nyuuzyou/wikis huggingface dataset:
Constant to Power of Number of Distinct Prime Divisors is Multiplicative Function

== Theorem ==
Let $c \in \R$ be a constant.
Let $f: \N \to \R$ denotes the mapping defined as:
:$\forall n \in \N: \map f n = c^k$
where $k$ is number of distinct primes that divide $n$.
Then $f$ is multiplicative.
== Proof ==
Let $r, s \in \Z$ such that $r \perp s$.
Let $r$ be composed of $p$ distinct primes:
:$r_1, r_2, \ldots r_p$
Thus:
:$\map f r = c^p$
Let $s$ be composed of $q$ distinct primes:
:$s_1, s_2, \ldots s_q$
Thus:
:$\map f s = c^q$
As $r \perp s$, all the $r_k$ and $s_k$ are distinct.
Thus $r s$ is composed of:
:the $p$ distinct primes $r_1, r_2, \ldots r_p$
and:
:the $q$ distinct primes $s_1, s_2, \ldots s_q$
which is a total of $p + q$ distinct primes.
Thus:
Hence the result.
== Also see ==
* Möbius Function is Multiplicative, a similar result.
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $29 \ \text{(c)}$
Category:Multiplicative Functions
Category:Number Theory
Constructible Length with Compass and Straightedge

== Theorem ==
Let $L$ be a line segment in a Eucldiean space.
Let the length of $L$ be $d$.
Let $L'$ be a line segment of length $d'$ constructed from $L$ using a compass and straightedge construction.
Then:
:$d' = q d$
where $q$ is an algebraic number whose degree is a power of $2$.
== Proof ==
== Sources ==
* : Chapter $\text {B}.18$: Algebraic and Transcendental Numbers. $e$ is Transcendental
Category:Compass and Straightedge Constructions
Category:Euclidean Geometry
Constructible numbers

== Definition ==
Given two distinct complex numbers, $z$, \(w\), plotted as points in the complex plane, we may construct a straight line through \(z\) and \(w\), or construct a circle whose center is \(z\) and whose circumference passes through \(w\). Denote such a line as \(L(z, w)\) and such a circle as \(C(z, w)\). The constructible numbers are defined recursively:
* \(0\) is a constructible number
* \(1\) is a constructible number
* If \(\alpha\), \(\beta\), \(\gamma\), and \(\delta\) are constructible numbers, then
** if there are finitely many intersection points of \(L(\alpha, \beta)\) with \(L(\gamma, \delta)\), then those intersection points are constructible numbers
** if there are finitely many intersections of \(C(\alpha, \beta)\) with \(L(\gamma, \delta)\), then those intersection points are constructible numbers
** if there are finitely many intersections of \(C(\alpha, \beta)\) with \(C(\gamma, \delta)\), then those intersection points are constructible numbers
Construction of Apotome is Unique

== Theorem ==
Let $D$ be the domain $\left\{{x \in \R_{>0} : x^2 \in \Q}\right\}$, the rationally expressible numbers.
Let $a, b \in D$ be two rationally expressible numbers such that $a - b$ is an apotome.
Then, there exists only one $x \in D$ such that $a - b + x$ and $a$ are commensurable in square only.
== Proof ==
:
Let $AB$ be an apotome.
Let $BC$ be added to $AB$ so that $AC$ and $CB$ are rational straight lines which are commensurable in square only.
It is to be proved that no other rational straight line can be added to $AB$ which is commensurable in square only with the whole.
Suppose $BD$ can be added to $AB$ so as to fulfil the conditions stated.
Then $AD$ and $DB$ are rational straight lines which are commensurable in square only.
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 $AD^2 + DB^2$ and $AC^2 + CB^2$ are both rational.
Therefore $AD^2 + DB^2 - AC^2 + CB^2$ is rational.
Therefore $2 \cdot AD \cdot DB - 2 \cdot AC \cdot CB$ is rational.
But from :
:$2 \cdot AD \cdot DB$ and $2 \cdot AC \cdot CB$ are both medial.
From :
:a medial area does not exceed a medial area by a rational area.
Therefore no other rational straight line can be added to $AB$ which is commensurable in square only with the whole.
Therefore only one rational straight line can be added to $AB$ which is commensurable in square only with the whole.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Construction of Circle from Segment/Proof 1

