avewright/proofwiki-math huggingface dataset:
Theorem: Angles on Equal Arcs are Equal

In equal circles, angles standing on equal arcs are equal to one another, whether at the center or at the circumference of those circles.
{{:Euclid:Proposition/III/27}}

Proof:
Let $ABC$ and $DEF$ be equal circles.
Let $\angle BGC$ and $\angle EHF$ stand on the equal arcs $BC$ and $EF$.
:500px
Suppose $\angle BGC \ne \angle EHF$.
Then one of them is bigger.
Suppose $\angle BGC > \angle EHF$.
On the straight line $BG$, construct $\angle BGK$ equal to $\angle EHF$.
From Equal Angles in Equal Circles, these angles stand on equal arcs.
So arcs $BK = EF$.
But $EF = BC$ and so $BK = BC$, which is impossible.
So $\angle BGC$ and $\angle EHF$ are not unequal, therefore $\angle BGC = \angle EHF$.
From the Inscribed Angle Theorem, $2 \angle BAC = \angle BGC$ and $2 \angle EDF = \angle EHF$.
Therefore $2 \angle BAC = \angle EHF$.
{{qed}}
{{Euclid Note|27|III|{{EuclidNoteConverse|prop=26|title=Equal Angles in Equal Circles}}}}
Theorem: Bisection of Arc

It is possible to bisect an arc of a circle.
{{:Euclid:Proposition/III/30}}

Proof:
Join $AD$ and $BD$.
We have that $AC = CB$ and $CD$ is common.
We also have that $\angle ACD = \angle BCD$ as they are both right angles.
So from Triangle Side-Angle-Side Congruence $\triangle ACD = \triangle BCD$ and so $AD = BD$.
But from Straight Lines Cut Off Equal Arcs in Equal Circles, the arc $AD$ equals the arc $BD$.
That is, the arc $AB$ has been bisected at $D$.
{{qed}}
{{Euclid Note|30|III}}
Theorem: Relative Sizes of Angles in Segments

In a circle:
:the angle in a semicircle is right
:the angle in a segment greater than a semicircle is acute
:the angle in a segment less than a semicircle is obtuse.
Further:
:the angle of a segment greater than a semicircle is obtuse
:the angle of a segment less than a semicircle is acute.
{{:Euclid:Proposition/III/31}}

Proof:
:300px
Let $ABCD$ be a circle whose diameter is $BC$ and whose center is $E$.
Join $AB$, $AC$, $AD$, $DC$ and $AE$.
Let $BA$ be produced to $F$.
Since $BE = EA$, from Isosceles Triangle has Two Equal Angles it follows that $\angle ABE = \angle BAE$.
Since $CE = EA$, from Isosceles Triangle has Two Equal Angles it follows that $\angle ACE = \angle CAE$.
So from $\angle BAC = \angle ABE + \angle ACE = \angle ABC + \angle ACB$.
But from Sum of Angles of Triangle Equals Two Right Angles $\angle FAC = \angle ABC + \angle ACB$.
So $\angle BAC = \angle FAC$, and so from Book I Definition 10 each one is a right angle.
So the angle in the semicircle $BAC$ is a right angle.
{{qed|lemma}}
From Two Angles of Triangle are Less than Two Right Angles, in $\triangle ABC$, $\angle ABC + \angle BAC$ is less than two right angles.
As $\angle BAC$ is a right angle, it follows that $\angle ABC$ is less than a right angle.
It is also the angle in a segment $ABC$ greater than a semicircle.
Therefore the angle in a segment greater than a semicircle is acute.
{{qed|lemma}}
We have that $ABCD$ is a cyclic quadrilateral.
From Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles, $\angle ABC + \angle ADC$ equals two right angles.
As $\angle ABC$ is less than a right angle, it follows that $\angle ADC$ is greater than a right angle.
But $\angle ADC$ is the angle in a segment $ADC$ less than a semicircle.
Therefore the angle in a segment less than a semicircle is obtuse.
{{qed|lemma}}
Since the angle contained by the straight lines $BA$ and $AC$ is a right angle, the angle contained by the arc $ABC$ and the straight line $AC$ is greater than a right angle.
So the angle of a segment greater than a semicircle is obtuse.
{{qed|lemma}}
Since the angle contained by the straight lines $AC$ and $AF$ is a right angle, the angle contained by the arc $ADC$ and the straight line $AC$ is less than a right angle.
So the angle of a segment less than a semicircle is acute.
{{qed}}
{{Euclid Note|31|III}}
Theorem: Tangent-Chord Theorem

Let $EF$ be a tangent to a circle $ABCD$, touching it at $B$.
Let $BD$ be a chord of $ABCD$.
Then:
:the angle in segment $BCD$ equals $\angle DBE$
and: 
:the angle in segment $BAD$ equals $\angle DBF$.

