nyuuzyou/wikis huggingface dataset:
Construction of Equilateral Triangle

== Theorem ==
On a given straight line segment, it is possible to construct an equilateral triangle.
== Construction ==
:
Let $AB$ be the given straight line segment.
We construct a circle $BCD$ with center $A$ and radius $AB$.
We construct a circle $ACE$ with center $B$ and radius $AB$.
From $C$, where the circles intersect, we draw a line segment to $A$ and to $B$ to form the straight line segments $AC$ and $BC$.
Then $\triangle ABC$ is the equilateral triangle required.
== Proof ==
As $A$ is the center of circle $BCD$, it follows from that $AC = AB$.
As $B$ is the center of circle $ACE$, it follows from that $BC = AB$.
So, as $AC = AB$ and $BC = AB$, it follows from that $AC = BC$.
Therefore $AB = AC = BC$.
Therefore $\triangle ABC$ is equilateral.
== Sources ==
* : Book $\text{I}$. Propositions
Category:Equilateral Triangles
Construction of Excircle to Triangle

== Theorem ==
It is possible to construct an excircle to a triangle tangent to any of its three sides.
== Construction ==
:
Let $ABC$ be the given triangle.
Let us be required to construct the excircle tangent to $BC$.
Let $AB$ and $AC$ be produced beyond $B$ and $C$ to $X$ and $Y$ respectively.
Let $\angle CBX$ and $\angle BCY$ be bisected by $BE$ and $CE$ and let these lines join at $E$.
From $E$ construct the perpendiculars $EP, EQ, ER$ to $AY, AX, BC$ respectively.
Construct the circle with radius $ER$ and center $E$.
This is the required excircle.
== Proof ==
We have that $\angle XBE = \angle CBE$ and $\angle BQE = \angle BRE$, a right angle, and $BE$ is common.
So from Triangle Angle-Side-Angle Congruence we have that $\triangle EQB = \triangle ERB$.
So $EQ = ER$.
For the same reason $ER = EP$.
So $EQ = ER = EP$.
So the circle drawn with radius $ER$ will pass through $P, Q$ and $R$.
From Line at Right Angles to Diameter of Circle it follows that $AX, AY, BC$ are tangent to the circle $PQR$.
Hence the result.
Category:Excircles of Triangles
Construction of Fifth Apotome

== Theorem ==
== Proof ==
:
Let $A$ be a rational straight line.
Let $CG$ be commensurable in length with $A$.
Therefore, by definition, $CG$ is also rational.
Let $DF$ and $FE$ be square numbers such that $DE$ does not have to either $DF$ or $FE$ the ratio that a square number has to another square number.
Using , let it be contrived that:
:$FE : ED = CG^2 : GB^2$
Therefore by :
:$GB^2$ is commensurable with $GC^2$.
Therefore $GB^2$ is rational.
Therefore $BG$ is rational.
We have that $DE : EF$ is not the ratio that a square number has to another square number.
Therefore $BG^2 : GC^2$ is not the ratio that a square number has to another square number.
Therefore from :
:$BG$ is incommensurable in length with $GC$.
Both $BG$ and $GC$ are rational.
Therefore $BG$ and $GC$ rational straight lines which are commensurable in square only.
Therefore $BC$ is an apotome.
It remains to be shown that $BC$ is a fifth apotome.
Let $H$ be a straight line such that $H^2 = BG^2 - GC^2$.
We have that:
:$DE : EF = BG^2 : GC^2$ 
So by :
:$ED : EF = BG^2 : H^2$
So $ED$ does not have to $EF$ the ratio that a square number has to another square number.
Therefore neither does $GB^2$ have to $H^2$ the ratio that a square number has to another square number.
Therefore from :
:$BG$ is incommensurable in length with $H$.
We have that:
:$BG^2 = GC^2 + H^2$
Therefore $GB^2$ is greater than $GC^2$ by the square on a straight line which is incommensurable in length with $GB$.
Also, the annex $CG$ is commensurable in length with the rational straight line $A$.
Therefore, by definition, $BC$ is a fifth apotome.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Construction of Fifth 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 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 :
:$EF$ 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.
Since:
:$CA : AB = EF^2 : FG^2$
while:
:$AB > CA$
Therefore:
:$GF^2 > FE^2$
Let:
:$EF^2 + H^2 = GF^2$
for some $H$.
From :
:$AB : BC = GF^2 : H^2$
But $AB : BC$ is not the ratio that a square number has to a square number.
Therefore $FG^2 : H^2$ is not the ratio that a square number has to a square number.
Therefore by :
:$FG$ is incommensurable in length with $H$.
Therefore $FG^2 > EF^2$ by the square on a straight line which is incommensurable in length with $FG$.
We have that:
:$EF$ and $FG$ are rational straight lines which are commensurable in square only
and:
:the lesser term $EF$ is commensurable in length with $D$.
Therefore $EG$ is a fifth binomial straight line.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Construction of Figure Similar to One and Equal to Another

