avewright/proofwiki-math huggingface dataset:
Theorem: Angle Bisector Theorem

Let $\triangle ABC$ be a triangle.
Let $D$ lie on the base $BC$ of $\triangle ABC$.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$AD$ is the angle bisector of $\angle BAC$}}
{{item|(2):|$BD : DC {{=}} AB : AC$}}
{{end-itemize}}
where $BD : DC$ denotes the ratio between the lengths $BD$ and $DC$.

Proof:
=== $(1)$ implies $(2)$ ===
:300px
Let $CE$ be drawn through $C$ parallel to $DA$.
Let $BA$ be produced so as to meet $CE$ at $E$.
From {{EuclidPropLink|book = I|prop = 29|title = Parallelism implies Equal Alternate Angles}} we have that:
:$\angle ACE = \angle CAD$
But {{hypothesis}}:
:$\angle CAD = \angle BAD$
and so:
:$\angle DAB = \angle ACE$
From {{EuclidPropLink|book = I|prop = 29|title = Parallelism implies Equal Corresponding Angles}}:
:$\angle BAD = \angle AEC$
But from above $\angle ACE = \angle BAD$, so:
:$\angle ACE = \angle AEC$
So from {{EuclidPropLink|book = I|prop = 6|title = Triangle with Two Equal Angles is Isosceles}}:
:$AC = AE$
Since $AD \parallel EC$, from {{EuclidPropLink|book = VI|prop = 2|title = Parallel Transversal Theorem}}:
:$BD : DC = BA : AE$
But $AE = AC$, so:
:$BD : DC = AB : AC$
{{Qed|lemma}}
Theorem: Equiangular Triangles are Similar

Let two triangles have the same corresponding angles.
Then their corresponding sides are proportional.
Thus, by definition, such triangles are similar.
{{:Euclid:Proposition/VI/4}}

Proof:
Let $\triangle ABC, \triangle DCE$ be equiangular triangles such that:
{{begin-eqn}}
{{eqn | l = \angle ABC
      | r = \angle DCE
}}
{{eqn | l = \angle BAC
      | r = \angle CDE
}}
{{eqn | l = \angle ACB
      | r = \angle CED
}}
{{end-eqn}}
:300px
Let $BC$ be placed in a straight line with $CE$.
From Two Angles of Triangle are Less than Two Right Angles $\angle ABC + \angle ACB$ is less than two right angles.
As $\angle ACB = \angle DEC$, it follows that $\angle ABC + \angle DEC$ is also less than two right angles.
So from the Parallel Postulate, $BA$ and $ED$, when produced, will meet.
Let this happen at $F$.
We have that $\angle ABC = \angle DCE$.
So from Equal Corresponding Angles implies Parallel Lines:
:$BF \parallel CD$
Again, we have that $\angle ACB = \angle CED$.
Again from Equal Corresponding Angles implies Parallel Lines:
:$AC \parallel FE$
Therefore by definition $\Box FACD$ is a parallelogram.
Therefore from Opposite Sides and Angles of Parallelogram are Equal $FA = DC$ and $AC = FD$.
Since $AC \parallel FE$, it follows from Parallel Transversal Theorem that:
:$BA : AF = BC : CE$
But $AF = CD$ so:
:$BA : AF = BC : CE$
From Proportional Magnitudes are Proportional Alternately:
:$AB : BC = DC : CE$
Since $CD \parallel BF$, from Parallel Transversal Theorem:
:$BC : CE = FD : DE$
But $FD = AC$ so $BC : CE = AC : DE$.
So from Proportional Magnitudes are Proportional Alternately, $BC : CA = CE : ED$.
It then follows from Equality of Ratios Ex Aequali that $BA : AC = CD : DE$.
{{qed}}
{{Euclid Note|4|VI}}
Theorem: Triangles with Proportional Sides are Similar

Let two triangles have corresponding sides which are proportional.
Then their corresponding angles are equal.
Thus, by definition, such triangles are similar.
{{:Euclid:Proposition/VI/5}}

