avewright/proofwiki-math huggingface dataset:
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:
{{begin-eqn}}
{{eqn | l = m \times \paren {a - b}
      | r = m \times \paren {a + \paren {- b} }
      | c = {{Defof|Real Subtraction}}
}}
{{eqn | r = m \times a + m \times \paren {- b}
      | c = {{Real-number-axiom|D}}
}}
{{eqn | r = m \times a + \paren {- m \times b}
      | c = Multiplication by Negative Real Number
}}
{{eqn | r = m \times a - m \times b
      | c = {{Defof|Real Subtraction}}
}}
{{end-eqn}}
{{qed}}
Theorem: Ratios of Equal Magnitudes

{{:Euclid:Proposition/V/7}}
That is:
:$a = b \implies a : c = b : c$
:$a = b \implies c : a = c : b$

Proof:
Let $A, B$ be equal magnitudes and let $C$ be any other arbitrary magnitude.
We need to show that $A : C = B : C$ and $C : A = C : B$.
:400px
Let equimultiples $D, E$ of $A, B$ be taken, and another arbitrary multiple $F$ of $C$.
We have that $D$ is the same multiple of $A$ that $E$ is of $B$, while $A = B$.
Therefore $D = E$.
But $F$ is another arbitrary magnitude.
Therefore:
:$D > F \implies E > F$
:$D = F \implies E = F$
:$D < F \implies E < F$
We have that $D, E$ are equimultiples of $A, B$ while $F$ is another arbitrary multiple of $C$.
So from {{EuclidDefLink|V|5|Equality of Ratios}}, $A : C = B : C$.
With the same construction we can show that $D = E$, while $F$ is some other magnitude.
Therefore:
:$F > D \implies F > E$
:$F = D \implies F = E$
:$F < D \implies F < E$
But $F$ is a multiple of $C$, while $D, E$ are equimultiples of $A, B$.
So from {{EuclidDefLink|V|5|Equality of Ratios}}, $C : A = C : B$.
{{qed}}
{{Euclid Note|7|V|{{EuclidNoteConverse|book = V|prop = 9|title = Magnitudes with Same Ratios are Equal}}}}
Theorem: Relative Sizes of Ratios on Unequal Magnitudes

{{:Euclid:Proposition/V/8}}
That is:
:$a > b \implies a : c > b : c$
:$a > b \implies c : a < c : b$

Proof:
Let $AB, C$ be unequal magnitudes, and let $AB$ be greater.
Let $D$ be another arbitrary magnitude.
We are to show that $AB$ has to $D$ a greater ratio than $C$ has to $D$, and $D$ has to $C$ a greater ratio than it has to $AB$.
We have that $AB > C$, so let $BE = C$.
Then from {{EuclidDefLink|V|4|Existence of Ratio}}, the lesser of the magnitudes $AE, EB$, if multiplied, will eventually be greater than $D$.
There are two cases.
;First, let $AE < EB$.
:410px
Let $AE$ be multiplied by some number, and let $FG$ be a multiple of it which is greater than $D$.
Then whatever multiple $FG$ is of $AE$, let $GH$ be made the same multiple of $EB$ and $K$ of $C$.
Let $L := 2D, M := 3D, N := 4D \ldots$ until one of these multiples is greater than $K$.
Suppose $M \le K$ while $N > K$.
WE have that $FG$ is the same multiple of $AE$ that $GH$ is of $EB$.
So from Multiplication of Numbers is Left Distributive over Addition, $FG$ is the same multiple of $AE$ that $FHK$ is of $AB$.
But $FG$ is the same multiple of $AE$ that $K$ is of $C$.
Therefore $FH, K$ are equimultiples of $AB, C$.
Again, we have that $GH$ is the same multiple of $EB$ that $K$ is of $C$, and $EB = C$.
Therefore $GH = K$.
But $K \ge M$ and so $GH \ge M$.
Also, $FG > D$ so $FH > D + M$.
But $D + M = N$ by the construction of $M$ and $N$, and $FH > M + D$.
So $FH > N$ while $K \le N$.
Also, $FH, K$ are equimultiples of $AB, C$ while $N$ is a multiple of some arbitrary $D$.
So from Ratios of Equal Magnitudes $AB$ has a greater ratio to $D$ than $C$ has to $D$.
Next, note that with the same construction, we can show similarly that $N > K$ while $N \le FH$.
Also we have that $N$ is a multiple of $D$, while $FH, K$ are other equimultiples of $AB, C$.
Therefore $D$ has a greater ratio to $C$ than $D$ has to $AB$.
{{qed|lemma}}
;Second, let $AE > EB$.
From {{EuclidDefLink|V|4|Existence of Ratio}}, $EB$, if multiplied, will eventually be greater than $D$.
:410px
Let it be so multiplied, and let $GH$ be a multiple of $EB$ and greater than $D$.
Whatever multiple $GH$ is of $EB$, let $FG$ be made the same multiple of $AE$, and $K$ of $C$.
Then we can prove similarly that $FH, K$ are equimultiples of $AB, C$.
Similarly, let $M, N$ be consecutive multiples of $D$ such that $M \le FG$ and $N > FG$.
But $GH > D$, therefore $FH > D + M$, that is, $FH > N$.
Now $K \le N$ inasmuch as $FG$ also, which is greater than $GH$, that is, than $K$, is not in excess of $N$.
Hence the result.
{{qed}}
{{Euclid Note|8|V|{{EuclidNoteConverse|book=V|prop=10|title=Relative Sizes of Magnitudes on Unequal Ratios}}}}
Theorem: Magnitudes with Same Ratios are Equal

{{:Euclid:Proposition/V/9}}
That is:
:$a : c = b : c \implies a = b$
:$c : a = c : b \implies a = b$