Proof:
:280px
Draw $BA$ perpendicular to $EF$ through $B$.
Let $C$ be selected on the circle on the arc $BD$.
Join $AD, DC, CB$.
From Right Angle to Tangent of Circle goes through Center, the center of the circle lies on $AB$.
By definition, then, $AB$ is a diameter of the circle.
From Relative Sizes of Angles in Segments, it follows that $\angle ADB$ is a right angle.
Therefore from Sum of Angles of Triangle Equals Two Right Angles $\angle BAD + \angle ABD$ equals a right angle.
But $\angle ABF$ is also a right angle.
So $\angle ABF = \angle BAD + \angle ABD$
Subtracting $\angle ABD$ from each, it follows that $\angle DBF = \angle BAD$.
{{qed|lemma}}
Next we have that $ABCD$ is a cyclic quadrilateral.
From Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles $\angle BAD + \angle BCD$ equals two right angles.
But from Two Angles on Straight Line make Two Right Angles, so does $\angle DBE + \angle DBF$.
But as $\angle BAD = \angle DBF$ it follows that $\angle BCD = \angle DBE$.
{{qed}}
Theorem: Construction of Segment on Given Line Admitting Given Angle

On any given line segment, it is possible to describe a segment of a circle which admits an angle equal to any given rectilineal angle.
{{:Euclid:Proposition/III/33}}

Proof:
=== Proof for Acute Angle ===
We have that $AF = FB$, $\angle AFG = \angle BFG$ (both are right angles) and $AG$ is common.
So from Triangle Side-Angle-Side Congruence it follows that $AG = BG$.
So the circle $ABE$ whose center is at $G$ and whose radius is $GB$ also passes through $A$.
As $AE$ passes through the  center $G$, $AE$ is a diameter of the circle $ABE$.
As $AD$ is perpendicular to $AE$, it follows from Line at Right Angles to Diameter of Circle that $AD$ is tangent to circle $ABE$.
From the Tangent-Chord Theorem, $\angle DAB = \angle AEB$.
But $\angle DAB = \angle C$ by construction.
Hence the result.
{{qed|lemma}}
Theorem: Tangent Secant Theorem

Let $D$ be a point outside a circle $ABC$.
Let $DB$ be tangent to the circle $ABC$.
Let $DA$ be a straight line which cuts the circle $ABC$ at $A$ and $C$.
Then $DB^2 = AD \cdot DC$.
{{:Euclid:Proposition/III/36}}

Proof:
Let $DA$ pass through the center $F$ of circle $ABC$.
Join $FB$.
From Radius at Right Angle to Tangent, $\angle FBD$ is a right angle.
:300px
We have that $F$ bisects $AC$ and that $CD$ is added to it.
So we can apply Square of Sum less Square and see that:
:$AD \cdot DC + FC^2 = FD^2$
But $FC = FB$ and so:
:$AD \cdot DC + FB^2 = FD^2$
But from Pythagoras's Theorem we have that $FD^2 = FB^2 + DB^2$ and so:
:$AD \cdot DC + FB^2 = FB^2 + DB^2$
from which it follows that:
:$AD \cdot DC = DB^2$
which is what we wanted to show.
{{qed|lemma}}
Now let $DA$ be such that it does not pass through the center $E$ of circle $ABC$.
Draw $EF$ perpendicular to $DA$ and draw $EB, EC, ED$.
:360px
From Radius at Right Angle to Tangent, $\angle EBD$ is a right angle.
From Conditions for Diameter to be Perpendicular Bisector, $EF$ bisects $AC$.
So $AF = FC$.
So we can apply Square of Sum less Square and see that:
:$AD \cdot DC + FC^2 = FD^2$
Let $FE^2$ be added to each:
:$AD \cdot DC + FC^2 + FE^2 = FD^2 + FE^2$
Now $\angle DFE$ is a right angle and so by Pythagoras's Theorem we have:
:$FD^2 + FE^2 = ED^2$
:$FC^2 + FE^2 = EC^2$
This gives us:
:$AD \cdot DC + EC^2 = ED^2$
But $EC = EB$ as both are radii of the circle $ABC$.
Next note that $\angle EBD$ is a right angle and so by Pythagoras's Theorem we have:
:$ED^2 = EB^2 + DB^2$
which gives us:
:$AD \cdot DC + EB^2 = EB^2 + DB^2$
from which it follows that:
:$AD \cdot DC = DB^2$
which is what we wanted to show.
{{qed}}
Theorem: Identity Element of Natural Number Multiplication is One

