nyuuzyou/wikis huggingface dataset:
Straight Line Commensurable with Apotome of Medial Straight Line

== Theorem ==
== Proof ==
:
Let $AB$ be an apotome of a medial straight line.
Let $CD$ be commensurable in length with $AB$.
It is to be demonstrated that:
:$CD$ is an apotome of a medial straight line
and:
:the order of $CD$ is the same as the order of $AB$.
Let $BE$ be the annex of $CD$.
Therefore by definition of apotome of a medial straight line:
:$AE$ and $EB$ are medial straight lines which are commensurable in square only.
From , let it be contrived that:
:$BE : DF = AB : CD$
From :
:$AE : CF = AB : CD$
and so from :
:$AE : CF = BE : DF$
But $AB$ is commensurable in length with $CD$.
Therefore from :
:$AE$ is commensurable in length with $CF$
and:
:$BE$ is commensurable in length with $DF$.
We have that $AE$ and $EB$ are medial straight lines which are commensurable in square only.
From :
: $CF$ and $FD$ are medial straight lines.
Therefore from :
: $CF$ and $FD$ are medial straight lines which are commensurable in square only.
Thus $CD$ is an apotome of a medial straight line.
It remains to be shown that the order of $CD$ is the same as the order of $AB$.
We have that:
:$AE : CF = BE : DF$
So from :
:$AE : EB = CF : FD$
Therefore:
:$AE^2 : AE \cdot EB = CF^2 : CF \cdot FD$
But $AE^2$ is commensurable with $CF^2$.
Therefore from:
::
and:
::
we have that:
:$AE \cdot EB$ is commensurable with $CF \cdot FD$.
Therefore by :
:if $AE \cdot EB$ is rational, then $CF \cdot FD$ is rational.
From :
:if $AE \cdot EB$ is medial, then $CF \cdot FD$ is medial.
Thuis $CD$ is an apotome of a medial straight line of the same as the order as $AB$.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Straight Line Commensurable with Bimedial Straight Line is Bimedial and of Same Order

== Theorem ==
== Proof ==
:
Let $AB$ be bimedial.
Let $CD$ be commensurable in length with $AB$.
It is to be shown that $CD$ is bimedial, and that the order of $CD$ is the same as the order of $AB$.
Let $AB$ be divided into its medials by $E$.
Let $AE$ be the greater medial.
By definition, $AE$ and $EB$ are medial straight lines which are commensurable in square only.
Using , let it be contrived that:
:$AB : CD = AE : CF$
Therefore by :
:$EB : FD = AB : CD$
But $AB$ is commensurable in length with $CD$.
Therefore from :
:$AE$ is commensurable in length with $CF$
and:
:$EB$ is commensurable in length with $FD$.
But by hypothesis $AE$ and $EB$ are medial.
Therefore by :
: $CF$ and $FD$ are medial.
From :
:$AE : CF = EB : FD$
Therefore by :
:$AE : EB = CF : FD$
But by hypothesis $AE$ and $EB$ are commensurable in square only.
Therefore by :
:$CF$ and $FD$ are commensurable in square only.
But $CF$ and $FD$ are medial.
Therefore, by definition, $CD$ is bimedial.
It remains to be demonstrated that $CD$ is of the same order as $AB$.
We have that:
:$AE : EB = CF : FD$
Therefore:
:$AE^2 : AE \cdot EB = CF^2 : CF \cdot FD$
Therefore by :
:$AE^2 : CF^2 = AE \cdot EB : CF \cdot FD$
But:
:$AE^2$ is commensurable with $CF^2$.
Therefore $AE \cdot EB$ is commensurable with $CF \cdot FD$.
Suppose $AB$ is a first bimedial.
Then $AE \cdot EB$ is rational.
It follows that $CF \cdot FD$ is rational.
Thus by definition $CD$ is a first bimedial.
Suppose otherwise that $AB$ is a second bimedial.
Then $AE \cdot EB$ is medial.
It follows that $CF \cdot FD$ is medial.
Thus by definition $CD$ is a second bimedial.
Thus $CD$ is of the same order as $AB$.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Straight Line Commensurable with Major Straight Line is Major

