nyuuzyou/wikis huggingface dataset:
Construction of Medial Straight Lines Commensurable in Square Only containing Rational Rectangle whose Square Differences Commensurable with Greater

== Theorem ==
== Proof ==
Let $A$ and $B$ be rational straight lines which are commensurable in square only.
Let $A^2 > B^2$ such that:
:$A^2 = B^2 + \rho^2$
where $\rho$ is a straight line which is commensurable in length with $A$.
This can be done using Construction of Rational Straight Lines Commensurable in Square Only whose Square Differences Commensurable with Greater.
Let $C$ be a straight line such that $C^2 = A B$.
From Medial is Irrational, $A B$ is medial.
Therefore $C^2$ is medial.
So by definition $C$ is medial.
Let $C D = B^2$.
By hypothesis $B^2$ is rational.
Therefore $C D$ is rational.
We have that:
:$A : B = A^2 : A B$
But:
:$C^2 = A B$
and:
:$C D = B^2$
and so:
:$A : B = C^2 : C D$
But as:
:$C^2 : C D = C : D$
it follows that:
:$A : B = C : D$
But $A$ is commensurable in square only with $B$.
Therefore by Commensurability of Elements of Proportional Magnitudes:
:$C$ is commensurable in square only with $D$.
We have that $C$ is medial.
Therefore by Straight Line Commensurable with Medial Straight Line is Medial, $D$ is medial.
We have that:
:$A : B = C : D$
and:
: $A^2 = B^2 + \rho^2$
where $\rho$ is a straight line which is commensurable in length with $A$.
Therefore Commensurability of Squares on Proportional Straight Lines:
:$C^2 = D^2 + \sigma^2$
where $\sigma$ is a straight line which is commensurable in length with $C$.
Therefore two medial straight lines $C$ and $D$ have been found which are commensurable in square only and which contain a rational rectangle.
Further, $C^2$ is greater than $D^2$ by the square on a straight line commensurable in length with $C$.
Similarly it can be proved that the square on $C$ exceeds the square on $D$ by the square on a straight line incommensurable in length with $C$, when the square on $A$ is greater than the square on $B$ by the square on a straight line incommensurable in length with $A$.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Construction of Minor is Unique

== Theorem ==
== Proof ==
:
Let $AB$ be a minor.
Let $BC$ be added to $AB$ such that:
: $AC$ and $CB$ are incommensurable in square
: $AC^2 + CB^2$ is rational
: $2 \cdot AC \cdot CB$ is a medial rectangle.
It is to be proved that no other straight line can be added to $AB$ which fulfils these conditions.
Suppose $BD$, different from $BC$, can be added to $AB$ such that:
:$AD$ and $DB$ are incommensurable in square
:$AD^2 + DB^2$ is rational
:$2 \cdot AD \cdot DB$ is a medial 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 :$AD^2 + DB^2$ 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 by definition of medial area, this is impossible.
The result follows.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Construction of Outer Measure