Proof:
Let $\triangle ABC, \triangle DEF$ be triangles whose sides are proportional, so that:
:$ AB : BC = DE : EF$
:$ BC : CA = EF : FD$
:$ BA : AC = ED : DF$
We need to show that
:$\angle ABC = \angle DEF$
:$\angle BCA = \angle EFD$
:$\angle BAC = \angle EDF$
:400px
On the straight line $EF$, and at the points $E, F$ on it, construct $\angle FEG = \angle ABC$ and $\angle EFG = \angle ACB$.
From Sum of Angles of Triangle Equals Two Right Angles, the remaining angle at $A$ equals the remaining angle at $G$.
Therefore $\triangle ABC$ is equiangular with $\triangle GEF$.
From Equiangular Triangles are Similar, the sides about the equal angles are proportional, and those are corresponding sides which subtend the equal angles.
So:
:$AB : BD = GE : EF$
But by hypothesis:
:$AB : BC = DE : EF$
So from Equality of Ratios is Transitive
:$DE : EF = GE : EF$
So each of $DE, GE$ has the same ratio to $EF$.
So from Magnitudes with Same Ratios are Equal:
:$DE = GE$
For the same reason:
:$DF = GF$
So we have that $DE = EG$, $EF$ is common and $DF = FG$.
So from Triangle Side-Side-Side Congruence:
:$\triangle DEF = \triangle GEF$
That is:
:$\angle DEF = \angle GEF, \angle DFE = \angle GFE, \angle EDF = \angle EGF$
As $\angle GEF = \angle ABC$ it follows that:
:$\angle ABC = \angle DEF$
For the same reason $\angle ACB = \angle DFE$ and $\angle BAC = \angle EDF$.
Hence the result.
{{Qed}}
{{Euclid Note|5|VI}}
Theorem: Triangles with One Equal Angle and Two Sides Proportional are Similar

Let two triangles have two corresponding sides which are proportional.
Let the angles adjacent to both of these sides be equal.
Then all of their corresponding angles are equal.
Thus, by definition, such triangles are similar.
{{:Euclid:Proposition/VI/6}}

Proof:
Let $\triangle ABC, \triangle DEF$ be triangles having $\angle BAC = \angle EDF$ and two sides proportional, so that $BA : AC = ED : DF$.
We need to show that $\triangle ABC$ is equiangular with $\triangle DEF$, such that $\angle ABC = \angle DEF$ and $\angle ACB = \angle DFE$.
:350px
Let $\angle FDG$ be constructed equal to either $\angle BAC$ or $\angle EDF$, and $\angle DFG = \angle ACB$.
From Sum of Angles of Triangle Equals Two Right Angles:
:$\angle ABC = \angle DGF$
Therefore $\triangle ABC$ is equiangular with $\triangle DGF$.
From Equiangular Triangles are Similar:
:$BA : AC = GD : DF$
But by hypothesis:
:$BA : AC = ED : DF$
So from Equality of Ratios is Transitive:
:$ED : DF = GD : DF$
From Magnitudes with Same Ratios are Equal:
:$ED = DG$
But $DF$ is common to $\triangle DEF$ and $\triangle DGF$, and $\angle EDF = \angle GDF$.
So from Triangle Side-Angle-Side Congruence:
:$\triangle DEF = \triangle DGF$
But $\angle DFG = \angle ACB$ and so:
:$\angle ABC = \angle DFE$
But by hypothesis:
:$\angle BAC = \angle EDF$
Therefore by Sum of Angles of Triangle Equals Two Right Angles:
:$\angle ABC = \angle DEF$
Hence the result.
{{qed}}
{{Euclid Note|6|VI}}
Theorem: Triangles with One Equal Angle and Two Other Sides Proportional are Similar

Let two triangles be such that one of the angles of one triangle equals one of the angles of the other.
Let two corresponding sides which are adjacent to one of the other angles, be proportional.
Let the third angle in both triangles be either both acute or both not acute.
Then all of the corresponding angles of these triangles are equal.
Thus, by definition, such triangles are similar.
{{:Euclid:Proposition/VI/7}}