== Theorem ==
== Proof ==
:
Let $AB$ be a major.
Let $CD$ be commensurable in length with $AB$.
It is to be shown that $CD$ is also a major.
Let $AB$ be divided into its terms by $E$.
Let $AE$ be the greater term.
By definition, $AE$ and $EB$ are straight lines such that:
: $AE$ and $EB$ are incommensurable in square
: $AE^2 + EB^2$ is rational
: $AE \cdot EB$ is medial.
Using , let it be contrived that:
:$AB : CD = AE : CF$
Therefore by :
:$EB : FD = AB : CD$
Therefore by :
:$AE : CF = EB : FD$
But $AB$ is commensurable in length with $CD$.
Therefore from :
:$AE$ is commensurable in length with $CF$
and:
:$EB$ is commensurable in length with $FD$.
We have that:
:$AE : CF = EB : FD$
Therefore by :
:$AE : EB = CF : FD$
and by by :
:$AB : BE = CD : DF$
Therefore by :
:$AB^2 : BE^2 = CD^2 : DF^2$
Using a similar line of reasoning to the above:
:$AB^2 : AE^2 = CD^2 : CF^2$
and:
:$AB^2 : AE^2 + EB^2 = CD^2 : CF^2 + FD^2$
Therefore by :
:$AB^2 : CD^2 = AE^2 + EB^2 : CF^2 + FD^2$
But $AB^2$ is commensurable with $CD^2$.
Therefore $AE^2 + EB^2$ is commensurable with $CF^2 + FD^2$.
We have that $AE^2 + EB^2$ is rational.
Therefore $CF^2 + FD^2$ is rational.
Similarly, $2 \cdot AE \cdot EB$ is commensurable with $2 \cdot CF \cdot FD$.
By :
:$2 \cdot AE \cdot EB$ is medial.
Therefore $2 \cdot CF \cdot FD$ is medial.
Thus it has been demonstrated that $CF$ and $FD$ are straight lines such that:
: $CF$ and $FD$ are incommensurable in square
: $CF^2 + FD2$ is rational
: $CF \cdot FD$ is medial.
Thus by definition $CD$ is a major.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Straight Line Commensurable with Medial Straight Line is Medial

== Theorem ==
=== Porism ===
== Proof ==
:
Let $A$ be a medial straight line.
Let $B$ be a straight line which is commensurable in length with $A$.
Let $CD$ be a rational straight line.
Let a rectangle $CE$ be set up on $CD$ equal to the square on $A$.
Let the breadth of $CE$ be $DE$.
Then from Square on Medial Straight Line:
: $ED$ is a rational straight line which is commensurable in length with CD.
Let a rectangle $CF$ be set up on $CD$ equal to the square on $B$.
Let the breadth of $CF$ be $DF$.
As $A$ is commensurable in length with $B$, the square on $A$ is also commensurable with the square on $B$.
But as $CE = A^2$ and $CF = B^2$, $CE$ is commensurable with $CF$.
From Areas of Triangles and Parallelograms Proportional to Base:
:$EC : CF = ED : DF$
From Commensurability of Elements of Proportional Magnitudes:
:$ED$ is commensurable with $DF$.
But $ED$ is a rational straight line which is commensurable in length with CD.
By , $DF$ is a rational straight line.
By Commensurable Magnitudes are Incommensurable with Same Magnitude:
: $DF$ is incommensurable in length with $DC$.
Therefore $CD$ and $DF$ are rational and commensurable in square only.
But from Medial is Irrational, the side of the square equal to therectangle contained by $CD$ and $DF$ is medial.
Therefore $B$ is medial.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Straight Line Commensurable with Medial Straight Line is Medial/Porism

== Porism to Straight Line Commensurable with Medial Straight Line is Medial ==
== Proof ==
:And in the same way as was explained in the case of the rationals it follows, as regards medials, that a straight line commensurable in length with a medial straight line is called medial and commensurable with it not only in length but in square also, since, in general, straight lines commensurable in length are always commensurable in square also.
:But, if any straight line be commensurable in square with a medial straight line , then, if it is also commensurable in length with it, the straight lines are called, in this case too, medial and commensurable in length and in square, but, if in square only, they are called medial straight lines commensurable in square only.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Straight Line Commensurable with Minor Straight Line