== Theorem ==
== Proof ==
Let $ABC$ be the given segment of a circle whose base is $AC$.
Bisect $AC$ at $D$, draw $DB$ perpendicular to $AC$, and join $AB$.
First suppose that $ABC$ is such that $\angle ABD > \angle BAD$:
:
On $BA$, construct $\angle BAE$ equal to $\angle ABD$.
Join $BD$ through to $E$ and join $EC$.
Then $E$ is the center of the required circle.
Second, suppose that $ABC$ is such that $\angle ABD = \angle BAD$:
:
Then $D$ is the center of the required circle.
Finally, suppose that $ABC$ is such that $\angle ABD < \angle BAD$:
:
On $BA$, construct $\angle BAE$ equal to $\angle ABD$.
The point $E$, which falls on $BD$, is the center of the required circle.
=== Proof of Construction ===
First suppose that $ABC$ is such that $\angle ABD > \angle BAD$:
:
Since $\angle ABE = \angle BAE$, from Triangle with Two Equal Angles is Isosceles we have that $EB = EA$.
Since $AD = DC$ and $DE$ is common, and $\angle ADE$ is a right angle, by Triangle Side-Angle-Side Congruence we have that $\triangle ADE = \triangle CDE$.
Hence $AE = CE$ both of which are equal to $BE$ from above.
So from Condition for Point to be Center of Circle $E$ is the center of the required circle.
Second, suppose that $ABC$ is such that $\angle ABD = \angle BAD$:
:
From Triangle with Two Equal Angles is Isosceles we have that $AD = DB$ and so also equal to $DC$.
So from Condition for Point to be Center of Circle $D$ is the center of the required circle.
Incidentally, note that in this case segment $ABC$ is actually a semicircle.
Finally, suppose that $ABC$ is such that $\angle ABD < \angle BAD$:
:
The same proof applies:
Since $\angle ABE = \angle BAE$, from Triangle with Two Equal Angles is Isosceles we have that $EB = EA$.
Since $AD = DC$ and $DE$ is common, and $\angle ADE$ is a right angle, by Triangle Side-Angle-Side Congruence we have that $\triangle ADE = \triangle CDE$.
Hence $AE = CE$ both of which are equal to $BE$ from above.
So from Condition for Point to be Center of Circle $E$ is the center of the required circle.
== Sources ==
* : Book $\text{III}$. Propositions
Category:Construction of Circle from Segment
Construction of Circle from Segment/Proof 2

== Theorem ==
== Proof ==
:
Choose any point $C$ on the circumference.
Bisect $AC$ at $D$ and $BC$ at $E$ and construct a perpendicular $DF$ and $EF$ from each through the point of bisection.
The point of intersection $F$ is the center of the required circle.
$AFC$ and $CFB$ are isosceles triangles and so $AF, CF$ and $BF$ are all equal.
The result follows from Condition for Point to be Center of Circle.
== Sources ==
* : Book $\text{III}$. Propositions
Category:Construction of Circle from Segment
Construction of Components of First Bimedial

== Theorem ==
== Proof ==
Let $\rho$ and $\rho \sqrt k$ be rational straight lines which are commensurable in square only.
Their mean proportional is $\rho \sqrt [4] k$ which is medial.
Let $x$ be such that:
:$\rho : \rho \sqrt k = \rho \sqrt [4] k : x$
which gives:
:$x = \rho k^{3/4}$
We have that:
:$\rho \frown \!\! - \rho \sqrt k$
where $\frown \!\! -$ denotes commensurability in square only.
Thus:
:$\rho \sqrt [4] k \frown \!\! - \rho k^{3/4}$
From Straight Line Commensurable with Medial Straight Line is Medial it follows that $\rho k^{3/4}$ is also medial.
== Also see ==
* Definition:First Bimedial
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Construction of Components of Major