Proof:
Let $\triangle ABC, \triangle DEF$ be triangles such that:
:$\angle BAC = \angle EDF$
:two sides adjacent to $\angle ABC$ and $\angle DEF$ proportional, so that $AB : BC = DE : EF$
:$\angle ACB$ and $\angle DFE$ either both acute or both not acute.
We need to show that $\angle ABC = \angle DEF$ and $\angle BCA = \angle EFD$.
First assume that $\angle ACB$ and $\angle DFE$ are both acute.
:350px
Suppose $\angle ABC \ne \angle DEF$.
Then one of them is greater.
{{WLOG}}, let $\angle ABC > \angle DEF$.
On $AB$ and at point $B$, let $\angle ABG$ be constructed equal to $\angle DEF$.
Since $\angle BAG = \angle EDF$ and $\angle ABG = \angle DEF$, from Sum of Angles of Triangle Equals Two Right Angles:
:$\angle AGB = \angle DFE$
So $\triangle ABG$ is equiangular with $\triangle DEF$.
So from Equiangular Triangles are Similar:
:$AB : BG = DE : EF$
But {{hypothesis}}:
:$DE : EF = AB : BC$
Thus from Equality of Ratios is Transitive:
:$AB : BG = AB : BC$
So from Magnitudes with Same Ratios are Equal:
:$BC = BG$
So from Isosceles Triangle has Two Equal Angles:
:$\angle BCG = \angle BGC$
We have {{hypothesis}} that $\angle BCG$ is less than a right angle.
So $\angle BGC$ is less than a right angle.
So from Two Angles on Straight Line make Two Right Angles $\angle AGB$ is greater than a right angle.
But this was proved equal to $\angle DFE$.
But {{hypothesis}} $\angle DFE$ is less than a right angle.
From this contradiction we conclude that:
:$\angle ABC = \angle DEF$
We also have:
:$\angle BAC = \angle DEF$
So from Sum of Angles of Triangle Equals Two Right Angles:
:$\angle ACB = \angle DFE$
So, as we wanted to show, $\triangle ABC$ is equiangular with $\triangle DEF$.
Now, suppose $\angle ACB$ and $\angle DFE$ are both not acute:
:350px
From the same construction we show similarly that $BC = BG$.
So, from Isosceles Triangle has Two Equal Angles:
:$\angle BCG = \angle BGC$
But $\angle BCG$ is not less than a right angle.
So neither is $\angle BGC$ less than a right angle.
So in $\triangle BGC$, $\angle BCG + \angle BGC$ are not less than two right angles.
From Two Angles of Triangle are Less than Two Right Angles, this is impossible.
So from this contradiction we conclude that:
:$\angle ABC = \angle DEF$
Hence, similarly to above, $\triangle ABC$ is equiangular with $\triangle DEF$.
{{qed}}
{{Euclid Note|7|VI}}
Theorem: Zeckendorf's Theorem

Every positive integer has a unique Zeckendorf representation.
That is:
Let $n$ be a positive integer.
Then there exists a unique increasing sequence of integers $\sequence {c_i}$ such that:
{{begin-eqn}}
{{eqn | q = \forall i \in \N
      | l = c_i
      | o = \ge
      | r = 2
      | c = 
}}
{{eqn | q = \forall i \in \N
      | l = c_{i + 1}
      | o = >
      | r = c_i + 1
      | c = 
}}
{{eqn | l = n
      | r = \sum_{i \mathop = 0}^k F_{c_i}
      | c = 
}}
{{end-eqn}}
where $F_m$ is the $m$th Fibonacci number.
For any given $n$, such a $\sequence {c_i}$ is unique.