== Theorem ==
== Proof ==
:
Let $AB$ be a minor straight line.
Let $CD$ be commensurable in length with $AB$.
It is to be demonstrated that $CD$ is a minor straight line.
Let $BE$ be the annex of $CD$.
Therefore by definition of minor straight line:
: $AE$ and $EB$ are incommensurable in square
: $AE^2 + EB^2$ is rational
: $2 \cdot AE \cdot EB$ is a medial rectangle.
From , let it be contrived that:
:$BE : DF = AB : CD$
From :
:$AE : CF = AB : CD$
and so from :
:$AE : CF = BE : DF$
We have that $AE$ and $EB$ are incommensurable in square.
Therefore from :
:$CF$ and $FD$ are incommensurable in square.
We have that:
:$AE : CF = BE : DF$
So from :
:$AE : EB = CF : FD$
So from :
:$AE^2 : EB^2 = CF^2 : FD^2$
Therefore from :
:$AE^2 + EB^2 : EB^2 = CF^2 + FD^2 : FD^2$
But $EB^2$ is commensurable with $FD^2$.
So from:
::
and:
::
we have that:
:$AE^2 + EB^2$ is commensurable with $CF^2 + FD^2$.
But by definition $AE^2 + EB^2$ is rational.
Therefore by :
:$CF^2 + FD^2$ is rational.
We have that:
:$AE : EB = CF : FD$
Therefore:
:$AE^2 : AE \cdot EB = CF^2 : CF \cdot FD$
Therefore $AE \cdot EB$ is commensurable with $CF \cdot FD$.
Therefore $CF \cdot FD$ is medial.
Thus $CF$ and $FD$ are such that:
: $CF$ and $FD$ are incommensurable in square
: $CF^2 + FD^2$ is rational
: $2 \cdot CF \cdot FD$ is a medial rectangle.
Hence the result.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Straight Line Commensurable with Side of Rational plus Medial Area

== Theorem ==
== Proof ==
:
Let $AB$ be the side of a rational plus a medial area.
Let $CD$ be commensurable in length with $AB$.
It is to be shown that $CD$ is also the side of a rational plus a medial area.
Let $AB$ be divided into its terms by $E$.
Let $AE > EB$.
By definition, $AE$ and $EB$ are straight lines such that:
: $AE$ and $EB$ are incommensurable in square
: $AE^2 + EB^2$ is medial
: $AE \cdot EB$ is rational.
Using , let it be contrived that:
:$AB : CD = AE : CF$
Therefore by :
:$EB : FD = AB : CD$
Therefore by :
:$AE : CF = EB : FD$
But $AB$ is commensurable in length with $CD$.
Therefore from :
:$AE$ is commensurable in length with $CF$
and:
:$EB$ is commensurable in length with $FD$.
We have that:
:$AE : CF = EB : FD$
Therefore by :
:$AE : EB = CF : FD$
and by by :
:$AB : BE = CD : DF$
Using a similar line of reasoning to :
: $CF$ and $FD$ are incommensurable in square
: $AE^2 + EB^2$ is commensurable with $CF^2 + FD^2$
: $AE \cdot EB$ is commensurable with $CF \cdot FD$.
We have that $AE^2 + EB^2$ is medial.
Therefore $CF^2 + FD^2$ is medial.
Similarly we have that $AE \cdot EB$ is rational.
Therefore $CF \cdot FD$ is rational.
Hence:
: $CF$ and $FD$ are incommensurable in square
: $CF^2 + FD^2$ is medial
: $CF \cdot FD$ is rational.
Thus by definition $CD$ is the side of a rational plus a medial area.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Straight Line Commensurable with Side of Sum of two Medial Areas