Let $1$ be the element one of $\N$.
Then $1$ is the identity element of multiplication:
:$\forall n \in \N: n \times 1 = n = 1 \times n$

Proof:
Firstly, by definition of multiplication:
{{begin-eqn}}
{{eqn | l = n \times 1
      | r = \paren {n \times 0} + n
}}
{{eqn | r = n
}}
{{end-eqn}}
Next, recall that multiplication is recursively defined as:
:$\forall m, n \in \N: \begin{cases}
m \times 0 & = 0 \\
m \times \paren {n + 1} & = m \times n + m
\end{cases}$
From the Principle of Recursive Definition, there is only one mapping $f$ satisfying this definition for $m = 1$; that is, such that:
:$\forall n \in \N: \begin {cases}
\map f 0 = 0 \\
\map f {n + 1} = \map f n + 1
\end{cases}$
Consider now $f'$ defined as $\map {f'} n = n$.
Then evidently $\map {f'} 0 = 0$.
Also:
{{begin-eqn}}
{{eqn | l = \map {f'} {n + 1}
      | r = n + 1
}}
{{eqn | r = \map {f'} n + 1
}}
{{end-eqn}}
showing that $f'$ also satisfies the definition for $1 \times n$.
Hence $n \times 1 = 1 \times n = n$ for all $n \in \N$.
That is, $1$ is the identity element for multiplication.
{{qed}}
Theorem: Secant Secant Theorem

Let $C$ be a point external to a circle $ABED$.
Let $CA$ and $CB$ be straight lines which cut the circle at $D$ and $E$ respectively.
Then:
:$CA \cdot CD = CB \cdot CE$

Proof:
:320px
Draw $CF$ tangent to the circle.
From the Tangent Secant Theorem we have that:
:$CF^2 = CA \cdot CD$
:$CF^2 = CB \cdot CE$
from which the result is obvious and immediate.
{{qed}}
Theorem: Converse of Tangent Secant Theorem

Let $D$ be a point outside a circle $ABC$.
Let $DA$ be a straight line which cuts the circle $ABC$ at $A$ and $C$.
Let $DB$ intersect the circle at $B$ such that $DB^2 = AD \cdot DC$.
Then $DB$ is tangent to the circle $ABC$.
{{:Euclid:Proposition/III/37}}

Proof:
:320px
Let $DE$ be drawn tangent to the circle $ABC$.
Let $F$ be the center of $ABC$ and join $FB, FD, FE$.
From Radius at Right Angle to Tangent, $\angle FED$ is a right angle.
We have that $DE$ is tangent to the circle $ABC$ and $DA$ cuts it.
So from the Tangent Secant Theorem $AD \cdot DC = DE^2$.
But we also have {{hypothesis}} that $AD \cdot DC = DB^2$.
So $DE^2 = DB^2$ and so $DE = DB$.
Also $FE = FB$ and so $DE, EF = DB, DF$ and $DF$ is common.
So from Triangle Side-Side-Side Congruence $\angle DEF = \angle DBF$.
But $\angle DEF$ is a right angle and so also is $\angle DBF$.
So from Line at Right Angles to Diameter of Circle $DB$ is tangent to the circle $ABC$.
{{qed}}
{{Euclid Note|37|III|{{EuclidNoteConverse|prop = 36|title = Tangent Secant Theorem}}}}
Theorem: Fitting Chord Into Circle

