nyuuzyou/wikis huggingface dataset:
Construction of Second Apotome

== Theorem ==
== Proof ==
:
Let $A$ be a rational straight line.
Let $GC$ be commensurable in length with $A$.
Therefore, by definition, $GC$ is also rational.
Let $DE$ and $EF$ be square numbers such that $DF = DE - EF$ is not square.
Then $ED : FD$ is not the ratio that a square number has to another square number.
Using , let it be contrived that:
:$FD : DE = CG^2 : GB^2$
Therefore by :
:$CG^2$ is commensurable with $GB^2$.
But $CG^2$ is rational.
Therefore $GB^2$ is rational.
Therefore $GB$ is rational.
We have that $GC^2 : GB^2$ is not the ratio that a square number has to another square number.
Therefore from :
:$GC$ is incommensurable in length with $GB$.
Both $GC$ and $GB$ are rational.
Therefore $GC$ and $GB$ rational straight lines which are commensurable in square only.
Therefore $BC$ is an apotome.
It remains to be shown that $BC$ is a second apotome.
Let $H$ be a straight line such that $H^2 = BG^2 - GC^2$.
We have that:
:$ED : DF = BG^2 : GC^2$ 
So by :
:$DE : EF = BG^2 : H^2$
But $DE$ and $EF$ are both square numbers.
So $DE$ has to $EF$ the ratio that a square number has to another square number.
Therefore $GB^2$ has to $H^2$ the ratio that a square number has to another square number.
Therefore from :
:$BG$ is commensurable in length with $H$.
We have that:
:$BG^2 = GC^2 + H^2$
Therefore $BG^2$ is greater than $GC^2$ by the square on a straight line which is commensurable in length with $BG$.
Also, the annex $CG$ is commensurable in length with the rational straight line $A$.
Therefore, by definition, $BC$ is a second apotome.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Construction of Second Apotome of Medial is Unique

== Theorem ==
== Proof ==
:
Let $AB$ be the first apotome of a medial straight line.
Let $BC$ be added to $AB$ such that:
: $AC$ and $CB$ are medial straight lines
: $AC$ and $CB$ are commensurable in square only
: $AC \cdot CB$ is a medial rectangle.
It is to be proved that no other medial straight line can be added to $AB$ which is commensurable in square only with the whole and which contains with the whole a medial rectangle.
Suppose $BD$ can be added to $AB$ so as to fulfil the conditions stated.
Then by definition of the first apotome of a medial straight line, $AD$ and $DB$ are such that:
: $AD$ and $DB$ are medial straight lines
: $AD$ and $DB$ are commensurable in square only
: $AD \cdot DB$ is a medial rectangle.
Let $EF$ be a rational straight line.
Let $EG = AC^2 + CB^2$ be applied to $EF$, producing $EM$ as breadth.
Let $HG = 2 \cdot AC \cdot CB$ be subtracted from $EG$ producing $HM$ as breadth.
Therefore from :
:the remainder $EL$ equals $AB^2$.
Thus $AB$ equals the "side" of $EL$.
Let $EI = AD^2 + DB^2$ be applied to $EF$, producing $EN$ as breadth.
But $EL = AB^2$.
Therefore the remainder $HI$ equals $2 \cdot AD \cdot DB$.
We have that $AC$ and $CB$ are medial straight lines.
Therefore $AC^2$ and $CB^2$ are also medial.
Also $AC^2 + CB^2 = EG$.
So by:
:
and:
::
it follows that:
:$EG$ is medial.
But $EG$ is applied to the rational straight line $EF$, producing $EM$ as breadth.
Therefore from :
:$EM$ is rational and incommensurable in length with $EF$.
We have that $AC \cdot CB$ is a medial rectangle.
So from :
:$2 \cdot AC \cdot CB$ is a medial rectangle.
But $2 \cdot AC \cdot CB = HG$.
Therefore $HG$ is medial.
Also, $HG$ has been applied to the rational straight line $EF$, producing $EM$ as breadth.
Therefore by :
:$HM$ is rational and incommensurable in length with $EF$.
Also, $AC$ and $CB$ are commensurable in square only.
Therefore $AC$ and $CB$ are incommensurable in length.
But:
:$AC : CB = AC^2 : AC \cdot CB$
Therefore from :
:$AC^2$ is incommensurable with $AC \cdot CB$.
But $AC^2 + CB^2$ is commensurable with $AC^2$.
Also from :
:$2 \cdot AC \cdot CB$ is commensurable with $AC \cdot CB$.
Therefore by :
:$AC^2 + CB^2$ incommensurable with $AC \cdot CB$.
We have that:
:$EG = AC^2 + CB^2$
and:
:$GH = 2 \cdot AC \cdot CB$
Therefore:
:$EG$ is incommensurable with $HG$.
But from :
:$EG : HG = EM : HM$
Therefore from :
:$EM$ is incommensurable in length with $MH$.
But both $EM$ and $MH$ are rational straight lines.
Therefore $EM$ and $MH$ are rational straight lines which are commensurable in square only.
Therefore $EH$ is an apotome, and $HM$ an annex to it.
Similarly it can be shown that $HN$ is also an annex to it.
Therefore we have different straight lines which are annexes to an apotome which are commensurable in square only to the whole.
From , this is impossible.
The result follows.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Construction of Second Binomial Straight Line