== Theorem ==
== Proof ==
:
Let $AB$ be the side of the sum of two medial areas.
Let $CD$ be commensurable in length with $AB$.
It is to be shown that $CD$ is also the side of the sum of two medial areas.
Let $AB$ be divided into its terms by $E$.
Let $AE > EB$.
By definition, $AE$ and $EB$ are straight lines such that:
: $AE$ and $EB$ are incommensurable in square
: $AE^2 + EB^2$ is medial
: $AE \cdot EB$ is medial.
Using , let it be contrived that:
:$AB : CD = AE : CF$
Therefore by :
:$EB : FD = AB : CD$
Therefore by :
:$AE : CF = EB : FD$
But $AB$ is commensurable in length with $CD$.
Therefore from :
:$AE$ is commensurable in length with $CF$
and:
:$EB$ is commensurable in length with $FD$.
We have that:
:$AE : CF = EB : FD$
Therefore by :
:$AE : EB = CF : FD$
and by by :
:$AB : BE = CD : DF$
Using a similar line of reasoning to :
: $CF$ and $FD$ are incommensurable in square
: $AE^2 + EB^2$ is commensurable with $CF^2 + FD^2$
: $AE \cdot EB$ is commensurable with $CF \cdot FD$.
We have that $AE^2 + EB^2$ is medial.
Therefore $CF^2 + FD^2$ is medial.
Similarly we have that $AE \cdot EB$ is medial.
Therefore $CF \cdot FD$ is medial.
Hence:
: $CF$ and $FD$ are incommensurable in square
: $CF^2 + FD^2$ is medial
: $CF \cdot FD$ is medial.
Thus by definition $CD$ is the side of the sum of two medial areas.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Straight Line Commensurable with that which produces Medial Whole with Medial Area

== Theorem ==
== Proof ==
:
Let $AB$ be a straight line which produces with a medial area a medial whole.
Let $CD$ be commensurable in length with $AB$.
It is to be demonstrated that $CD$ is a straight line which produces with a medial area a medial whole.
Let $BE$ be the annex of $CD$.
Therefore by definition of a straight line which produces with a rational area a medial whole:
: $AE$ and $EB$ are incommensurable in square
: $AE^2 + EB^2$ is medial
: $2 \cdot AE \cdot EB$ is a medial rectangle.
From , let it be contrived that:
:$BE : DF = AB : CD$
From :
:$AE : CF = AB : CD$
and so from :
:$AE : CF = BE : DF$
We have that $AE$ and $EB$ are incommensurable in square.
Therefore from :
:$CF$ and $FD$ are incommensurable in square.
We have that:
:$AE : CF = BE : DF$
So from :
:$AE : EB = CF : FD$
So from :
:$AE^2 : EB^2 = CF^2 : FD^2$
Therefore from :
:$AE^2 + EB^2 : EB^2 = CF^2 + FD^2 : FD^2$
But $EB^2$ is commensurable with $FD^2$.
So from:
::
and:
::
we have that:
:$AE^2 + EB^2$ is commensurable with $CF^2 + FD^2$.
But by definition $AE^2 + EB^2$ is medial.
Therefore by $CF^2 + FD^2$ is medial.
We have that:
:$AE : EB = CF : FD$
Therefore:
:$AE^2 : AE \cdot EB = CF^2 : CF \cdot FD$
Therefore $AE \cdot EB$ is commensurable with $CF \cdot FD$.
Therefore $CF \cdot FD$ is medial.
Thus $CF$ and $FD$ are such that:
: $CF$ and $FD$ are incommensurable in square
: $CF^2 + FD^2$ is medial
: $2 \cdot CF \cdot FD$ is a medial rectangle.
Hence the result.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Straight Line Commensurable with that which produces Medial Whole with Rational Area