== Theorem ==
Let $X$ be a set.
Let $\powerset X$ be the power set of $X$.
Let $\AA$ be a subset of $\powerset X$ which contains the empty set.
Let $\overline \R_{\ge 0}$ denote the set of positive extended real numbers.
Let $\gamma: \AA \to \overline \R_{\ge 0}$ be a mapping such that $\map \gamma \O = 0$.
Let $\mu^*: \powerset X \to \overline \R_{\ge 0}$ be the mapping defined as:
:$\ds \forall S \in \powerset X: \map {\mu^*} S = \inf \set {\sum_{n \mathop = 0}^\infty \map \gamma {A_n}: \forall n \in \N : A_n \in \AA, \ S \subseteq \bigcup_{n \mathop = 0}^\infty A_n}$
where the infimum of the empty set is $+\infty$.
Then $\mu^*$ is an outer measure on $X$.
== Proof ==
We check each of the criteria for an outer measure.
From the assumption $\map \gamma \O = 0$:
:$\map {\mu^*} \O = 0$
Any cover of a set is also a cover of a subset of it.
Thus $\mu^*$ is monotone by Infimum of Subset.
It remains to be shown that $\mu^*$ is countably subadditive.
Let $\sequence {S_n}$ be a sequence of subsets of $X$.
Let $\ds S := \bigcup_{n \mathop = 0}^\infty S_n$.
Suppose there does not exist a countable cover for $S_n$ by elements of $\AA$ for some $n \in \N$.
Then there does not exist a countable cover for $S$ by elements of $\AA$.
In this case, the result follows immediately.
Now suppose that for each $n \in \N$, there exists a countable cover for $S_n$ by elements of $\AA$.
Let $\epsilon \in \R_{>0}$ be an arbitrary (strictly) positive real number.
By definition of infimum, for each $n \in \N$, we can apply the axiom of countable choice to choose a countable cover $\CC_n \subseteq \AA$ of $S_n$ such that:
:$\ds \sum_{x \mathop \in \CC_n} \map \gamma x < \map {\mu^*} {S_n} + \frac \epsilon {2^{n + 1} }$
Let $\ds \CC = \bigcup_{n \mathop = 0}^\infty \CC_n$.
Then $\CC$ is a subset of $\AA$ and a cover for $S$.
Furthermore, $\CC$ is the countable union of countable sets.
By Countable Union of Countable Sets is Countable, $\CC$ is itself countable.
Therefore:
Since $\epsilon$ was arbitrary, the result follows.
== Consequences ==
It follows immediately that the induced outer measure is an outer measure.
== Sources ==
* 
Category:Outer Measures
Construction of Parallel Line

== Theorem ==
Given a straight line, and a given point not on that straight line, it is possible to draw a parallel to the given straight line.
== Construction ==
Let $A$ be the point, and let $BC$ be the infinite straight line.
Take a point $D$ at random on $BC$, and construct the segment $AD$.
Construct $\angle DAE$ equal to $\angle ADC$ on $AD$ at point $A$.
Extend $AE$ into an infinite straight line.
Then the line $AE$ is parallel to the given infinite straight line $BC$ through the given point $A$.
== Proof ==
The transversal $AD$ cuts the lines $BC$ and $AE$ and makes $\angle DAE = \angle ADC$.
From Equal Alternate Angles implies Parallel Lines it follows that $EA \parallel BC$.
== Sources ==
* : Book $\text{I}$. Propositions
Category:Parallel Lines
Construction of Parallelepiped Similar to Given Parallelepiped

== Theorem ==
== Proof ==
:
Let $AB$ be the given straight line.
Let $CD$ be the given parallelepiped.
Using :
:On the straight line $AB$ let a solid angle contained by the plane angles $\angle BAH, \angle HAK, \angle KAB$ be constructed equal to the solid angle at $C$ contained by the plane angles $\angle ECF, \angle ECG, \angle GCF$ such that:
::$\angle BAH = \angle ECF$
::$\angle KAH = \angle GCF$
::$\angle BAK = \angle ECG$
Using :
:Let it be contrived that:
::$EC : CG = BA : AK$
and:
::$GC : CF = KA : AH$
Therefore from :
:$EC : CF = BA : AH$
Let the parallelogram $HB$ and the parallelepiped $AL$ be completed.
We have that:
:$EC : CG = BA : AK$
and the sides about the equal angles $\angle BAK$ and $\angle ECG$ are proportional.
Therefore the parallelogram $GE$ is similar to the parallelogram $KB$.
For the same reason:
:the parallelogram $KH$ is similar to the parallelogram $GF$.
:the parallelogram $FE$ is similar to the parallelogram $HB$.
Therefore three parallelograms of the parallelepiped $CD$ are similar to the three parallelograms of the parallelepiped $AL$.
But the former three are equal and similar to the three opposite parallelograms.
Therefore the whole parallelepiped $CD$ is similar to the whole parallelepiped $AL$.
== Sources ==
* : Book $\text{XI}$. Propositions
Category:Parallelepipeds
Construction of Parallelogram Equal to Given Figure Exceeding a Parallelogram