== Theorem ==
== Proof ==
:
Let $AC$ and $CB$ be straight lines constructed such that $AB = AC + CB$ is itself a straight line.
Using , let:
: $AB : BC = m^2 : n^2$
where $m$ and $n$ are numbers such that $m^2 - n^2$ is not square.
Let $D$ be a rational straight line.
Let $EF$ be constructed commensurable in length with $D$.
Then $EF$ is also a rational straight line.
Using , let:
:$CA : AB = EF^2 : FG^2$
where $FG$ is a straight line constructed such that $EG = EF + FG$ is itself a straight line.
From :
: $EF^2 : FG^2 = m^2 : \left({m^2 - n^2}\right)$
$EF$ and $FG$ are commensurable in square.
But from :
$EF$ and $FG$ are incommensurable in length.
Therefore $EF$ and $FG$ are rational straight lines which are commensurable in square only.
Therefore by definition $EG$ is a binomial.
We have that:
:$BA : AC = GF^2 : FE^2$
while:
:$BA > AC$
Therefore:
:$GF^2 > FE^2$
Let:
:$FE^2 + H^2 = GF^2$
for some $H$.
From :
:$AB : BC = FG^2 : H$
But $AB : BC$ has the ratio that a square number has to a square number.
Therefore by :
:$FG$ is commensurable in length with $H$.
Therefore $FG^2$ is greater than $FE^2$ by the square on a straight line commensurable with $EF$.
We have that:
:$EF$ and $FG$ are rational straight lines which are commensurable in square only
and:
:$EF$ is commensurable in length with $D$.
Therefore $EG$ is a second binomial straight line.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Construction of Segment on Given Circle Admitting Given Angle

== Theorem ==
== Construction ==
:
Let $ABC$ be the given circle, and let $D$ be the given rectilineal angle.
Let $EF$ be drawn tangent to $ABC$ at $B$.
Let $\angle FBC$ be constructed equal to $\angle D$.
Then $BC$ is a straight line which forms the base of a segment which admits an angle equal to $D$.
== Proof ==
Let $A$ be selected anywhere on the circle opposite the segment in question.
From the Tangent-Chord Theorem, $\angle BAC = \angle FBC$.
But $\angle FBC = \angle D$.
Hence the result.
== Sources ==
* : Book $\text{III}$. Propositions
Category:Circles
Construction of Sequence of Numbers with Given Ratios