== Theorem ==
== Proof ==
:
Let $AB$ be a straight line which produces with a rational area a medial whole.
Let $CD$ be commensurable in length with $AB$.
It is to be demonstrated that $CD$ is a straight line which produces with a rational area a medial whole.
Let $BE$ be the annex of $CD$.
Therefore by definition of a straight line which produces with a rational area a medial whole:
: $AE$ and $EB$ are incommensurable in square
: $AE^2 + EB^2$ is medial
: $2 \cdot AE \cdot EB$ is a rational rectangle.
From , let it be contrived that:
:$BE : DF = AB : CD$
From :
:$AE : CF = AB : CD$
and so from :
:$AE : CF = BE : DF$
We have that $AE$ and $EB$ are incommensurable in square.
Therefore from :
:$CF$ and $FD$ are incommensurable in square.
We have that:
:$AE : CF = BE : DF$
So from :
:$AE : EB = CF : FD$
So from :
:$AE^2 : EB^2 = CF^2 : FD^2$
Therefore from :
:$AE^2 + EB^2 : EB^2 = CF^2 + FD^2 : FD^2$
But $EB^2$ is commensurable with $FD^2$.
So from:
::
and:
::
we have that:
:$AE^2 + EB^2$ is commensurable with $CF^2 + FD^2$.
But by definition $AE^2 + EB^2$ is medial.
Therefore by $CF^2 + FD^2$ is medial.
We have that:
:$AE : EB = CF : FD$
Therefore:
:$AE^2 : AE \cdot EB = CF^2 : CF \cdot FD$
Therefore $AE \cdot EB$ is commensurable with $CF \cdot FD$.
Therefore $CF \cdot FD$ is rational.
Thus $CF$ and $FD$ are such that:
: $CF$ and $FD$ are incommensurable in square
: $CF^2 + FD^2$ is medial
: $2 \cdot CF \cdot FD$ is a rational rectangle.
Hence the result.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Straight Line Perpendicular to Plane from Point is Unique

== Theorem ==
== Proof ==
:
Suppose it were possible to set up two straight lines $AB$ and $AC$ perpendicular to the plane of reference and on the same side.
Let a plane be drawn through $BA$ and $AC$.
From :
:let the common section be the straight line $DAE$.
Therefore the straight lines $AB$, $AC$ and $DAE$ are all in the same plane.
We have that $AC$ is perpendicular to the plane of reference.
So from :
:$AC$ is perpendicular to all the straight lines which meet it and are in the plane of reference.
But $DAE$ meets $AC$ and is in the plane of reference.
Therefore $\angle CAE$ is a right angle.
For the same reason, $\angle BAE$ is a right angle.
Therefore $\angle BAE = \angle CAE$.
But both are in the same plane, which is impossible.
Hence the result.
== Sources ==
* : Book $\text{XI}$. Propositions
Category:Planes
Category:Perpendiculars
Straight Line Segment is Shortest Path between Two Points/Lemma 1

== Lemma to Straight Line Segment is Shortest Path between Two Points ==
Let $p, q \in \R$ be real numbers such that $\sqrt {1 + p^2} = \sqrt {1 + q^2}$.
Then:
:$1 + \size p = 1 + \size q$
== Proof ==
By definition of absolute value of a real number:
:$\sqrt {p^2} = \size p$
Hence:
Category:Straight Line Segment is Shortest Path between Two Points
Straight Line Segment is Shortest Path between Two Points/Lemma 2

== Lemma to Straight Line Segment is Shortest Path between Two Points ==
Let:
:$\forall x \in \closedint a b: \map f x = \map g x$
Then:
:$\ds \int_a^b \map f x \rd x = \int_a^b \map g x \rd x$
== Proof ==
Follows directly from the definition of the definite integral.
Category:Straight Line Segment is Shortest Path between Two Points
Straight Line cannot be in Two Planes

== Theorem ==
== Proof ==
:
Suppose it were possible to have a straight line in more than one plane.
Let a part $AB$ of the straight line $ABC$ be in the plane of reference, and another part $BC$ be in a plane more elevated.
There will then be in the plane of reference some straight line $BD$ continuous with $AB$ in a straight line.
Therefore $AB$ is a common segment of the two straight lines $ABC$ and $ABD$.
Suppose a circle is described with center $B$ and radius $AB$.
Then the diameters would cut off unequal arcs of the circle.
Hence the result.
== Sources ==
* : Book $\text{XI}$. Propositions
Category:Straight Lines
Category:Planes
Straight Line cut in Extreme and Mean Ratio plus its Greater Segment