== Theorem ==
=== Lemma ===
== Proof ==
:
From :
Let $AB$ and $BC$ be rational straight lines which are commensurable in square only such that:
: $AB^2 = BC^2 + \rho^2$
such that $\rho$ is incommensurable in length with $AB$.
Let $BC$ be bisected at $D$.
From Construction of Parallelogram Equal to Given Figure Exceeding a Parallelogram:
Let a parallelogram be applied to $AB$ equal to the square on either of $BD$ or $DC$, and deficient by a square.
Let this parallelogram be the rectangle contained by $AE$ and $EB$.
Let the semicircle $AFB$ be drawn with $AB$ as the diameter.
Let $EF$ be drawn perpendicular to $AB$.
Join $AF$ and $FB$.
We have that $AB > BC$ such that $AB^2 = BC^2 + \rho^2$ such that $\rho$ is incommensurable in length with $AB$.
We also have that the rectangle contained by $AE$ and $EB$ equals the parallelogram on $AB$ equal to $\dfrac {BC} 4$.
Thus from Condition for Incommensurability of Roots of Quadratic Equation:
:$AE$ is incommensurable in length with $EB$.
We have that:
:$AE : BE = AB \cdot AE : AB \cdot BE$
From :
:$AB \cdot AE = AF^2$
and:
:$AB \cdot BE = BF^2$
Therefore $AF^2$ and $BF^2$ are incommensurable.
By definition, $AF$ and $BF$ are therefore incommensurable in square.
As $AB$ is a rational straight line, it follows by definition that $AB^2$ is a rational area.
From Pythagoras's Theorem:
:$AB^2 = \left({AF + FB}\right)^2$
Thus $\left({AF + FB}\right)^2$ is also a rational area.
Therefore $AF + FB$ is rational.
From :
:$AE \cdot EB = EF^2$
and by hypothesis:
:$AE \cdot EB = BD^2$
then:
:$FE = BD$
Therefore:
: $BC = 2 FE$
Thus $AB \cdot BC$ is commensurable with $AB \cdot EF$.
But from Medial is Irrational:
:$AB \cdot BC$ is medial.
Therefore by :
:$AB \cdot EF$ is also medial.
But from :
:$AB \cdot EF = AF \cdot FB$
Therefore $AF \cdot FB$ is also medial.
But it has been proved that $AF + FB$ is rational.
Therefore we have found two rational straight lines which are incommensurable in square whose sum of squares is rational, but such that the rectangle contained by them is medial.
== Also see ==
* Definition:Major (Euclidean)
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Construction of Components of Major/Lemma

== Lemma to Construction of Components of Major ==
== Proof ==
:
Let $\triangle ABC$ be a right-angled triangle where $\angle A$ is right.
Let the perpendicular $AD$ be drawn.
From :
:triangles $\triangle ABC$, $\triangle ABD$ and $\triangle ADC$ are all similar to each other.
From :
:$CB : BA = BA : BD$
Therefore from :
:the recangle contained by $CB$ and $BD$ equals the square on $AB$.
For the same reason:
:the recangle contained by $BC$ and $CD$ equals the square on $AC$.
From :
:$BD : DA = AD : DC$
Therefore from :
:the recangle contained by $BD$ and $DC$ equals the square on $AD$.
From :
:$CB : CA = BA : AD$
From :
:the recangle contained by $BC$ and $AD$ equals the recangle contained by $BA$ and $AC$.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Construction of Components of Second Bimedial

== Theorem ==
== Proof ==
Let $\rho$, $\rho \sqrt k$ and $\rho \sqrt \lambda$ be rational straight lines which are commensurable in square only.
Take the mean proportional $\rho \sqrt [4] k$ of $\rho$ and $\rho \sqrt k$, which is medial.
Let $x$ be such that:
:$\rho \sqrt k : \rho \sqrt \lambda = \rho \sqrt [4] k : x$
which gives:
:$x = \dfrac {\rho \sqrt \lambda} {\sqrt [4] k}$
We have that:
:$\rho \sqrt k \frown \!\! - \rho \sqrt \lambda$
where $\frown \!\! -$ denotes commensurability in square only.
Thus:
:$\rho \sqrt [4] k \frown \!\! - x$
From Straight Line Commensurable with Medial Straight Line is Medial it follows that $\rho k^{3/4}$ is also medial.
Then:
== Also see ==
* Definition:Second Bimedial
== Sources ==
* : Book $\text X$. Propositions
Category:Euclidean Number Theory
Construction of Components of Side of Rational plus Medial Area