== Theorem ==
== Construction ==
Let $ABC$ be the given rectilineal figure to which the figure to be constructed is to be similar, and let $D$ be the rectilineal figure to which it must be equal (in area).
:
Using Construction of Parallelogram on Given Line equal to Triangle in Given Angle, we apply the parallelogram $BE$ to the straight line $BC$ equal in area to $\triangle ABC$.
Using Construction of Parallelogram in Given Angle equal to Given Polygon we apply the parallelogram $CM$ to the straight line $CE$ equal in area to $D$, where $\angle FCE = \angle CBL$.
Now using Construction of Mean Proportional‎ we construct the straight line $GH$ such that:
: $BC : GH = GH : CF$
Using Construction of Similar Polygon we construct $\triangle KGH$ to be similar to $\triangle ABC$ and similarly situated.
Then $\triangle KGH$ is the required rectilineal figure, similar to $\triangle ABC$ and equal in size to $D$.
== Proof ==
We have that $BC$ is in a straight line with $CF$ and that $LE$ is in a straight line with $EM$.
From the porism to Ratio of Areas of Similar Triangles:
: $BC : CF = \triangle ABC : \triangle KGH$
But from Areas of Triangles and Parallelograms Proportional to Base:
: $BC : CF = \Box BE : \Box EF$
So:
: $\triangle ABC : \triangle KGH = \Box BE : \Box EF$
So from Proportional Magnitudes are Proportional Alternately:
: $\triangle ABC : \Box BE = \triangle KGH : \Box EF$
But:
: $\triangle ABC = \Box BE$
and so:
: $\triangle KGH = \Box EF$
But $\Box EF = D$, and so $\triangle KGH = D$
Also we have that $\triangle KGH$ is similar to $\triangle ABC$.
Hence the result.
== Sources ==
* : Book $\text{VI}$. Propositions
Category:Polygons
Construction of First Apotome

== Theorem ==
== Proof ==
:
Let $A$ be a rational straight line.
Let $BG$ be commensurable in length with $A$.
Therefore, by definition, $BG$ is also rational.
Let $DE$ and $EF$ be square numbers such that $FD = DE - EF$ 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:
:$ED : DF = BG^2 : CG^2$
Therefore by :
:$BG^2$ is commensurable with $CG^2$.
But $BG^2$ is rational.
Therefore $GC^2$ is rational.
Therefore $GC$ is rational.
We have that $ED : FD$ is not the ratio that a square number has to another square number.
Therefore $BG^2 : GC^2$ is not the ratio that a square number has to another square number.
Therefore from :
:$BG$ is incommensurable in length with $GC$.
Both $BG$ and $GC$ are rational.
Therefore $BG$ and $GC$ rational straight lines which are commensurable in square only.
Therefore $BC$ is an apotome.
It remains to be shown that $BC$ is a first apotome.
Let $H$ be a straight line such that $H^2 = BG^2 - GC^2$.
We have that:
:$ED : FD = BG^2 : GC^2$ 
So by :
:$DE : EF = GB^2 : H^2$
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 whole $BG$ is commensurable in length with the rational straight line $A$.
Therefore, by definition, $BC$ is a first apotome.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Construction of First 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 rational 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 rational 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 rational rectangle.
From :
:$AD^2 + DB^2 - 2 \cdot AD \cdot DB = AC^2 + CB^2 - 2 \cdot AC \cdot CB = AB^2$
Therefore:
:$AD^2 + DB^2 - AC^2 + CB^2 = 2 \cdot AD \cdot DB - 2 \cdot AC \cdot CB$
But $AD \cdot DB$ and $AC \cdot CB$ are both rational.
Therefore $2 \cdot AD \cdot DB - 2 \cdot AC \cdot CB$ is rational.
Therefore $AD^2 + DB^2 - AC^2 + CB^2$ is rational.
But from:
:
and:
::
both $AD^2 + DB^2$ and $AC^2 + CB^2$ is medial.
By this cannot happen.
Therefore only one medial straight line can be added to $AB$ which is commensurable in square only with the whole and which contains with the whole a rational rectangle.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Construction of First 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:
:$BA : AC = EF^2 : FG^2$
where $FG$ is a straight line constructed such that $EG = EF + FG$ is itself a straight line.
But:
:$AB : AC = m^2 : \left({m^2 - n^2}\right)$
where $\left({m^2 - n^2}\right)$ is not square.
So 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:
:$BA : AC = EF^2 : FG^2$
while:
:$BA > AC$
Therefore:
:$EF^2 > FG^2$
Let:
:$FG^2 + H^2 = EF^2$
for some $H$.
From :
:$AB : BC = EF^2 : H$
But $AB : BC$ has the ratio that a square number has to a square number.
Therefore by :
: $EF$ is commensurable in length with $H$.
Therefore $EF^2$ is greater than $FG^2$ by the square on a straight line commensurable with $EF$.
We have that:
:$EF$ and $FG$ are rational straight lines
and:
:$EF$ is commensurable in length with $D$.
Therefore $EF$ is a first binomial straight line.
== Sources ==
* : Book $\text{X}$. Propositions
::in which a mistake appears.
Category:Euclidean Number Theory
Construction of First Binomial Straight Line/Mistake