== Theorem ==
== Proof ==
Let the given ratios in least numbers be:
:$a : b$, $c : d$, $e : f$
From , we can find:
:$g = \lcm set {b, c}$
Let:
:$g = r b$
:$g = s c$
Then let:
:$h = r a$
:$k = s d$
Now either $e \divides k$ or $e \nmid k$.
First, suppose $e \divides k$.
Let $k = t e$.
Then let $l = t f$.
We have:
:$h = r a$
:$g = r b$
thus from :
:$a : h = b : g$
So from :
:$a : b = h : g$
Similarly:
:$c : d = g : k$
and further:
:$e : f = k : l$
Thus $h, g, k, l$ are continuously proportional:
:in the ratio of $a : b$
:in the ratio of $c : d$
:in the ratio of $e : f$.
It is necessary to prove that these are the least natural numbers with this property.
Suppose $h, g, k, l$ are not the least natural numbers in the ratios of $a : b$, $c : d$, $e : f$.
Let these numbers be $n, o, m, p$.
Then:
:$a : b = n : o$
while $a$ and $b$ are least.
By :
:$b \divides o$
:$c \divides o$
and so $o$ is a common multiple of $b$ and $c$.
Therefore by :
:$\lcm \set {b, c} \divides o$
But $g = \lcm \set {b, c}$ from above.
Therefore $g \divides o$, the greater the less, which is impossible.
Therefore there are no numbers $n, o, m, p$ less than $h, g, k, l$ which have the properties in question.
Now suppose $e \nmid k$.
Let $m = \lcm \set {e, k}$.
Thus let:
:$m = u k$
:$n = u h$
:$o = u g$
and:
:$m = v e$
:$p = v f$
Thus from :
:$h : n = g : o$
From :
:$h : g = n : o$
Similarly, let:
But:
:$h : g = a : b$
and so:
:$a : b = n : o$
For the same reason:
:$c : d = o : m$
Since:
:$m = v e$
:$p = v f$
from :
:$e : m = f : p$
and so from :
:$e : f = m : p$
Therefore $n, o, m, p$ are continuously proportional:
:in the ratio of $a : b$
:in the ratio of $c : d$
:in the ratio of $e : f$.
It is necessary to prove that these are the least natural numbers with this property.
Suppose $n, o, m, p$ are not the least natural numbers in the ratios of $a : b$, $c : d$, $e : f$.
Let these numbers be $n', o', m', p'$.
Then:
:$a : b = n' : o'$
while $a$ and $b$ are least.
By :
:$b \divides o'$
:$c \divides o'$
and so $o'$ is a common multiple of $b$ and $c$.
Therefore by :
:$\lcm \set {b, c} \divides o'$
But $g = \lcm \set {b, c}$ from above.
Therefore $g \divides o'$, the greater the less, which is impossible.
Therefore there are no numbers $n', o', m', p'$ less than $n, o, m, p$ which have the properties in question.
== Sources ==
* : Book $\text{VIII}$. Propositions
Category:Euclidean Number Theory
Construction of Similar Polygon

== Theorem ==
On any given straight line it is possible to construct a polygon similar to any given polygon.
== Construction ==
Let $AB$ be the given straight line and let $CDEF$ be the given polygon.
Thus we need to construct a polygon similar to $CDEF$ on $AB$.
:
Join $DF$ and use Construction of Equal Angle to construct $\angle GAB = \angle FCD$ and $\angle ABG = \angle CDF$.
On the straight line $BG$ we similarly use Construction of Equal Angle to construct $\angle BGH = \angle DFE$ and $\angle GBH = \angle FDE$.
Then $ABHG$ is a polygon similar to $CDEF$.
== 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 , $CDEF$ is similar $ABHG$.
== Sources ==
* : Book $\text{VI}$. Propositions
Category:Polygons
Construction of Similarly Cut Straight Line

== Theorem ==
== Construction ==
Let $AB$ be the given uncut straight line.
Let $AC$ be the straight line cut at $D$ and $E$.
Let $AB$ and $AC$ be placed so as to contain any angle.
Join $CB$ and construct $DF$ and $EG$ parallel to $BC$.
Then construct $DHK$ parallel to $AB$.
Then $FG$ is the required cut of $AB$.
:
== Proof ==
We have that each of $FH$ and $HB$ is a parallelogram.
So from Opposite Sides and Angles of Parallelogram are Equal:
:$DH = FG$ and $HK = GB$
We have that $HE \parallel KC$, and $KC$ is one of the sides of $\triangle DKC$.
So from Parallel Transversal Theorem:
:$CE : ED = KH : HD$
But $KH = BG$ and $HD = GF$.
So:
:$CE : ED = BG : GF$
Again, we have $FD \parallel GE$, and $GE$ is one of the sides of $\triangle AGE$.
So from Parallel Transversal Theorem:
:$ED : DA = GF : FA$
So $AB$ has been cut at $F$ and $G$ similarly to $AC$ at $D$ and $E$.
== Sources ==
* : Book $\text{VI}$. Propositions
Category:Lines
Construction of Sixth Apotome