== Theorem ==
== Proof ==
:
From :
Let $AB$ and $BC$ be medial straight lines which are commensurable in square only such that:
: $AB^2 = BC^2 + \rho^2$
such that $\rho$ is incommensurable in length with $AB$.
Let the semicircle $ADB$ be drawn with $AB$ as the diameter.
Let $BC$ be bisected at $E$.
From Construction of Parallelogram Equal to Given Figure Exceeding a Parallelogram:
Let a parallelogram be applied to $AB$ equal to the square on either of $BE$ or $EC$, and deficient by a square.
Let this parallelogram be the rectangle contained by $AF$ and $FB$.
From :
:$AF$ is incommensurable in length with $FB$.
Let $FD$ be drawn perpendicular to $AB$.
Join $AD$ and $DB$.
We have that $AF$ is incommensurable in length with $FB$.
So from :
:$BA \cdot AF$ is incommensurable with $AB \cdot BF$.
From :
:$BA \cdot AF = AD^2$
and:
:$AB \cdot BF = DB^2$
Therefore $AD^2$ and $DB^2$ are incommensurable.
As $AB$ is medial, it follows by definition that $AB^2$ is a medial area.
From Pythagoras's Theorem:
:$AB^2 = \left({AD + DB}\right)^2$
Thus $\left({AD + DB}\right)^2$ is also a medial area.
Therefore $AF + FB$ is medial.
As $BC = 2 DF$:
:$AB \cdot BC = 2 AB \cdot FD$
But $AB \cdot BC$ is a rational area.
Therefore from :
:$AB \cdot FD$ is a rational area.
But from :
:$AB \cdot FD = AD \cdot DB$
Thus $AD \cdot DB$ is a rational area.
Therefore we have found two straight lines which are incommensurable in square whose sum of squares is medial, but such that the rectangle contained by them is rational.
== Also see ==
* Definition:Side of Rational plus Medial Area
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Construction of Components of Side of Sum of Medial Areas

== Theorem ==
== Proof ==
:
From :
Let $AB$ and $BC$ be medial straight lines which are commensurable in square only such that:
: $(1): \quad$ $AB$ and $BC$ contain a medial rectangle
: $(2): \quad AB^2 = BC^2 + \rho^2$
such that $\rho$ is incommensurable in length with $AB$.
Let the semicircle $ADB$ be drawn with $AB$ as the diameter.
Let $BC$ be bisected at $E$.
From Construction of Parallelogram Equal to Given Figure Exceeding a Parallelogram:
Let a parallelogram be applied to $AB$ equal to the square on either of $BE$ or $EC$, and deficient by a square.
Let this parallelogram be the rectangle contained by $AF$ and $FB$.
Let $FD$ be drawn perpendicular to $AB$.
Join $AD$ and $DB$.
From :
:$AF$ is incommensurable in length with $FB$.
So from :
:$BA \cdot AF$ is incommensurable with $AB \cdot BF$.
From :
:$BA \cdot AF = AD^2$
and:
:$AB \cdot BF = DB^2$
Therefore $AD^2$ and $DB^2$ are incommensurable.
... this far
As $AB$ is medial, it follows by definition that $AB^2$ is a medial area.
From Pythagoras's Theorem:
:$AB^2 = \left({AD + DB}\right)^2$
Thus $\left({AD + DB}\right)^2$ is also a medial area.
Therefore $AD + DB$ is medial.
Since $AF \cdot FB = BE^2$ and also $AF \cdot FB = DF^2$:
:$BE^2 = DF^2$
As $BC = 2 DF$:
:$AB \cdot BC = 2 AB \cdot FD$
But $AB \cdot BC$ is a medial area.
Therefore from :
:$AB \cdot FD$ is a medial area.
But from :
:$AB \cdot FD = AD \cdot DB$
Thus $AD \cdot DB$ is a medial area.
Since:
:$AB$ is incommensurable in length with $BC$
and
:$CB$ is commensurable in length with $BE$
from :
:$AB$ is incommensurable in length with $BE$.
By :
:$AB^2 = AB \cdot BE$
From Pythagoras's Theorem:
:$AD^2 + DB^2 = AB^2$
But from :
:$AB \cdot FD = AD \cdot DB$
and:
:$AD \cdot DB = AD \cdot BE$
Therefore $AD^2 + DB^2$ is incommensurable with $AD \cdot DB$.
Therefore we have found two straight lines $AD$ and $DB$ which are incommensurable in square whose sum of squares is medial, but such that the rectangle contained by them is medial.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Construction of Conic Section