== Theorem ==
== Construction ==
Let $AB$ be the given straight line, $C$ the given rectilineal figure to which the figure to be applied to $AB$ is to be equal, and $\Box D$ that to which the excess is required to be similar.
We need to apply to $AB$ a parallelogram equal to the area of $C$ exceeding it by a parallelogrammic figure similar to $\Box D$.
:
Let $AB$ be bisected at $E$.
Describe on $EB$ the parallelogram $\Box BF$ similar and similarly situated to $\Box D$.
From Construction of Figure Similar to One and Equal to Another, let $\Box GH$ be constructed at once equal to $\Box BF + C$ and similarly situated to $D$.
Let $FL$ and $FE$ be produced to $M$ and $N$ so that $FLM = KH$ and $FEN = KG$, and complete the parallelogram $\Box MN$, whose diameter is $FO$.
Construct the parallelogram $\Box AO$.
Then $\Box AO$ is the required parallelogram.
== Proof ==
Let $KH$ correspond to $FL$ and $KG$ to $FE$.
Since $GH > FB$, we have that $KH > FL$ and $KG > FE$.
Since $FLM = KH$ and $FEN = KG$, it follows that $\Box MN = \Box GH$ and also $\Box MN$ is similar to $\Box GH$.
But $\Box GH$ is similar to $\Box EL$.
So from Similarity of Polygons is Equivalence Relation‎‎, $\Box MN$ is similar to $\Box EL$.
Therefore from Parallelogram Similar and in Same Angle has Same Diameter $\Box EL$ is about the same diameter with $\Box MN$.
Let that diameter $FO$ be drawn, and the figure described.
Since $\Box GH = \Box EL + C$, while $\Box MN = \Box GH$, it follows that $\Box MN = \Box EL + C$.
Subtract $\Box EL$ from each.
Then the gnomon $XWV$ is equal to $C$.
Since $AE = EB$, from Parallelograms with Equal Base and Same Height have Equal Area, $\Box AN = \Box NB$.
Thus, from Complements of Parallelograms are Equal, $\Box AN = \Box LP$.
Add $\Box EO$ to each.
Then $\Box AO$ equals the gnomon $XWV$.
But the gnomon $XWV$ equals $C$.
So $\Box AO = C$.
Hence the result.
== Sources ==
* : Book $\text{VI}$. Propositions
Category:Parallelograms
Construction of Parallelogram Equal to Given Figure Less a Parallelogram