== Theorem ==
== Proof ==
:
Let $A$ be a rational straight line.
Let $E$, $BC$ and $CD$ be numbers set out which do not have pairwise between them the ratio that a square number has to another square number.
Let $CB : BD$ also not be the ratio that a square number has to a square number.
Using , let it be contrived that:
:$E : BC = A^2 : FG^2$
and:
:$BC : CD = FG^2 : GH^2$
We have that:
:$E : BC = A^2 : FG^2$.
From :
:$A^2$ is commensurable with $FG^2$.
But $A^2$ is rational.
Therefore $FG^2$ is rational.
Therefore $FG$ is rational.
We have that $E : BC$ is not the ratio that a square number has to another square number.
Therefore $A^2 : FG^2$ is not the ratio that a square number has to another square number.
Therefore from :
:$A$ is incommensurable in length with $FG$.
We have that:
:$BC : CD = FG^2 : GH^2$
From :
:$FG^2$ is commensurable with $GH^2$.
But $FG^2$ is rational.
Therefore $GH^2$ is rational.
Therefore $GH$ is rational.
We have that $BC : CD$ is not the ratio that a square number has to another square number.
Therefore $FG^2 : GH^2$ is not the ratio that a square number has to another square number.
Therefore from :
:$FG$ is incommensurable in length with $GH$.
Both $FG$ and $GH$ are rational.
Therefore $FG$ and $GH$ rational straight lines which are commensurable in square only.
Therefore $FH$ is an apotome.
It remains to be shown that $FH$ is a sixth apotome.
We have that:
:$E : BC = A^2 : FG^2$.
and:
:$BC : CD = FG^2 : GH^2$
Therefore from :
:$E : CD = A^2 : HG^2$
But $E : CD$ is not the ratio that a square number has to another square number.
Therefore neither is $A^2 : GH^2$ the ratio that a square number has to another square number.
Therefore neither $FG$ and $GH$ is incommensurable in length with the rational straight line $A$.
Now let $K^2 = FG^2 - GH^2$.
We have that:
:$BC : CD = FG^2 GH^2$
Therefore by :
:$BC : BD = FG^2 : K^2$
But $BC$ does not have to $BD$ the ratio that a square number has to another square number.
Therefore $FG^2$ does not have to $K^2$ the ratio that a square number has to another square number.
Therefore by :
:$FG$ is incommensurable in length with $K$.
So:
:$FG^2$ is greater than $GH^2$ by the square on $K$.
:$FG^2$ is greater than $GH^2$ by the square on a straight line which is incommensurable in length with $FG$.
So neither $FG$ nor $GH$ is commensurable in length with the rational straight line $A$.
Therefore, by definition, $FH$ is a sixth apotome.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Construction of Sixth Binomial Straight Line

== Theorem ==
== Proof ==
:
Let $AC$ and $CB$ be straight lines constructed such that $AB = AC + CB$ is itself a straight line.
Let neither $AB : AC$ nor $AB : BC$ be the ratio which a square number has to a square number.
Let $D$ be a number which is not square.
Let $D$ have to neither $AB$ nor $AC$ the ratio that a square number has to another square number.
Let $E$ be a rational straight line.
Using , let:
:$D : AB = E^2 : FG^2$
where $FG$ is a straight line.
From :
:$E^2$ is commensurable with $FG^2$.
We have that $E$ is rational.
Therefore $FG$ is also rational.
Since:
: $D$ does not have to $AB$ the ratio that a square number has to another square number
then:
: $E^2$ does not have to $FG^2$ the ratio that a square number has to another square number.
From :
:$E$ is incommensurable in length with $FG$.
Using , let:
:$BA : AC = FG^2 : GH^2$
where $GH$ is a straight line constructed such that $FH = FG + GH$ is itself a straight line.
From :
:$FG^2$ is commensurable with $GH^2$.
We have that $FG$ is rational.
Therefore $GH$ is also rational.
Since:
: $BA$ does not have to $AC$ the ratio that a square number has to another square number
then:
: $FG^2$ does not have to $GH^2$ the ratio that a square number has to another square number.
From :
:$FG$ is incommensurable in length with $GH$.
Therefore $FG$ and $GH$ are rational straight lines which are commensurable in square only.
Therefore by definition $FH$ is a binomial.
We have that:
:$D : AB = E^2 : FG^2$
and
:$BA : AC = FG^2 : GH^2$
From :
:$D : AC = E^2 : GH^2$
But $D$ does not have to $AC$ the ratio that a square number has to another square number.
Therefore, neither does $E$ have to $GH$ the ratio that a square number has to another square number.
Therefore from :
:$E$ is incommensurable in length with $GH$.
We have that:
:$BA : AC = FG^2 : GH^2$
Therefore:
: $FG^2 > GH^2$
Let:
: $FG^2 = GH^2 + K^2$
for some $K$.
From :
:$AB : BC = FG : K$
But $AB$ does not have to $AC$ the ratio that a square number has to another square number.
Therefore, neither does $FG^2$ have to $k^2$ the ratio that a square number has to another square number.
Therefore from :
:$FG$ is incommensurable in length with $K$.
Therefore $FG^2$ is greater than $GH^2$ by the square on a straight line which is incommensurable in length with $FG$.
But $FG$ and $GH$ are rational straight lines which are commensurable in square only.
Also, neither $FG$ nor $GH$ is commensurable in length with $E$.
Therefore $FH$ is a sixth binomial straight line.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Construction of Smith Number from Prime Repunit