== Theorem ==
Consider a right circular cone $\CC$ with opening angle $2 \alpha$ whose apex is at $O$.
Consider a plane $\PP$ which intersects $\CC$, not passing through $O$, at an angle $\beta$ to the axis of $\CC$.
Then the points of intersection form one of the following:
:a circle
:an ellipse
:a hyperbola
:a parabola.
== Proof ==
When $\beta$ is a right angle, the points of intersection form a circle, by definition of transverse section.
Otherwise, let the plane $OAA'$ through the axis of $\CC$ perpendicular to $\PP$ intersect $\PP$ in the line $AA'$.
Let $P$ be an arbitrary point on the intersection of $\PP$ with $\CC$.
Let $PM$ be constructed perpendicular to $AA'$.
$PM$ is then perpendicular to the plane $OAA'$.
The transverse section through $P$ then contains $PM$ and cuts $OAA'$ in $TT'$, which is a diameter of that transverse section.
The diagram below is the cross-section through $\CC$ corresponding to the plane $OAA'$.
The point $P$ is imagined to be perpendicular to the plane whose projection onto $OAA'$ coincides with $M$.
:
Because the transverse section is a circle, we have by the Intersecting Chords Theorem:
:$PM^2 = TM \cdot MT'$
Also:
:$\dfrac {TM} {AM} = \dfrac {\map \sin {\beta + \alpha} } {\cos \alpha}$
:$\dfrac {MT'} {MA'} = \dfrac {\map \sin {\beta - \alpha} } {\cos \alpha}$
and so:
:$PM^2 = k \cdot AM \cdot MA'$
where $k$ is the constant:
:$k = \dfrac {\map \sin {\beta + \alpha} \map \sin {\beta - \alpha} } {\cos \alpha}$
When $\beta > \alpha$, $\PP$ intersects one nappe of $\CC$.
In this case the intersection is a closed curve which is the ellipse, and $k$ is positive.
The plane passing through $O$ which is parallel to $\PP$ intersects $\CC$ at $O$ only.
When $\beta < \alpha$, $\PP$ intersects both nappes of $\CC$.
In this case the intersection consists of two open curves which together form the hyperbola, and $k$ is negative.
The plane passing through $O$ which is parallel to $\PP$ intersects $\CC$ in two straight lines which themselves intersect at $O$.
When $\beta = \alpha$, $\PP$ intersects one nappes of $\CC$.
In this case the intersection consists of one open curves which is the parabola, and $k$ is zero.
The plane passing through $O$ which is parallel to $\PP$ intersects $\CC$ in one straight line, which is a generatrix of $\CC$.
Hence:
:$MT' = 2 OA \sin \alpha$
which is constant.
Also:
:$TM = 2 AM \sin \alpha$
and so:
:$PM^2 = 4 OA \sin^2 \alpha \cdot AM$
That is:
:$PM^2 : AM$ is constant.
== Sources ==
* : Chapter $\text {IV}$. The Ellipse: $1$.
Category:Conic Sections
Construction of Cube within Given Sphere

== Theorem ==
== Proof ==
:
Let $AB$ be the diameter of the given sphere.
Let $AB$ be cut at $C$ where $AC = 2 \cdot CB$.
Let $ADB$ be a semicircle on the diameter $AB$.
Let $CD$ be drawn from $C$ perpendicular to $AB$.
Let $DB$ be joined.
Let the square $EFGH$ be set out whose sides are all equal to $DB$.
From :
:let $EK, FL, GM, HN$ be drawn from $E, F, G, H$ respectively perpendicular to the plane of the square $EFGH$.
Let $EK, FL, GM, HN$ be made equal to one of the straight lines $EF, FG, GH, HE$.
Let $KL, LM, MN, NK$ be joined.
The cube $FN$ can be seen to have been constructed which is contained by six squares.
Next it is to be demonstrated that the cube $FN$ can be inscribed in the given sphere.
Let $KG, EG$ be joined.
We have that $KE$ is perpendicular to the plane of the square $EFGH$.
So from :
:$KE$ is perpendicular to the straight line $EG$.
Therefore $\angle KEG$ is a right angle.
Therefore also if $FK$ is joined, then $GF$ is perpendicular to $FK$.
For this reason, the semicircle described on $GK$ will also pass through $F$.
Similarly it will also pass through the remaining vertices of the cube.
Let $GK$ remain fixed, and let that semicircle be carried around and restored to the same position from which it began.
Thus the cube $FN$ has been inscribed in a sphere.
We have that:
:$GF = FE$
and that $\angle F$ is a right angle.
Therefore from :
:$EG^2 = 2 EF^2$
But:
:$EF = EK$
Therefore:
:$EG^2 = 2 \cdot EK^2$
Therefore from :
:$GK^2 = EG^2 + EK^2 = 3 EK^2$
We have that:
:$AB = 3 BC$
while:
:$AB : BC = AB^2 : BD^2$
Therefore:
:$AB^2 = 3 \cdot BD^2$
But it has been proved that:
:$GK^2 = 3 EK^2$
while:
:$KE = DB$
Therefore:
:$KG = AB$
But $AB$ is the diameter of the given sphere.
Therefore:
:the cube $FN$ can be inscribed in the given sphere
and:
:the square on the diameter of the given sphere is three times the square on the side of the cube.
== Sources ==
* : Book $\text{XIII}$. Propositions
Category:Cubes
Category:Spheres
Construction of Equal Angle