== Theorem ==
== Construction ==
Let:
: $AB$ be the given straight line
: $C$ be the given rectilineal figure to which the figure to be applied to $AB$ is to be equal, not being greater than the parallelogram described on the half of $AB$ and similar to the defect
and:
: $D$ be the parallelogram to which the defect required is to be similar.
We are to construct a parallelogram on $AB$ equal to $C$ and deficient by a parallelogrammic figure similar to $D$.
:
Let $AB$ be bisected at $E$.
Using Construction of Similar Polygon, let $\Box EBFG$ be described similar to $D$ and similarly situated.
Let the parallelogram $\Box AG$ be completed.
If $AG = C$ then the job is done.
Otherwise let $\Box HE > C$.
By Construction of Figure Similar to One and Equal to Another, let $\Box KLMN$ be constructed equal to the excess by which $GB$ is greater than $C$ and similar and similarly situated to $D$.
Let $GO = KL$ and $GP = LM$, and complete the parallelogram $\Box OGPQ$.
Construct the lines $RT \parallel AB$ and $PS \parallel GE$ passing through $Q$.
Then the parallelogram $ST$ is the required figure.
== Proof ==
Let $AG = C$.
Then $\Box AG$ has been applied to $AB$ equal in area to $C$ and deficient by $\Box GB$ similar to $D$.
Otherwise let $\Box HE > C$.
Now $HE = GB$, so:
: $GB > C$
We have that $\Box KLMN$ is constructed equal to the excess by which $GB$ is greater than $C$ and similar and similarly situated to $D$.
But $D$ is similar to $GB$ and so from Similarity of Polygons is Equivalence Relation‎ $KM$ is also similar to $GB$.
Let $KL$ correspond to $GE$ and $LM$ to $GF$.
Since $\Box GB = C + \Box KLMN$ it follows that:
: $\Box GB > \Box KLMN$
Therefore $GE > KL$ and $GF > KM$.
We have $GO$ constructed equal to $KL$ and $GP$ equal to $KM$.
Therefore $\Box OGPQ = \Box KLMN$ and $\Box OGPQ$ is similar to $\Box KLMN$.
Therefore from Similarity of Polygons is Equivalence Relation‎‎ $\Box GQ$ is similar to $\Box GB$.
So from Parallelogram Similar and in Same Angle has Same Diameter, $GQ$ is about the same diameter with $GB$.
Let $GQB$ be this diameter and let the figure be described.
Since $\Box BG = C + \Box KM$, and in them $\Box GQ = \Box KM$, it follows that the gnomon $UWV$ equals the remainder $C$.
From [[Complements of Parallelograms are Equal]:
: $\Box PR = \Box OS$.
Let $\Box QB$ be added to each.
Then:
: $\Box PB = \Box OB$
But since $AE = EB$, it follows from Parallelograms with Equal Base and Same Height have Equal Area that:
: $\Box OB = \Box TE$
Therefore:
: $\Box TE = \Box PB$
Let $\Box OS$ be added to each.
Therefore the whole of $\Box TS$ equals the area of the gnomon $UWV$.
But $UWV$ was shown to be equal to $C$.
Hence the result.
== Sources ==
* : Book $\text{VI}$. Propositions
Category:Parallelograms
Construction of Parallelogram equal to Triangle in Given Angle

== Theorem ==
A parallelogram can be constructed in a given angle the same size as any given triangle.
== Proof ==
:
Let $ABC$ be the given triangle, and $D$ the given angle.
Bisect $BC$ at $E$, and join $AE$.
Construct $AG$ parallel to $EC$.
Construct $\angle CEF$ equal to $\angle D$.
Construct $CG$ parallel to $EF$.
Then $FEGC$ is a parallelogram.
Since $BE = EC$, from Triangles with Equal Base and Same Height have Equal Area, $\triangle ABE = \triangle AEC$.
So $\triangle ABC$ is twice the area of $\triangle AEC$.
But from Parallelogram on Same Base as Triangle has Twice its Area, $FECG$ is also twice the area of $\triangle AEC$.
So $FECG$ has the same area as $\triangle ABC$, and has the given angle $D$.
== Sources ==
* : Book $\text{I}$. Propositions
Category:Triangles
Category:Parallelograms
Construction of Parallelogram in Given Angle equal to Given Polygon