== Source Work ==
The mistake can be seen in the edition as published by Dover: ISBN 0-486-60089-0.
== Mistake ==
:Therefore $EF$ is a first binomial straight line.
See the construction:
:
$EG$ is the binomial which the rest of the proof is to demonstrate is a first binomial.
This line therefore should read:
:Therefore $EG$ is a first binomial straight line.
Category:Mistakes/Euclid: The Thirteen Books of The Elements: Volume 3
Construction of Fourth Apotome

== Theorem ==
== Proof ==
:
Let $A$ be a rational straight line.
Let $BG$ be commensurable in length with $A$.
Therefore, by definition, $BG$ is also rational.
Let $DF$ and $FE$ be square numbers such that $DE$ does not have to either $DF$ or $FE$ the ratio that a square number has to another square number.
Using , let it be contrived that:
:$DE : EF = BG^2 : GC^2$
Therefore by :
:$BG^2$ is commensurable with $GC^2$.
But $BG^2$ is rational.
Therefore $GC^2$ is rational.
Therefore $GC$ is rational.
We have that $DE : EF$ is not the ratio that a square number has to another square number.
Therefore $BG^2 : GC^2$ is not the ratio that a square number has to another square number.
Therefore from :
:$BG$ is incommensurable in length with $GC$.
Both $BG$ and $GC$ are rational.
Therefore $BG$ and $GC$ rational straight lines which are commensurable in square only.
Therefore $BC$ is an apotome.
It remains to be shown that $BC$ is a fourth apotome.
Let $H$ be a straight line such that $H^2 = BG^2 - GC^2$.
We have that:
:$DE : EF = BG^2 : GC^2$ 
So by :
:$ED : EF = GB^2 : H^2$
So $ED$ does not have to $EF$ the ratio that a square number has to another square number.
Therefore neither does $GB^2$ have to $H^2$ the ratio that a square number has to another square number.
Therefore from :
:$BG$ is incommensurable 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 incommensurable in length with $BG$.
Also, the whole $BG$ is commensurable in length with the rational straight line $A$.
Therefore, by definition, $BC$ is a fourth apotome.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Construction of Fourth 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 rational straight line.
Let $EF$ be constructed commensurable in length with $D$.
Then $EF$ is also a rational straight line.
Using , let:
:$AB : AC = EF^2 : FG^2$
where $FG$ is a straight line constructed such that $EG = EF + FG$ is itself a straight line.
From :
:$EF$ and $FG$ are commensurable in square.
Therefore $FG$ is also a rational straight line.
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.
Since:
:$AB : AC = EF^2 : FG^2$
while:
:$BA > AC$
Therefore:
:$EF^2 > FG^2$
Let:
:$FG^2 + H^2 = EF^2$
for some $H$.
From :
:$AB : BC = EF^2 : H^2$
But $AB : BC$ is not the ratio that a square number has to a square number.
Therefore $EF^2 : H^2$ is not the ratio that a square number has to a square number.
Therefore by :
:$EF$ is incommensurable in length with $H$.
Therefore $EF^2 > GF^2$ by the square on a straight line which is incommensurable in length 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 fourth binomial straight line.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Construction of Fourth Proportional Straight Line