== Theorem ==
== Proof ==
:
Let the line $AB$ be cut in extreme and mean ratio at the point $C$.
Let $AC$ be the greater segment.
Let $AD = AC$.
It is to be demonstrated that $DB$ is cut in extreme and mean ratio at the point $A$ where $BA > AD$.
Let the square $AE$ be described on $AB$.
Let the figure be drawn as above.
We have that $AB$ is cut in extreme and mean ratio at $C$.
Therefore from:
:
and:
:
it follows that:
:$AB \cdot BC = AC^2$
We have that:
:$CE = AB \cdot BC$
and:
:$CH = AC^2$
THerefore:
:$CH = HC$
But:
:$HE = CE$
and:
:$DH = HC$
Therefore:
:$DH = HE$
Therefore:
:$DK = AE$
Thus:
:$DK = BD \cdot DA$
That is:
:$BD \cdot DA = AB^2$
Therefore:
:$DB : BA = BA : AD$
and:
:$DB > BA$
Therefore by :
:$BA > AD$
Hence the result as stated.
== Sources ==
* : Book $\text{XIII}$. Propositions
Category:Golden Mean
Straight Line has Zero Curvature

== Theorem ==
A straight line has zero curvature.
== Proof ==
From Equation of Straight Line in Plane: Slope-Intercept Form, a straight line has the equation:
:$y = m x + c$
Differentiating twice $x$:
By definition, the curvature of a curve is defined as:
:$\kappa = \dfrac {y} {\paren {1 + y'^2}^{3/2} }$
But we have that:
:$y := \dfrac {\d^2 y} {\d x^2} = 0$
and so, as in general $y' := \dfrac {\d y} {\d x} = m \ne 0$:
:$\kappa = \dfrac 0 {\paren {1 + m^2}^{3/2} }$
the curvature is zero.
== Sources ==
* : Chapter $\text I$: Introduction and Definitions. Elimination. Graphical Representation: Examples for solution: $(7)$
Category:Straight Lines
Category:Curvature
Straight Line making Larger Angle with Perpendicular to Straight Line is Longer

== Theorem ==
Let $AB$ be a straight line.
Let $C$ be a point which is not on $AB$.
Let $D$ be a point on $AB$ such that $CD$ is perpendicular to $AB$.
Let $E, F$ be points on $AB$ such that $\angle DCE > \angle DCF$.
Then $CE > CF$. 
== Proof ==
== Sources ==
* : Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.21 \ \text{(iii)}$
Category:Lines
Category:Angles
Straight Line which cuts Harmonic Pencil forms Harmonic Range

== Theorem ==
Let $AB$ and $PQ$ be line segments on a straight line such that $\tuple {AB, PQ}$ is a harmonic range.
Let $O$ be a point which is not on the straight line $AB$.
Let $\map O {AB, PQ}$ be the harmonic pencil formed from $O$ and $\tuple {AB, PQ}$.
Let a straight line intersect $OA$, $OB$, $OP$ and $OQ$ at $A'$, $B'$, $P'$ and $Q'$ respectively.
:
Then $A'$, $B'$, $P'$ and $Q'$ themselves form a harmonic range $\tuple {A'B', P'Q'}$.
== Proof ==
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $19$. Harmonic ranges and pencils
Category:Harmonic Ranges
Straight Lines Cut Off Equal Arcs in Equal Circles

== Theorem ==
== Proof ==
Let $ABC$ and $DEF$ be equal circles.
Let $AB, DE$ be equal straight lines cutting off arcs $ACB$ and $DFE$ as the greater, and $AGB, DHE$ as lesser.
:
Let $K$ and $L$ be the centers of the circles $ABC$ and $DEF$ respectively.
Let $AK, KB, DL, LE$ be joined.
Since the circles are equal, so are their radii.
So $AK = DL, KB = LE$.
As $AB = DE$ , from Triangle Side-Side-Side Congruence it follows that $\angle AKB = \angle DLE$.
But from Angles on Equal Arcs are Equal the arc $AGB$ equals the arc $DHE$.
As the whole circles $ABC$ and $DEF$ are equal, the arc $ACB$ which remains also equals arc $DFE$.
== Sources ==
* : Book $\text{III}$. Propositions
Category:Circles
Straight Lines Subtending Two Consecutive Angles in Regular Pentagon cut in Extreme and Mean Ratio