Proof:
First note that every Fibonacci number $F_n$ is itself a '''Zeckendorf representation''' of itself, where the sequence $\sequence {c_i}$ contains $1$ term.
Theorem: Sum of Non-Consecutive Fibonacci Numbers

Let $S$ be a non-empty set of distinct non-consecutive Fibonacci numbers not containing $F_0$ or $F_1$.
Let the largest element of $S$ be $F_j$.
Then:
:$\ds \sum_{F_i \mathop \in S} F_i < F_{j + 1}$
That is, the sum of all the elements of $S$ is strictly less than the next largest Fibonacci number.
That is, given some increasing sequence $\sequence {c_i}$ satisfying $c_i \ge 2$ and $c_{i + 1} \ge c_i + 1$:
:$\ds F_{c_k + 1} > \sum_{i \mathop = 0}^k F_{c_i}$

Proof:
The proof proceeds by induction on $j$ for $j \ge 2$.
For all $j \in \N_{>0}$, let $\map P j$ be the proposition:
:$\ds \sum_{F_i \mathop \in S} F_i < F_{j + 1}$
Let the term '''allowable set''' be used to mean a non-empty set of distinct non-consecutive Fibonacci numbers not containing $F_0$ or $F_1$.
Theorem: Construction of Third Proportional Straight Line

Given any two straight lines of given length $a$ and $b$, it is possible to construct a third straight line of length $c$ such that $a : b = b : c$.
{{:Euclid:Proposition/VI/11}}

Proof:
:300px
We have that:
:$BC \parallel DE$
So from Parallel Transversal Theorem:
:$AB : BD = AC : CE$
But $BD = AC$ and so:
:$AB : AC = AC : CE$
as required.
{{qed}}
{{Euclid Note|11|VI|It is a special case of {{EuclidPropLink|book = VI|prop = 12|title = Construction of Fourth Proportional Straight Line}}.}}
Theorem: Construction of Fourth Proportional Straight Line

Given three straight lines of lengths $a, b, c$, it is possible to construct a fourth straight line of length $d$ such that $a : b = c : d$.
{{:Euclid:Proposition/VI/12}}

Proof:
:450px
From Parallel Transversal Theorem:
:$DG : GE = DH : HF$
But:
:$DG = A, GE = B, DH = C$
Hence the result.
{{qed}}
{{Euclid Note|12|VI}}
Theorem: Construction of Mean Proportional

Given any two straight lines of length $a$ and $b$ it is possible to find a straight line of length $c$ such that $a : c = c : b$.
{{:Euclid:Proposition/VI/13}}

Proof:
:250px
From Relative Sizes of Angles in Segments, $\angle ADC$ is a right angle.
So from the porism to Perpendicular in Right-Angled Triangle makes two Similar Triangles, $DB$ is the mean proportional between $AB$ and $BC$.
{{qed}}
{{Euclid Note|13|VI|Also see {{EuclidPropLink|book=II|prop=2|title=Construction of Square equal to Given Polygon}} for what amounts to an application of this technique.<br/>Note that nowhere in {{BookLink|The Elements|Euclid}} is the term '''mean proportional''' specifically defined.}}
Theorem: Sides of Equal and Equiangular Parallelograms are Reciprocally Proportional

{{:Euclid:Proposition/VI/14}}
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.
:300px
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:
: $DB : BE = GB : BF$
{{qed|lemma}}
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:
: $\Box AB = \Box BC$
{{qed}}
{{Euclid Note|14|VI}}
Theorem: Sides of Equiangular Triangles are Reciprocally Proportional

{{:Euclid:Proposition/VI/15}}
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.
:250px
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:
: $CA : AD = EA : AB$
{{qed|lemma}}
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:
: $\triangle ABC = \triangle ADE$
{{qed}}
{{Euclid Note|15|VI}}
Theorem: Rectangles Contained by Proportional Straight Lines

{{:Euclid:Proposition/VI/16}}
Note: in the above, ''equal'' is to be taken to mean ''of equal area''.

