== Proof ==
== Sources ==
* : $\text V$. Trigonometry: Formulae $(49)$
Category:Sides of Orthic Triangle of Acute Triangle
Category:Orthic Triangles
Category:Acute Triangles
Sides of Orthic Triangle of Acute Triangle/Proof

== Theorem ==
== Proof ==
:
Let $H$ be the orthocenter of $\triangle ABC$.
Let $R$ be the circumradius of $\triangle ABC$.
The same argument can be applied to $DF$ and $DE$.
Hence the result.
== Sources ==
* : $\text V$. Trigonometry: The pedal triangle
Category:Sides of Orthic Triangle of Acute Triangle