== Theorem ==
A parallelogram can be constructed in a given angle the same size as any given polygon.
== Proof ==
:
Let $ABCD$ be the given polygon, and let $E$ be the given angle.
Join $DB$, and construct the parallelogram $FGHK$ equal in size to $\triangle ABD$, in $\angle HKF = \angle E$.
Then construct the parallelogram $GLMH$ equal in area to $\triangle BCD$ on the line segment $GH$, in $\angle GHM = \angle E$.
We now need to show that $KFLM$ is the required parallelogram.
By , $\angle HKF = \angle GHM$ as both are equal to $\angle E$.
Add $\angle KHG$ to each, so as to make $\angle FKH + \angle KHG = \angle KHG + \angle GHM$.
From Parallelism implies Supplementary Interior Angles:
:$\angle FKH + \angle KHG$
Therefore $\angle KHG + \angle GHM$ equal two right angles.
So from Two Angles making Two Right Angles make Straight Line, $KH$ is in a straight line with $HM$.
From Parallelism implies Equal Alternate Angles:
:$\angle MHG = \angle HGF$
Add $\angle HGL$ to each, so as to make $\angle MHG + \angle HGL = \angle HGF + \angle HGL$.
From Parallelism implies Supplementary Interior Angles:
:$\angle MHG + \angle HGL$
Therefore $\angle HGF + \angle HGL$ equal two right angles.
So from Two Angles making Two Right Angles make Straight Line, $FG$ is in a straight line with $GL$.
From Parallelism is Transitive Relation, as $KF \parallel HG$ and $HG \parallel ML$, it follows that $KF \parallel ML$.
Similarly, from common notion 1, $KF = ML$.
As $KM$ and $FL$ join them at their endpoints, $KM \parallel FL$ and $KM = FL$ from Lines Joining Equal and Parallel Straight Lines are Parallel.
Therefore $KFLM$ is a parallelogram.
But the area of $KFLM$ equals the combined areas of $FGHK$ and $GLMH$, which are equal to the combined areas of $\triangle ABD$ and $\triangle BCD$.
Therefore from , $KFLM$ has the same area as the polygon $ABCD$, in the angle $E$
== Sources ==
* : Book $\text{I}$. Propositions
* : Chapter $\text I$: $\S 1$. Area of a Circle
Category:Parallelograms
Category:Polygons
Construction of Parallelogram on Given Line equal to Triangle in Given Angle

== Theorem ==
== Proof ==
:
Let $AB$ be the given line segment, $C$ be the given triangle and let $D$ be the given angle.
From Construction of Parallelogram equal to Triangle in Given Angle, construct the parallelogram $BEFG$ equal to $C$ such that $\angle EBG = \angle D$.
Place it so that $BE$ is in a straight line with $AB$.
Let $FG$ be produced to $H$, and draw $AH$ parallel to $BG$.
Then join $HB$.
We have that $HF$ falls on the parallel lines $AH$ and $EF$.
So from Parallelism implies Supplementary Interior Angles, $\angle AHF$ and $\angle HFE$ are supplementary.
So $\angle BHG + \angle GFE$ is less than two right angles.
From the Fifth Postulate, $HB$ and $FE$ will meet if produced. So let them meet at $K$.
Draw $KL$ parallel to $EA$, and produce $HA$ and $GB$ to the points $L$ and $M$.
Then $HLKF$ is a parallelogram.
From Complements of Parallelograms are Equal, $ABML$ has the same area as $BEFG$.
But $BEFG$ has the same area as $\triangle C$.
So by common notion 1, so does $ABML$.
From Two Straight Lines make Equal Opposite Angles, $\angle GBE = \angle ABM$, while $\angle GBE = \angle D$.
Therefore $ABML$ is the required parallelogram.
== Sources ==
* : Book $\text{I}$. Propositions
Category:Parallelograms
Category:Triangles
Construction of Part of Line

== Theorem ==
== Construction ==
:
Let $AB$ be the given straight line.
From $A$ draw $AC$ making any angle with $AB$.
On $AC$, take any point $D$ and make $AC$ the same multiple of $AD$ that $AB$ is to be of the required part which is to be cut from it.
Join $BC$ and draw $DE$ parallel to it.
Then $AE$ is the required part of $AB$.
== Proof ==
We have that $ED$ is parallel to one of the sides of $\triangle ABC$.
So from Parallel Transversal Theorem:
:$CD : DA = BE : EA$
From Magnitudes Proportional Separated are Proportional Compounded:
:$CA : AD = BA : AE$
From Proportional Magnitudes are Proportional Alternately:
:$CA : BA = AD : AE$
From Ratio Equals its Multiples:
:$AD : AE = n \cdot AD : n \cdot AE$
where $n$ is the number of times $AD$ is contained in $AC$.
From Equality of Ratios is Transitive:
:$AC : AB = n \cdot AD : n \cdot AE$
So from Relative Sizes of Components of Ratios it follows that:
:$AB = n \cdot AE$
== Sources ==
* : Book $\text{VI}$. Propositions
Category:Lines
Construction of Permutations