Proof:
Let the four straight lines $AB, CD, E, F$ be proportional, that is, $AB : CD = E : F$.
What we need to show is that the rectangle contained by $AB$ and $F$ is equal in area to the rectangle contained by $CD$ and $E$.
:400px
Let $AG, CH$ be drawn perpendicular to $AB$ and $CD$.
Let $AG = F$, $CH = E$.
Complete the parallelograms $BG$ and $DH$.
We have that $AB : CD = E : F$, while $E = CH$ and $F = AG$.
So in $\Box BG$ and $\Box DH$ the sides about the equal angles are reciprocally proportional.
But from Sides of Equal and Equiangular Parallelograms are Reciprocally Proportional:
: $\Box BG = \Box DH$ (in area).
We also have that:
: $\Box BG$ is the rectangle contained by $AB$ and $F$
: $\Box DH$ is the rectangle contained by $CD$ and $E$
Hence the result.
{{qed|lemma}}
Now suppose that the rectangle contained by $AB$ and $F$ is equal in area to the rectangle contained by $CD$ and $E$.
We use the same construction, and note that $\Box BG = \Box DH$ (in area).
But they are equiangular, as all angles are equal to a right angle.
So from Sides of Equal and Equiangular Parallelograms are Reciprocally Proportional:
: $AB : CD = CH : AG$
But $E = CH$ and $F = AG$.
So:
: $AB : CD = E : F$
{{qed}}
{{Euclid Note|16|VI|It is a special case of {{EuclidPropLink|book=VI|prop=14|title=Sides of Equal and Equiangular Parallelograms are Reciprocally Proportional}}.}}
Theorem: Construction of Similar Polygon

On any given straight line it is possible to construct a polygon similar to any given polygon.
{{:Euclid:Proposition/VI/18}}

Proof:
From Sum of Angles of Triangle Equals Two Right Angles $\angle CFD = \angle AGB$.
So $\triangle FCD$ is equiangular with $\triangle GAB$.
So from Equiangular Triangles are Similar, $\triangle FCD$ is similar to $\triangle GAB$.
So $FD : GB = FC : GA = CD : AB$.
Similarly from Sum of Angles of Triangle Equals Two Right Angles $\angle GHB = \angle FED$.
So $\triangle FDE$ is equiangular with $\triangle GBH$.
So from Equiangular Triangles are Similar, $\triangle FDE$ is similar to $\triangle GBH$.
So $FD : GB = FE : GH = ED : HB$.
Thus we have that:
:$FC : AG = CD : AB = FE : GH = ED : HB$
Since $\angle CFD = \angle AGB$ and $\angle DFE = \angle BGH$, we have that
:$\angle CFE = \angle CFD + \angle DFE = \angle AGB + \angle BGH = \angle AGH$
For the same reason:
:$\angle CDE = \angle CDF + \angle FDE = \angle ABG + \angle GBH = \angle ABH$
So $CDEF$ is equiangular with $ABHG$.
As has been shown, the sides of these polygons are proportional about their equal angles.
So from {{EuclidDefLink|VI|1|Similar Rectilineal Figures}}, $CDEF$ is similar $ABHG$.
{{qed}}
{{Euclid Note|18|VI|In {{EuclidPropLink|book=VI|prop=20|title=Similar Polygons are composed of Similar Triangles}}, it is shown by dividing any polygon into triangles, any two similar polygons are composed of similar triangles.<br/>Thus the construction as given here can be seen directly to extend to polygons with any number of sides.}}
Theorem: Triangles with Two Equal Angles are Similar

Two triangles which have two corresponding angles which are equal are similar.

Proof:
Let $\triangle ABC$ and $\triangle DEF$ be triangles such that $\angle ABC = \angle DEF$ and $\angle BAC = \angle EDF$.
Then from Sum of Angles of Triangle Equals Two Right Angles $\angle ACB$ is equal to two right angles minus $\angle ABC + \angle BAC$.
Also from Sum of Angles of Triangle Equals Two Right Angles $\angle DFE$ is equal to two right angles minus $\angle DEF + \angle EDF$.
That is, $\angle DFE$ is equal to two right angles minus $\angle ABC + \angle BAC$.
So $\angle DFE = \angle ACB$ and so all three corresponding angles of $\triangle ABC$ and $\triangle DEF$ are equal.
The result follows from Equiangular Triangles are Similar.
{{qed}}
Theorem: Ratio of Areas of Similar Triangles

{{:Euclid:Proposition/VI/19}}
That is, the ratio of the areas of the similar triangles is the square of the ratio of the corresponding sides.