== Theorem ==
== Construction ==
:
Let $A$ be the given point on the given straight line $AB$, and let $\angle DCE$ be the given rectilinear angle, in which points $D$ and $E$ are any points on the straight lines bounding the angle (one on each side).
We can then construct $\triangle AFG$ such that $CE = AF$, $CD = AG$, and $DE = GF$, with $F$ on $AB$.
$\angle GAF$ is the required angle.
== Proof ==
Since all three sides of the triangles are equal, the interior angles of the triangles are also equal.
Thus, $\angle GAF = \angle ECD$, with $\angle GAF$ at the point $A$ on the straight line $AB$.
This theorem is one of the two attributed to (the other being ).
== Sources ==
* : Book $\text{I}$. Propositions
Category:Angles
Construction of Equal Straight Line

== Theorem ==
At a given point, it is possible to construct a straight line segment of length equal to that of any given straight line segment.
The given point will be an endpoint of the constructed straight line segment.
== Construction ==
:
Let $A$ be a given point.
Let $BC$ be the given straight line segment.
We draw a line segment from $A$ to $B$ to form the straight line segment $AB$.
We construct an equilateral triangle $\triangle ABD$ on $AB$.
We extend the straight lines $DA$ and $DB$ to $E$ and $F$ respectively.
We construct a circle $CGH$ with center $B$ and radius $BC$.
 
We construct a circle $GKL$ with center $D$ and radius $DG$.
The line $AL$ is the straight line segment required.
== Proof ==
As $B$ is the center of circle $CGH$, it follows from that $BC = BG$.
As $D$ is the center of circle $GKL$, it follows from that $DL = DG$.
As $\triangle ABD$ is an equilateral triangle, it follows that $DA = DB$.
Therefore, by , $AL = BG$.
As $AL = BG$ and $BC = BG$, it follows from that $AL = BC$.
Therefore, at the given point $A$, the required straight line segment $AL$ has been placed equal in length to $BC$.
== Sources ==
* : Book $\text{I}$. Propositions
Category:Lines
Construction of Equal Straight Lines from Unequal

== Theorem ==
Given two unequal straight line segments, it is possible to cut off from the greater a straight line segment equal to the lesser.
== Construction ==
:
Let $AB$ and $C$ be the given straight line segments.
Let $AB$ be the greater of them.
At point $A$, we place $AD$ equal to $C$.
We construct a circle $DEF$ with center $A$ and radius $AD$.
The straight line segment $AE$ is the required line.
== Proof ==
As $A$ is the center of circle $DEF$, it follows from that $AE = AD$.
But $C$ is also equal to $AD$.
So, as $C = AD$ and $AD = AE$, it follows from Common Notion 1 that $AE = C$.
Therefore, given the two straight line segments $AB$ and $C$, from the greater of these $AB$, a length $AE$ has been cut off equal to the lesser $C$.
== Sources ==
* : Book $\text{I}$. Propositions
Category:Lines
Construction of Equilateral Polygon with Even Number of Sides in Outer of Concentric Circles

== Theorem ==
== Proof ==
:
Let $ABCD$ and $EFGH$ be two concentric circles whose centers are at $K$.
Let $ABCD$ be the larger.
It is required that a regular polygon with an even number of sides be inscribed in $ABCD$ such that it does not touch $EFGH$.
Let the straight line $BKD$ be drawn through the center $K$.
From the point $G$ let $GA$ be drawn perpendicular to $BD$ and carried through to $ABCD$ at $C$.
From :
:$AC$ is tangent to $EFGH$.
Let the arc $BAD$ of $ABCD$ be bisected repeatedly.
By :
:this can be done until an arc remains less than $AD$.
Let such remaining arc be $LD$.
Let $LM$ be drawn perpendicular to $BD$ and carried through to $ABCD$ at $N$.
Let $LD$ and $DN$ be joined.
From 
:$LD = DN$
We have that $LN \parallel AC$.
We also have that $AC$ is tangent to $EFGH$.
Therefore $LN$ does not touch $EFGH$.
Therefore $LD$ and $DN$ are far from touching $EFGH$.
We can then fit straight lines equal to $LD$ into $ABCD$ as chords going all the way round the circle.
Thus a regular polygon with an even number of sides has been inscribed in $ABCD$ such that it does not touch $EFGH$.
== Sources ==
* : Book $\text{XII}$. Propositions
Category:Concentric Circles
Category:Polygons