== Theorem ==
Let $R_n$ be a repunit which is prime where $n \ge 3$.
Then $3304 \times R_n$ is a Smith number.
$3304$ is not the only number this works for, neither is it the smallest, but it serves as an example of the technique.
== Proof ==
Let $\map S n$ denote the sum of the digits of a positive integer $n$.
Let $\map {S_p} n$ denote the sum of the digits of the prime decomposition of $n$.
Then $\map S n = \map {S_p} n$ $n$ is a Smith number.
Let $n \ge 3$.
We have that:
:$3304 = 2 \times 2 \times 2 \times 7 \times 59$
and so for a prime repunit $R_n$:
:$3304 \times R_n = 2 \times 2 \times 2 \times 7 \times 59 \times \underbrace {111 \ldots 11}_{n \text { ones} }$
and so:
:$\map {S_p} {3304 \times R_n} = 2 + 2 + 2 + 7 + 5 + 9 + n \times 1 = n + 27$
Note that:
This gives:
Hence the result.
== Sources ==
* 
* : $4,937,775$
Category:Smith Numbers
Category:Repunits
Construction of Smith Number from Prime Repunit/Mistake

== Source Work ==
:
:The Dictionary
::$4,937,775$
== Mistake ==
:Denote the number whose digits are $n$ units by $R_n$.
:If $R_n$ is prime, then $3304 \times R_n$ is a Smith number.
:$3304$ is not the only effective multiplier in this construction, merely the smallest.
== Correction ==
Smaller such numbers have been discovered since, for example, $1540$.
Research is ongoing.
For the second edition of this book, removed this statement.
 
== Sources ==
* : $4,937,775$
Category:Mistakes/Curious and Interesting Numbers
Construction of Solid Angle equal to Given Solid Angle

== Theorem ==
== Proof ==
:
Let $AB$ be the given straight line.
Let $A$ be the given point on $AB$.
Let the angle at $D$, contained by the plane angles $\angle EDC, \angle EDF$ and $\angle FDC$, be the given solid angle.
Let an arbitrary point $F$ be taken on $DF$.
Using :
:Let $FG$ be drawn perpendicular to the plane through $ED$ and $DC$, meeting the plane at $G$.
Let $DG$ be joined.
Using :
:Let $\angle BAL$ be constructed on the straight line $AB$ and at the point $A$ equal to $\angle EDC$, and $\angle BAK$ equal to $\angle EDG$.
Let $AK = DG$.
Using :
:Let $KH$ be set up from the point $K$ perpendicular to the plane through $BA$ and $AL$.
Let $KH = GF$.
Let $HA$ be joined.
It is to be demonstrated that the solid angle at $A$ contained by the plane angles $\angle BAL \angle BAH$ and $\angle HAL$ equals the angle at $D$, contained by the plane angles $\angle EDC, \angle EDF$ and $\angle FDC$.
Let $AB$ and $DE$ be cut off equal to each other.
Let $HB, KB, FE, GE$ be joined.
We have that $FG$ is perpendicular to the plane of reference.
So by :
:$FG$ is perpendicular to all straight lines which meet it and are in the plane of reference.
Therefore $\angle FGD$ and $\angle FGE$ are right angles.
For the same reason, $\angle HKA$ and $\angle HKB$ are right angles.
We have that $KA$ and $AB$ are equal to $GD$ and $DE$ respectively.
From :
:$KB = GE$
But also:
:$KH = GF$
Therefore from :
:$HB = FE$
We have that $AK$ and $KH$ are equal to $DG$ and $GF$ respectively.
From :
:$AH = FD$
But also:
:$AB = DE$
Therefore $HA$ and $AB$ are equal to $DF$ and $DE$.
We also have that $HB = FE$.
Therefore from :
:$\angle BAH = \angle EDF$
For the same reason:
:$\angle HAL = \angle FDC$
and:
:$\angle BAL = \angle EDC$
Hence the result.
== Sources ==
* : Book $\text{XI}$. Propositions
Category:Solid Angles
Construction of Solid Angle from Three Plane Angles any Two of which are Greater than Other Angle/Lemma