Into a given circle, it is possible to fit a chord equal to a given line segment which is not greater than the diameter of the circle.
{{:Euclid:Proposition/IV/1}}

Proof:
As $C$ is the center of $EFG$, $CG = CE$ and so $CG = D$.
But $CG$ is a chord of $ABC$.
{{Qed}}
{{Euclid Note|1|IV}}
Theorem: Construction of Isosceles Triangle whose Base Angle is Twice Apex

It is possible to construct an isosceles triangle such that each of the angles at the base is twice that at the apex.
{{:Euclid:Proposition/IV/10}}

Proof:
Join $CD$.
Circumscribe circle $ACD$ about $\triangle ACD$.
As $AC = BD$ we have that $AB \cdot BC = BD^2$.
We have that $B$ is outside the circle $ACD$.
From the converse of the Tangent Secant Theorem it follows that $BD$ is tangent to circle $ACD$.
Then from the Tangent-Chord Theorem:
:$\angle BDC = \angle DAC$
Add $\angle CDA$ to both:
:$\angle CDA + \angle BDC = \angle BDA = \angle CDA + \angle DAC$.
But from Sum of Angles of Triangle Equals Two Right Angles we have that:
:$(1) \quad \angle BCD = \angle CDA + \angle DAC$
So $\angle BDA = \angle BCD$.
But since $AD = AB$, from Isosceles Triangle has Two Equal Angles $\angle BDA = \angle CBD$.
So $\angle BDA = \angle BCD = \angle CBD$.
Since $\angle DBC = \angle BCD$, from Triangle with Two Equal Angles is Isosceles we have $BD = DC$.
But by hypothesis $BD = CA$ and so $CA = CD$.
So from Isosceles Triangle has Two Equal Angles $\angle CDA = \angle DAC$.
So $\angle CDA + \angle DAC = 2 \angle DAC$.
But from $(1)$ we have that $\angle BCD = \angle CDA + \angle DAC$.
So $\angle BCD = 2 \angle CAD = 2 \angle BAD$.
But $\angle BCD = \angle BDA = \angle DBA$.
So $\angle ABD = \angle BAD = 2 \angle BDA$.
{{qed}}
{{Euclid Note|10|IV|Having established in the proof that $CD$ equals $BD$, the construction can be simplified by constructing the circle whose center is at $C$ and whose radius is $AC$, then identifying $D$ as the point at which circle $ACD$ meets circle $ABD$, instead of invoking the somewhat more cumbersome construction that fits $BD$ into the circle $ABD$.}}
Theorem: Inscribing Regular Pentagon in Circle

In a given circle, it is possible to inscribe a regular pentagon.
{{:Euclid:Proposition/IV/11}}

Proof:
We have that $\angle CDA = \angle DCA = 2 \angle CAD$.
As $\angle CDA$ and $ \angle DCA$ have been bisected, $\angle DAC = \angle ACE = \angle ECD = \angle CDB = \angle BDA$.
From Equal Angles in Equal Circles it follows that the arcs $AB, BC, CD, DE, EA$ are all equal.
Hence from Equal Arcs of Circles Subtended by Equal Straight Lines, the straight lines $AB, BC, CD, DE, EA$ are all equal.
So the pentagon $ABCDE$ is equilateral.
Now since the arc $AB$ equals arc $DE$, we can add $BCD$ to each.
So the arc $ABCD$ equals arc $BCDE$.
So from Angles on Equal Arcs are Equal $\angle BAE = \angle AED$.
For the same reason, $\angle BAE = \angle AED = \angle ABC = \angle BCD = \angle CDE$.
So the pentagon $ABCDE$ is equiangular.
{{qed}}
{{Euclid Note|11|IV}}
Theorem: Circumscribing Regular Pentagon about Circle

About a given circle, it is possible to circumscribe a regular pentagon.
{{:Euclid:Proposition/IV/12}}