== Theorem ==
The ${}^n P_n$ permutations of $n$ objects can be generated algorithmically.
By Number of Permutations, that number is given by:
:${}^n P_n = n!$
where $n!$ denotes the factorial of $n$.
This will be demonstrated to hold.
== Proof 1 ==
== Proof 2 ==
Category:Construction of Permutations
Category:Permutations (Ordered Selections)
Category:Counting Arguments
Construction of Permutations/Proof 1

== Theorem ==
== Proof ==
The following is an inductive method of creating all the permutations of $n$ objects.
=== Base Case ===
There is clearly one way to arrange one object in order.
=== Inductive Hypothesis ===
We assume that we have constructed all $n!$ permutations of $n$ objects.
=== Induction Step ===
, let a set $S_n$ of $n$ objects be $\set {1, 2, \ldots, n}$.
Take a permutation of $S_n$:
:$a_1 \, a_2 \, a_3 \, \ldots \, a_n$
Now we take the number $n + 1$.
We can form $n + 1$ permutations from this one by putting $n + 1$ in all places possible:
:$a_{n + 1} \, a_1 \, a_2 \, a_3 \, \ldots \, a_n, \quad a_1 \, a_{n + 1} \, a_2 \, a_3 \, \ldots \, a_n, \quad a_1 \, a_2 \, a_{n + 1} \, a_3 \, \ldots \, a_n, \quad \ldots, \quad a_1 \, a_2 \, a_3 \, \ldots \, a_n \, a_{n + 1}$
It is clear that all permutations of $n + 1$ objects can be obtained in this manner, and no permutation is obtained more than once.
As there are ${}^n P_n$ permutations on $n$ objects, there are $\paren {n + 1} \times {}^n P_n$ permutations on $n + 1$ objects.
Hence by induction, and the recursive definition of the factorial:
:${}^n P_n = n!$
== Example ==
== Sources ==
* : $\S 1.2.5$: Permutations and Factorials
Category:Construction of Permutations
Construction of Permutations/Proof 1/Example

== Example of Construction of Permutations ==
From the permutation of $4$ objects $3 1 2 4$, the following permutations of $5$ objects can be made as follows:
:$5 3 1 2 4$
:$3 5 1 2 4$
:$3 1 5 2 4$
:$3 1 2 5 4$
:$3 1 2 4 5$
== Sources ==
* : $\S 1.2.5$: Permutations and Factorials: Exercise $3$
Category:Construction of Permutations
Construction of Permutations/Proof 2

== Theorem ==
== Proof ==
The following is an inductive method of creating all the permutations of $n$ objects.
=== Base Case ===
There is clearly one way to arrange one object in order.
=== Inductive Hypothesis ===
We assume that we have constructed all $n!$ permutations of $n$ objects.
=== Induction Step ===
, let a set $S_n$ of $n$ objects be $\left\{{1, 2, \ldots, n}\right\}$.
Take a permutation of $S_n$:
:$a_1 \, a_2 \, a_3 \, \ldots \, a_n$
Construct the array:
:$a_1 \, a_2 \, a_3 \, \ldots \, a_n \, \dfrac 1 2, \quad a_1 \, a_2 \, a_3 \, \ldots \, a_n \, \dfrac 3 2, \quad a_1 \, a_2 \, a_3 \, \ldots \, a_n \, \dfrac 5 2, \quad \ldots, \quad a_1 \, a_2 \, a_3 \, \ldots \, a_n \, \left({n + \dfrac 1 2}\right)$
This will contain $n + 1$ objects.
Now rename the elements of each permutation using the numbers $\left\{ {1, 2, \ldots, n}\right\}$ preserving order.
It is clear that all permutations of $n + 1$ objects can be obtained in this manner, and no permutation is obtained more than once.
As there are ${}^n P_n$ permutations on $n$ objects, there are $\left({n + 1}\right) \times {}^n P_n$ permutations on $n + 1$ objects.
Hence by induction, and the recursive definition of the factorial:
:${}^n P_n = n!$
== Example ==
== Sources ==
* : $\S 1.2.5$: Permutations and Factorials
Category:Construction of Permutations
Construction of Permutations/Proof 2/Example