Proof:
Let $\triangle ABC$ and $\triangle DEF$ be similar, such that $\angle ABC = \angle DEF$ and $AB : BC = DE : EF$ such that $BC$ corresponds to $EF$.
:400px
Let $BG$ be constructed such that $EF : BG = BC : EF$, and join $AG$.
From Proportional Magnitudes are Proportional Alternately $AB : DE = BC : EF$.
So from Equality of Ratios is Transitive $AB : DE = EF : BG$.
So in $\triangle ABC$ and $\triangle DEF$ the sides about the equal angles are reciprocally proportional.
From Sides of Equiangular Triangles are Reciprocally Proportional, the area of $\triangle ABG$ equals the area of $\triangle DEF$.
Now we have that $BC : EF = EF : BG$.
So from {{EuclidDefLink|V|9|Duplicate Ratio}} $BC$ has to $BG$ a ratio duplicate to that which $CB$ has to $EF$.
But from Areas of Triangles and Parallelograms Proportional to Base, $CB : BG = \triangle ABC : \triangle ABG$.
So $\triangle ABC$ has to $\triangle ABG$ a ratio duplicate to that which $BC$ has to $EF$.
But $\triangle ABC = \triangle DEF$.
So $\triangle ABC$ has to $\triangle DEF$ a ratio duplicate to that which $BC$ has to $EF$.
{{qed}}
Theorem: Similarity of Polygons is Equivalence Relation

Let $A, B, C$ be polygons.
If $A$ and $B$ are both similar to $C$, then $A$ is similar to $B$.
{{:Euclid:Proposition/VI/21}}
It is also worth noting that:
:$A$ is similar to $A$, and so similarity between polygons is reflexive.
:If $A$ is similar to $B$, then $B$ is similar to $A$, and so similarity between polygons is symmetric.
Hence the relation of similarity between polygons is an equivalence relation.

Proof:
:500px
We have that $A$ is similar to $C$.
From {{EuclidDefLink|VI|1|Similar Rectilineal Figures}}, it is equiangular with it and the sides about the equal angles are proportional.
We also have that $B$ is similar to $C$.
Again, from {{EuclidDefLink|VI|1|Similar Rectilineal Figures}}, it is equiangular with it and the sides about the equal angles are proportional.
So by definition $A$ is similar to $B$.
The statements of reflexivity and symmetry are shown similarly.
It follows that if $A$ is similar to $B$, and $B$ is similar to $C$, then $A$ is similar to $C$.
Thus similarity between polygons is transitive.
Hence the result, by definition of equivalence relation.
{{qed}}
{{Euclid Note|21|VI|{{AuthorRef|Euclid}} himself did not have the concept of an equivalence relation.<br/>However, the extra statements leading to the main result are sufficiently straightforward to justify adding the full proof here.}}
Theorem: Integral Multiple of an Algebraic Number

Let $K$ be a number field and $\alpha \in K$.
Then there exists a positive $n \in \Z$ such that $n \alpha \in \OO_K$.
In this context, $ \OO_K$ denotes the algebraic integers in $K$.

Proof:
If $\alpha = 0$ then any integer works and the proof is finished.
Let $\alpha \ne 0$.
Let $\map f x = x^d + a_{d - 1} x^{d - 1} + \dotsb + a_0$ be the minimal polynomial of $\alpha$ over $\Q$. 
Suppose that $a_i = \dfrac {b_i} {c_i}$ is a reduced fraction for each $i$ such that $a_i \ne 0$.
Let $n$ be the least common multiple of the $c_i$, of which there must be at least one by our assumptions. 
Consider the polynomial:
{{begin-eqn}}
{{eqn | l = \map g x
      | r = n^d \map f {\frac x n}
      | c = 
}}
{{eqn | r = n^d \paren {\frac {x^d} {n^d} + a_{d - 1} \frac {x^{d - 1} } {n^{d - 1} } + \dotsb + a_0}
      | c = 
}}
{{eqn | r = x^d + n a_{d - 1} x^{d - 1} + \dotsb + n^d a_0
      | c = 
}}
{{end-eqn}}
Note that $g$ is a monic polynomial with coefficients in $\Z$ by our choice of $n$.
Furthermore, by construction, we see that $n \alpha$ is a root of $g$ and is therefore an algebraic integer.
{{qed}}