== Theorem ==
== Proof ==
:
Let $ABCDE$ be a regular pentagon.
Let the straight lines $AC$ and $BE$ subtend two vertices $A$ and $B$ taken in order.
Let $AC$ and $BE$ cut each other at $H$.
It is to be demonstrated that $AC$ and $BE$ are cut in extreme and mean ratio at the point $H$, and that the greater segment equals the side of $ABCDE$.
Let the circle $ABCDE$ circumscribe the pentagon $ABCDE$.
We have that:
:$EA = AB$
and:
:$AB = BC$
and they contain equal angles.
Therefore from :
:$AC = BE$
and:
:$\triangle ABC = \triangle ABE$
Therefore:
:$\triangle BAC = \triangle ABE$
From :
:$\angle AHE = 2 \cdot \angle BAH$
We have that the arc $EDC$ is twice the arc $DC$.
Therefore from:
:
and:
:
it follows that:
:$\angle EAC = 2 \cdot \angle BAC$
Therefore:
:$\angle HAE = \angle AHE$
Hence from :
:$HE = EA$
That is:
:$HE = AB$
We have that:
:$BA = AE$
and:
:$\angle ABE = \angle AEB$
But it has been proved that:
:$\angle ABE = \angle BAH$
Therefore:
:$\angle BEA = \angle BAH$
We also have that $\angle ABE$ is common to $\triangle ABE$ and $\triangle ABH$.
Therefore from :
:$\angle BAE = \angle AHB$
Therefore $\triangle ABE$ has the same corresponding angles with $\triangle ABH$.
Therefore from :
:$EB : BA = AB : BH$
But:
:$BA = EH$
Therefore:
:$BE : EH = EH : HB$
and:
:$BE > EH$
Therefore:
:$EH > HB$
Therefore $BE$ is cut in extreme and mean ratio at the point $H$, and that the greater segment equals the side of $ABCDE$.
Similarly it is shown that $AC$ is also cut in extreme and mean ratio at the point $H$, and that the greater segment equals the side of $ABCDE$.
== Sources ==
* : Book $\text{XIII}$. Propositions
* : $1 \cdotp 61803 \, 39887 \, 49894 \, 84820 \, 45868 \, 34365 \, 63811 \, 77203 \, 09179 \, 80576 \ldots$
* : $1 \cdotp 61803 \, 39887 \, 49894 \, 84820 \, 45868 \, 34365 \, 63811 \, 77203 \, 09179 \, 80576 \ldots$
Category:Pentagons
Category:Golden Mean
Straight Lines cut in Same Ratio by Parallel Planes

== Theorem ==
== Proof ==
:
Let the two straight lines $AB$ and $CD$ be cut by the parallel planes $CH$, $KL$ and $MN$ at the points $A, E, B$ and $C, F, D$.
It needs to be demonstrated that:
:$AE : EB = CF : FD$
Let $AC$, $BD$ and $AD$ be joined.
Let $AD$ meet $KL$ at $O$.
Let $ED$ and $OF$ be joined.
We have that the parallel planes $KL$ and $MN$ are cut by the plane $EBDO$.
So by :
:their common sections $EO$ and $BD$ are parallel.
Similarly, the parallel planes $GH$ and $KL$ are cut by the plane $AOFC$.
So by :
:their common sections $AC$ and $OF$ are parallel.
We have that the straight line $EO$ is parallel to $BD$.
But $BD$ is one of the sides of $\triangle ABD$.
Therefore from :
:$AE : EB = AO : OD$
Similarly, the straight line $OF$ is parallel to $AC$.
But $AC$ is one of the sides of $\triangle ADC$.
Therefore from :
:$AO : OD = CF : FD$
From :
:$AE : EB = CF : FD$
== Sources ==
* : Book $\text{XI}$. Propositions
Category:Parallel Planes
Category:Straight Lines