Proof:
Join $FB, FK, FC, FL, FD$.
From Radius at Right Angle to Tangent, $HK, KL$ etc. are perpendicular to the radii they touch.
So $\angle KCF$ and $\angle LCF$ are right angles.
For the same reason, $\angle KBF$ and $\angle LDF$ are right angles.
By Pythagoras's Theorem, $FK^2 = FC^2 + CK^2$
For the same reason, $FK^2 = FB^2 + BK^2$.
As $FB = FC$ it follows that $BK = CK$.
From Triangle Side-Side-Side Congruence it follows that $\triangle FCK = \triangle FBK$ and so $\angle FKC = \angle FKB$ and $\angle KFC = \angle KFB$.
So $\angle BFC = 2 \angle KFC$ and $\angle BKC = 2 \angle FKC$.
For the same reason $\angle CFD = 2 \angle CFL$ and $\angle DLC = 2 \angle FLC$.
We have that the arc $BC$ equals the arc $CD$.
So from Angles on Equal Arcs are Equal, $\angle BFC = \angle CFD$.
As $\angle BFC = 2 \angle KFC$ and $\angle CFD = 2 \angle CFL$, it follows that $\angle KFC = \angle CFL$.
But $\angle FCK = \angle FCL$.
So from Triangle Angle-Side-Angle Congruence:
:$\triangle FCK = \triangle FLC$
So $KC = CL$ and $\angle FKC = \angle FLC$.
Since $KC = CL$ it follows that $KL = 2 KC$.
For the same reason $HK = 2 BK$ and $BK = KC$.
So $HK = KL$.
Similarly each of the straight lines $HG, GM, ML$ are all equal to $HK$ and $KL$.
So the pentagon $GHKLM$ is equilateral.
We have that:
:$\angle FKC = \angle FLC$
:$\angle HKL = 2 \angle FKC$
:$\angle KLM = 2 \angle FLC$
So:
:$\angle HKL = \angle KLM$
Similarly each of $\angle KHG, \angle HGM, \angle GML$ are equal to $\angle HKL$ and $\angle KLM$.
So the pentagon $GHKLM$ is equiangular.
{{qed}}
{{Euclid Note|12|IV}}
Theorem: Inscribing Circle in Regular Pentagon

In any given regular pentagon it is possible to inscribe a circle.
{{:Euclid:Proposition/IV/13}}

Proof:
Join the straight lines $FB, FA, FE$.
We have that $BC = CD$, $CF$ is common and $\angle BCF = \angle DCF$.
So by Triangle Side-Angle-Side Congruence:
:$\triangle BCF = \triangle DCF$
and so:
:$BF = DF$
Thus $\angle CBF = \angle CDF$.
Since $\angle CDE = 2 \angle CDF$ and $\angle CDE = \angle CBF$, then $\angle CDF = \angle CBF$.
So $\angle ABF = \angle FBC$ and so $\angle ABC$ has been bisected by the straight line $BF$.
Similarly it can be shown that $\angle BAE, \angle AED$ have been bisected by the straight lines $FA, FE$ respectively.
Now join $FH, FK, FL$ from $F$ perpendicular to $BC, CD, DE$.
We have that:
:$\angle HCF = \angle KCF$
:$\angle FHC = \angle FKC$ (both are right angles)
:$FC$ is common and subtends one of the equal angles.
So from Triangle Side-Angle-Angle Congruence:
:$\triangle FHC = \triangle FKC$
and so:
:$FH = FK$
Similarly it is shown that $FL = FM = FG = FH = FK$.
Therefore the circle whose center is $F$ and radius is $FM$ (or $FG$) passes through all of the points $G, H, K, L, M$.
We have that the angles at each of those points is a right angle.
So from Line at Right Angles to Diameter of Circle, the circle $GHKLM$ is tangent to each of lines $AB, BC, CD, DE, EA$.
Hence the result.
{{qed}}
{{Euclid Note|13|IV}}
Theorem: Circumscribing Circle about Regular Pentagon

About any given regular pentagon it is possible to circumscribe a circle.
{{:Euclid:Proposition/IV/14}}