== Theorem ==
Given three straight lines of lengths $a, b, c$, it is possible to construct a fourth straight line of length $d$ such that $a : b = c : d$.
== Construction ==
Let $A, B, C$ be the three given straight lines.
Let $DE, DF$ be set out containing any angle $EDF$.
Let $DG = A, GE = B, DH = C$.
Join $GH$ and construct $EF$ parallel to $GH$.
Then $HF$ is the required straight line such that $A : B = C : HF$.
== Proof ==
:
From Parallel Transversal Theorem:
:$DG : GE = DH : HF$
But:
:$DG = A, GE = B, DH = C$
Hence the result.
== Sources ==
* : Book $\text{VI}$. Propositions
Category:Lines
Construction of Geometric Sequence in Lowest Terms

== Theorem ==
It is possible to find a geometric sequence of integers $G_n$ of length $n + 1$ with a given common ratio such that $G_n$ is in its lowest terms.
=== Porism ===
== Proof 1 ==
== Proof 2 ==
Category:Geometric Sequences of Integers
Category:Construction of Geometric Sequence in Lowest Terms
Construction of Geometric Sequence in Lowest Terms/Porism

== Porism to Construction of Geometric Sequence in Lowest Terms ==
== Proof ==
Apparent from the construction.
From Construction of Geometric Sequence in Lowest Terms, such a geometric sequence is of the form:
:$P = \tuple {q^n, p q^{n - 1}, p^2 q^{n - 2}, \ldots, p^{n - 1} q, p^n}$
Thus when $n = 2$:
:$P = \tuple {q^2, p q, p^2}$
and when $n = 3$:
:$P = \tuple {q^3, p q^2, p^2 q, p^3}$
== Sources ==
* : Book $\text{VIII}$. Propositions
Category:Construction of Geometric Sequence in Lowest Terms
Construction of Geometric Sequence in Lowest Terms/Proof 1

== Theorem ==
== Proof ==
Let $r = \dfrac a b$ be the given common ratio.
Let the required geometric sequence have a length of $4$.
Let $a^2 = c$.
Let $a b = d$.
Let $b^2 = e$.
Let:
:$a c = f$
:$a d = g$
:$a e = h$
and let:
:$b e = k$
As:
:$a^2 = c$
:$a b = d$
it follows from that:
:$\dfrac a b = \dfrac c d$
As:
:$a b = d$
:$b^2 = e$
it follows from that:
:$\dfrac a b = \dfrac d e$
As:
:$a c = f$
:$a d = g$
it follows from that:
:$\dfrac c d = \dfrac f g$
As:
:$a d = g$
:$a e = h$
it follows from that:
:$\dfrac d e = \dfrac g h$
As:
:$a e = h$
:$b e = k$
it follows from that:
:$\dfrac a b = \dfrac h k$
Putting the above together:
:$c, d, e$ are in geometric sequence with common ratio $\dfrac a b$
:$f, g, h, k$ are in geometric sequence with common ratio $\dfrac a b$
We have that $a$ and $b$ are the smallest numbers with the same ratio.
So by :
:$a \perp b$
where $\perp$ denotes coprimality.
We also have that:
:$a^2 = c, b^2 = e$
and:
:$a c = e, b e = k$
so by :
:$c \perp e$
:$f \perp k$
But from , these are the least of those with the same common ratio.
Hence the result.
== Sources ==
* : Book $\text{VIII}$. Propositions
Category:Construction of Geometric Sequence in Lowest Terms
Construction of Geometric Sequence in Lowest Terms/Proof 2

== Theorem ==
== Proof ==
Let the required length of the geometric sequence $P$ be $n$.
Let $r$ be the given common ratio.
From Common Ratio in Integer Geometric Sequence is Rational, $r$ is a rational number.
Let $r = \dfrac p q$ be in canonical form.
Thus, by definition:
:$p \perp q$
Let $a$ be the first term of $P$.
Then the sequence $P$ is:
:$P = \tuple {a, a \dfrac p q, a \dfrac {p^2} {q^2}, \ldots, a \dfrac {p^n} {q^n} }$
All the elements of $P$ are natural numbers, so, in particular:
:$a \dfrac {p^n} {q^n} \in \N$
From Powers of Coprime Numbers are Coprime:
:$p^n \perp q^n$
and so from Euclid's Lemma:
:$q^n \divides a$
Thus:
:$a = k q^n$
for some $k \in \N$, and so:
:$P = \tuple {k q^n, k p q^{n - 1}, k p^2 q^{n - 2}, \ldots, k p^{n - 1} q, k p^n}$
From Geometric Sequence with Coprime Extremes is in Lowest Terms:
:$k q^n \perp k p^n$
from which it follows that $k = 1$.
It follows that the required geometric sequence is:
:$P = \tuple {q^n, p q^{n - 1}, p^2 q^{n - 2}, \ldots, p^{n - 1} q, p^n}$
Category:Construction of Geometric Sequence in Lowest Terms
Construction of Golden Section