== Theorem ==
== Proof ==
:
Let the straight lines $AB$ and $LO$ be set out.
Le $AB > LO$.
Let the semicircle $ACB$ be described on $AB$.
:
Using :
:Let $AC = LO$ be fitted into the semicircle $ACB$.
Let $BC$ be joined.
From :
:$\angle ACB$ is a right angle.
Therefore from :
:$AB^2 = AC^2 + CB^2$
Hence:
:$AB^2 - AC^2 = CB^2$
But $AC = LO$.
Therefore:
:$AB^2 - LO^2 = CB^2$
So if we then cut of $OR = BC$:
:$AB^2 - LO^2 = OR^2$
== Sources ==
* : Book $\text{XI}$. Propositions
Category:Solid Angles
Construction of Square equal to Given Polygon

== Theorem ==
A square can be constructed the same size as any given polygon.
== Proof ==
:
Let $A$ be the given polygon.
Construct the rectangle $BCDE$ equal to the given polygon.
If it so happens that $BE = ED$, then $BCDE$ is a square, and the construction is complete.
Suppose $BE \ne ED$. Then WLOG suppose $BE > ED$.
Produce $BE$ from $E$ and construct on it $EF = ED$.
Bisect $BF$ at $G$.
Construct the semicircle $BHF$ with center $G$ and radius $GF$ (see diagram).
Produce $DE$ from $D$ to $H$.
From Difference of Two Squares, the rectangle contained by $BE$ and $EF$ together with the square on $EG$ is equal to the square on $GF$.
But $GF = GH$.
So the rectangle contained by $BE$ and $EF$ together with the square on $EG$ is equal to the square on $GH$.
From Pythagoras's Theorem, the square on $GH$ equals the squares on $GE$ and $EH$.
Then the rectangle contained by $BE$ and $EF$ together with the square on $EG$ is equal to the squares on $GE$ and $EH$.
Subtract the square on $GE$ from each.
Then the rectangle contained by $BE$ and $EF$ is equal to the square on $EH$.
So the square on $EH$ is equal to the rectangle $BCDE$.
So the square on $EH$ is equal to the given polygon, as required.
== Sources ==
* : Book $\text{II}$. Propositions
Category:Squares
Category:Polygons
Construction of Square equal to Rectangle

== Theorem ==
== Proof ==
:
Let $AB$ be the given straight line segment.
Construct the square $ABDC$ on $AB$.
Bisect $CA$ at $E$ and join $BE$.
Produce $CA$ to $F$ and make $EF = BE$.
Construct the square $AFGH$ on $AF$.
Produce $GH$ to $K$.
Then $H$ is the point at which $AB$ has been cut so as to make the rectangle contained by $AB$ and $BH$ is the same size as the square on $AH$.
The proof is as follows.
From Square of Sum less Square, the rectangle contained by $CF$ and $FA$ together with the square on $AE$ equals the square on $EF$.
But $EF = EB$, so the rectangle contained by $CF$ and $FA$ together with the square on $AE$ equals the square on $BE$.
By Pythagoras's Theorem, the squares on $AB$ and $AE$ equal the square on $BE$, because $\angle EAB$ is a right angle.
So the rectangle contained by $CF$ and $FA$ together with the square on $AE$ equals the squares on $AB$ and $AE$.
Subtract the square $AE$ from each.
Then the rectangle contained by $CF$ and $FA$ equals the square on $AB$.
Now the rectangle contained by $CF$ and $FA$ is $\Box CFGK$ because $AF = FG$.
Also, the square on $AB$ is $\Box ABDC$.
So $\Box CFGK = \Box ABDC$.
Subtract $AHKC$ from each.
Then $\Box FGHA = \Box HBDK$.
Now $\Box HBDK$ is the rectangle contained by $AB$ and $BH$, because $AB = BD$.
Also, $\Box FGHA$ is the square on $AH$.
Hence the result.
== Sources ==
* : Book $\text{II}$. Propositions
Category:Squares
Category:Rectangles