Proof:
From $F$ join the straight lines $FB, FA, FE$.
In a similar manner to Inscribing Circle in Regular Pentagon it can be shown that $\angle CBA, \angle BAE, \angle AED$ have been bisected by the straight lines $FB, FA, FE$ respectively.
Since $\angle BCD = CDE$ and $2 \angle FCD = \angle BCD$ and $2 \angle CDF = \angle CDE$, it follows that $\angle FCD = \angle CDF$.
From Triangle with Two Equal Angles is Isosceles it follows that $FC = FD$.
Similarly it can be shown that $FB = FA = FE = FC = FD$.
Therefore the circle $ABCDE$ circumscribes the regular pentagon $ABCDE$ as required.
{{Qed}}
{{Euclid Note|14|IV}}
Theorem: Inscribing Regular Hexagon in Circle

In a given circle, it is possible to inscribe a regular hexagon.
{{:Euclid:Proposition/IV/15}}

Proof:
Since $G$ is the center of circle $ABCDEF$, it follows that $GE = GD$.
Since $D$ is the center of circle $EGCH$, it follows that $DE = GD$.
So $GE = GD = DE$ and so $\triangle EGD$ is equilateral and so equiangular.
By Sum of Angles of Triangle Equals Two Right Angles, $\angle EGD$ is one third of two right angles.
Similarly for $\angle DGC$.
Since the straight line $CG$ on $EB$ makes $\angle EGC + \angle CGB$ equal to two right angles, $\angle CGB$ is also equal to one third of two right angles.
So $\angle EGD = \angle DGC = \angle CGB$.
By the Vertical Angle Theorem, $\angle BGA = \angle AGF = \angle FGE = \angle EGD = \angle DGC = \angle CGB$.
From Equal Angles in Equal Circles, the six arcs $AB, BC, CD, DE, EF, FA$ are all equal.
So the six straight lines $AB, BC, CD, DE, EF, FA$ are all equal.
So the hexagon $ABCDEF$ is equilateral.
Now since the arc $FA$ equals the arc $ED$, let the arc $ABCD$ be added to each.
So arc $FABCD$ equals arc $ABCDE$.
Now $\angle FED$ stands on arc $FABCD$ and $\angle AFE$ stands on arc $ABCDE$.
So by Angles on Equal Arcs are Equal $\angle FED = \angle AFE$.
In the same way we show that all the angles around the hexagon $ABCDEF$ are equal.
Therefore $ABCDEF$ is a regular hexagon.
{{qed}}
{{Euclid Note|15|IV}}
Theorem: Inscribing Regular 15-gon in Circle

In a given circle, it is possible to inscribe a regular 15-gon.
{{:Euclid:Proposition/IV/16}}

Proof:
Consider the circumference of circle $ABCD$ as divided into $15$ equal arcs.
Of these, there will be $5$ in the shorter arc $AC$, and $3$ in the shorter arc $AB$.
So there are $2$ in the shorter arc $BC$.
Once this has been bisected by the construction which produces $E$, we see that each of these parts is a copy of these $15$ equal arcs.
Then $BE$ and $EC$ are $\dfrac 1 {15}$ of the length of the circumference of $ABCD$.
Hence the result.
{{qed}}
{{Euclid Note|16|IV}}
Theorem: Multiplication of Numbers is Left Distributive over Addition

{{:Euclid:Proposition/V/1}}
That is, if $m a$, $m b$, $m c$ etc. be any equimultiples of $a$, $b$, $c$ etc., then:
:$m a + m b + m c + \cdots = m \paren {a + b + c + \cdots }$

Proof:
Let any number of magnitudes whatever $AB, CD$ be respectively equimultiples of any magnitudes $E, F$ equal in multitude.
Then we are to show that whatever multiple $AB$ is of $E$, then that multiple will $AB + CD$ be of $E + F$.
:520px
Since $AB$ is the same multiple of $E$ that $CD$ is of $F$, as many magnitudes as there are in $AB$ equal to $E$, so many also are there in $CD$ equal to $F$.
Let $AB$ be divided into the magnitudes $AG, GB$ equal to $E$, and $CH, HD$ equal to $F$.
Then the multitude of the magnitudes $AG, GB$ will be equal to the multitude of the magnitudes $CH, HD$.
Since $AG = E$ and $CH = F$ it follows that $AG = E$ and $AG + CH = E + F$.
For the same reason, $GB = E$ and $GB + HD = E + F$.
Therefore, as many magnitudes as there are in $AB$ equal to $E$, so many also are there in $AB + CD$ equal to $E + F$.
{{qed}}
Theorem: Multiplication of Numbers is Right Distributive over Addition