== Theorem ==
== Construction ==
:
Let $AB$ be the given finite straight line.
Construct the square $\Box BC$ on $AB$.
Using Construction of Parallelogram Equal to Given Figure Exceeding a Parallelogram, construct $\Box CD$ equal to $\Box BC$ and exceeding by the figure $\Box AD$ similar to $\Box BC$.
Then $E$ is the point at which $AB$ is cut so that $AB : AE = AE : EB$.
== Proof ==
As $\Box BC$ is a square, then $\Box AD$ is also a square.
Since $\Box CD = \Box BC$, we subtract $\Box CE$ from each.
Therefore $\Box BF = \Box AD$.
From Sides of Equal and Equiangular Parallelograms are Reciprocally Proportional‎, in $\Box BF$ and $\Box AD$ the sides about the equal angles are reciprocally proportional.
Therefore $FE : ED = AE : EB$.
But $FE = AB$ and $ED = AE$.
So $BA : AE = AE : EB$.
Also $AB > AE$ and so $AE > EB$.
Hence the result.
== Sources ==
* : Book $\text{VI}$. Propositions
* 
* 
Category:Golden Mean
Construction of Incommensurable Lines

== Theorem ==
=== Lemma ===
== Proof ==
Let $A$ be the assigned straight line.
Let $B$ and $C$ be two numbers which do not have to one another the ratio which a square number has to a square number.
That is, from the lemma, that $B$ and $C$ are not similar plane numbers.
From Magnitudes with Rational Ratio are Commensurable: Porism, let $D$ be constructed such that:
:$\dfrac {A^2} {D^2} = \dfrac B C$
Therefore from Magnitudes with Rational Ratio are Commensurable, the square on $A$ is commensurable with the square on $D$.
Since:
: $B$ does not have to one $C$ the ratio which a square number has to a square number
then:
: $A^2$ does not have to one $D^2$ the ratio which a square number has to a square number.
Therefore from Commensurability of Squares $A$ is incommensurable in length with $D$.
Let a mean proportional $E$ be taken between $A$ and $D$.
From :
:$\dfrac A D = \dfrac {A^2} {E^2}$
But $A$ is incommensurable in length with $D$.
Therefore $A^2$ is incommensurable with $E^2$.
Therefore $A$ is incommensurable in square with $E$.
Therefore two straight lines $D$ and $E$ have been found, such that:
:$D$ is incommensurable in length with $A$
:$E$ is incommensurable in square with $A$.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Construction of Incommensurable Lines/Lemma

== Lemma to Construction of Incommensurable Lines ==
== Proof ==
The propositions referred to are:
:
and its converse, which does not explicitly prove.
:For, if they have, they will be similar plane numbers: which is contrary to the hypothesis.
:Therefore numbers which are not similar plane numbers have not to one another the ratio which a square number has to a square number.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Construction of Integer Multiple of Line Segment

== Theorem ==
Let $AB$ be a line segment in the plane.
Let $AC$ be a line segment in the plane through a point $C$
Let $D$ be a point on $AC$ such that $AD = n AB$ for some $n \in \Z$.
Then $AD$ is constructible using a compass and straightedge construction.
== Proof ==
Let $AB$ be given.
Let the straight line through $AC$ be constructed.
Let $D_1$ be constructed on $AC$ such that $A D_1 = AB$ by constructing a circle whose center is at $A$ and whose radius is $B$.
Let $D_0$ be identified as the point $A$.
For each $k \in \set {1, 2, \ldots, n - 1}$, construct a circle whose center is at $D_k$ and whose radius is $D_{k - 1} D_k$.
We have that $D_k D_{k + 1} = D_{k - 1} D_k = AB$.
Thus $D_0 D_{k + 1}$ is ${k + 1}$ times $D_0 D_1 = AB$.
When $k = n - 1$ we have that $D_{k + 1} = D_n = D$ is $n$ times $AB$.
The diagram below illustrates the construction for $n = 4$.
:
Category:Euclidean Geometry
Category:Compass and Straightedge Constructions