== Example of Construction of Permutations ==
From the permutation of $4$ objects $3 1 2 4$, the following permutations of $5$ objects can be made as follows:
:$3 1 2 4 \dfrac 1 2$
:$3 1 2 4 \dfrac 3 2$
:$3 1 2 4 \dfrac 5 2$
:$3 1 2 4 \dfrac 7 2$
:$3 1 2 4 \dfrac 9 2$
Renaming according to size:
:$4 2 3 5 1$
:$4 1 3 5 2$
:$4 1 2 5 3$
:$3 1 2 5 4$
:$3 1 2 4 5$
== Sources ==
* : $\S 1.2.5$: Permutations and Factorials: Exercise $3$
Category:Construction of Permutations
Construction of Perpendicular Line

== Theorem ==
== Construction ==
== Proof ==
Since $DC = CE$ and $FC$ is common to both, and $DF = FE$, triangle $\triangle DCF$ equals triangle $\triangle ECF$.
Thus $\angle DCF = \angle ECF$.
So $CF$ is a straight line set up on a straight line making the adjacent angles equal to one another.
Thus it follows from that each of $\angle DCF$ and $\angle ECF$ are right angles.
So the straight line $CF$ has been drawn at right angles to the given straight line $AB$ from the given point $C$ on it.
== Sources ==
* : Book $\text{I}$. Propositions
Category:Construction of Perpendicular Line
Category:Perpendiculars
Category:Lines
Category:Angles
Construction of Perpendicular Line/Construction

== Construction of Perpendicular Line ==
== Construction ==
:
Let $AB$ be the given straight line segment, and let $C$ be the given point on it.
Let a point $D$ be taken on $AB$.
We cut off from $CB$ a length $CE$ equal to $DC$.
We construct an equilateral triangle $\triangle DEF$ on $DE$.
We draw the line segment $FC$.
Then $FC$ is the required perpendicular to $AB$.
== Sources ==
* : Book $\text{I}$. Propositions
Category:Construction of Perpendicular Line
Construction of Perpendicular using Rusty Compass

== Theorem ==
Let $AB$ be a line segment.
Using a straightedge and rusty compass, it is possible to construct a straight line at right angles to $AB$ from the endpoint $A$, without extending $AB$ past $A$.
== Construction ==
:
Mark $AC$ equal to the fixed radius of the rusty compass.
Construct two arcs whose centers are at $A$ and $C$ respectively of radius $AC$.
Let them meet at $D$.
Produce $CD$.
Measure $E$ from $D$ using the rusty compass so that $DE = CD$.
Then $AE$ is the required straight line at right angles to $AB$.
== Proof ==
As $DE = CD = DA$, the points $A$, $C$ and $E$ all lie on a circle of radius $AC$.
$CE$ is a straight line through the centers of circle $ACE$ and so is a diameter of circle $ACE$.
Hence by Thales' Theorem, $\angle CAE$ is a right angle
== Historical Note ==
== Sources ==
* 
* : Abul Wafa (): $43$
Category:Lines
Category:Angles
Category:Rusty Compass Constructions