{{:Euclid:Proposition/V/2}}
That is:
:$m a + n a + p a + \cdots = \paren {m + n + p + \cdots} a$

Proof:
Let a first magnitude, $AB$, be the same multiple of a second, $C$, that a third, $DE$, is of a fourth, $F$.
Let a fifth, $BG$, be the same multiple of $C$ that a sixth, $EH$, is of $F$.
:380px
We need to show that $AG = AB + BG$ is the same multiple of $C$ that $DH = DE + EH$ is of $F$.
We have that $AB$ is the same multiple of $C$ that $DE$ is of $F$.
It follows that as many magnitudes as there are in $AB$ equal to $C$, so many also are there in $DE$ equal to $F$.
For the same reason, as many as there are in $BG$ equal to $C$, so many also are there in $EH$ equal to $F$.
So as many as there are in the whole $AG$ equal to $C$, so many also are there in the whole $DH$ equal to $F$.
Therefore the sum of the first and fifth, $AG$, is the same multiple of the second, $C$, that the sum of the third and sixth, $DH$ is of the fourth, $F$.
{{qed}}
Theorem: Multiples of Terms in Equal Ratios

Let $a, b, c, d$ be quantities.
Let $a : b = c : d$ where $a : b$ denotes the ratio between $a$ and $b$.
Then for any numbers $m$ and $n$:
:$m a : n b = m c : n d$
{{:Euclid:Proposition/V/4}}

Proof:
Let a first magnitude $A$ have to a second magnitude $B$ the same ratio as a third $C$ to a fourth $D$.
Let equimultiples $E, F$ be taken of $A, C$, and let different equimultiples $G, H$ be taken of $B, D$.
We need to show that $E : G = F : H$.
:600px
Let equimultiples $K, L$ be taken of $E, F$ and other arbitrary equimultiples $M, N$ be taken of $G, H$.
We have that $E$ is the same multiple of $A$ that $L$ is of $C$.
So from Proposition $3$: Associative Law of Multiplication, $K$ is the same multiple of $A$ that $L$ is of $C$.
For the same reason, $M$ is the same multiple of $B$ that $N$ is of $D$.
We have that:
:$A$ is to $B$ as $C$ is to $D$
:of $A, C$ equimultiples $K, L$ have been taken
:of $B, D$ other equimultiples $M, N$ have been taken.
So from the definition of equality of ratios:
:if $K$ is in excess of $M$, $L$ is also in excess of $N$
:if $K$ is equal to $M$, $L$ is equal to $N$
:if $K$ is less than $M$, $L$ is less than $N$
But $K, L$ are equimultiples of $E, F$.
Therefore as $E$ is to $G$, so is $F$ to $H$.
{{qed}}
Theorem: Multiplication of Real Numbers is Left Distributive over Subtraction

{{:Euclid:Proposition/V/5}}
That is, for any numbers $a, b$ and for any integer $m$:
:$m a - m b = m \paren {a - b}$

Proof:
Let the magnitude $AB$ be the same multiple of the magnitude $CD$ that the part $AE$ subtracted is of the part $CF$ subtracted.
We need to show that the remainder $EB$ is also the same multiple of the remainder $FD$ that the whole $AB$ is of the whole $CD$.
:320px
Whatever multiple $AE$ is of $CF$, let $EB$ be made that multiple of $CG$.
We have that $AE$ is the same multiple of $CF$ that $AB$ is of $GC$.
So from Distributive Property, $AE$ is the same multiple of $CF$ that $AB$ is of $GF$.
By by assumption, $AE$ is the same multiple of $CF$ that $AB$ is of $CD$.
Therefore $AB$ is the same multiple of each of the magnitudes $GF, CD$.
Therefore $GF = CD$.
Let $CF$ be subtracted from each.
Then the remainder $GC$ is equal to the remainder $FD$.
Since:
: $AE$ is the same multiple of $CF$ that $EB$ is of $GC$
: $GC = DF$
it follows that $AE$ is the same multiple of $CF$ that $EB$ is of $CD$.
That is, the remainder $EB$ will be the same multiple of the remainder $FD$ that the whole $AB$ is of the whole $CD$.
{{qed}}