Proof:
:350px
Let each of $A$ and $B$ have the same ratio to $C$, i.e. $A : C = B : C$.
Suppose $A \ne B$.
Then from Relative Sizes of Ratios on Unequal Magnitudes $A : C \ne B : C$.
But $A : C = B : C$ so therefore it is not the case that $A \ne B$.
Therefore $A = B$.
Again, let $C$ have the same ratio to each of $A$ and $B$, i.e. $C : A = C : B$.
Suppose $A \ne B$.
Then from Relative Sizes of Ratios on Unequal Magnitudes $C : A \ne C : B$.
But $C : A = C : B$ so therefore it is not the case that $A \ne B$.
Therefore $A = B$.
{{Qed}}
{{Euclid Note|9|V|{{EuclidNoteConverse|book=V|prop=7|title=Ratios of Equal Magnitudes}}}}
Theorem: Relative Sizes of Magnitudes on Unequal Ratios

{{:Euclid:Proposition/V/10}}
That is:
:$a : c > b : c \implies a > b$
:$c : b > c : a \implies b < a$

Proof:
Let $A$ have to $C$ a greater ratio than $B$ has to $C$.
:350px
Suppose $A = B$.
Then from Ratios of Equal Magnitudes $A : C = B : C$.
But by hypothesis $A : C > B : C$, so $A \ne B$.
Suppose $A < B$.
Then from Relative Sizes of Ratios on Unequal Magnitudes it would follow that $A : C < B : C$.
But by hypothesis $A : C > B : C$.
Therefore it must be that $A > B$.
{{qed|lemma}}
Let $C$ have to $B$ a greater ratio than $C$ has to $A$.
Suppose $B = A$.
Then from Ratios of Equal Magnitudes $C : B = C : A$.
But by hypothesis $C : B > C : A$, so $B \ne A$.
Suppose $B > A$.
Then from Relative Sizes of Ratios on Unequal Magnitudes it would follow that $C : B < C : A$.
But by hypothesis $C : B > C : A$.
Therefore it must be that $B < A$.
{{qed}}
{{Euclid Note|10|V|{{EuclidNoteConverse|book=5|prop=8|title=Relative Sizes of Ratios on Unequal Magnitudes}}}}
Theorem: Equality of Ratios is Transitive

{{:Euclid:Proposition/V/11}}
That is:
:$A : B = C : D, C : D = E : F \implies A : B = E : F$

Proof:
Let $A : B = C : D$ and $C : D = E : F$.
:560px
Of $A, C, E$ let equimultiples $G, H, K$ be taken.
Of $B, D, F$ let other arbitrary equimultiples $L, M, N$ be taken.
We have that:
:$A : B = C : D$
:$G, H$ are equimultiples of $A, C$
:$L, M$ are equimultiples of $B, D$
So:
:$G > L \implies H > M$
:$G = L \implies H = M$
:$G < L \implies H < M$
Also, we have that:
:$C : D = E : F$
:$H, K$ are equimultiples of $C, E$
:$M, N$ are equimultiples of $D, F$
So:
:$H > M \implies K > N$
:$H = M \implies K = N$
:$H < M \implies K < N$
Also, we have that:
:$G, K$ are equimultiples of $A, E$
:$L, N$ are equimultiples of $B, F$
Therefore $A : B = E : F$.
{{qed}}
{{Euclid Note|11|V}}
Theorem: Sum of Components of Equal Ratios

{{:Euclid:Proposition/V/12}}
That is:
:$a_1 : b_1 = a_2 : b_2 = a_3 : b_3 = \cdots \implies \left({a_1 + a_2 + a_3 + \cdots}\right) : \left({b_1 + b_2 + b_3 + \cdots}\right)$

Proof:
Let any number of magnitudes $A, B, C, D, E, F$ be proportional, so that:
:$A : B = C : D = E : F$
etc.
:450px
Of $A, C, E$ let equimultiples $G, H, K$ be taken, and of $B, D, F$ let other arbitrary equimultiples $L, M, N$ be taken.
We have that $A : B = C : D = E : F$.
Therefore:
:$G > L \implies H > M, K > N$
:$G = L \implies H = M, K = N$
:$G < L \implies H < M, K < N$
So, in addition: 
:$G > L \implies G + H + K > L + M + N$
:$G = L \implies G + H + K = L + M + N$
:$G < L \implies G + H + K < L + M + N$
It follows from Multiplication of Numbers is Left Distributive over Addition that $G$ and $G + H + K$ are equimultiples of $A$ and $A + C + E$.
For the same reason, $L$ and $L + M + N$ are equimultiples of $B$ and $B + D + F$.
The result follows from {{EuclidDefLink|V|5|Equality of Ratios}}.
{{qed}}
{{Euclid Note|12|V}}
Theorem: Relative Sizes of Proportional Magnitudes

{{:Euclid:Proposition/V/13}}
That is:
:$a : b = c : d, c : d > e : f \implies a : b > e : f$

Proof:
Let a first magnitude $A$ have to a second $B$ the same ratio as a third $C$ to a fourth $D$.
Let the third $C$ have to the fourth $D$ a greater ratio than a fifth $E$ has to a sixth $F$.
:450px
We have that $C : D > E : F$.
From {{EuclidDefLink|V|7|Greater Ratio}}, there will be some equimultiples of $C, E$ and other arbitrary equimultiples of $D, F$ such that the multiple of $C$ is in excess of the multiple of $D$, while the multiple of $E$ is not in excess of the multiple of $F$.
Let these equimultiples be taken.
Let $G, H$ be equimultiples of $C, E$, and $K, L$ be other arbitrary equimultiples of $D, F$, so that $G > K$ but $H \le L$.
Whatever multiple $G$ is of $C$, let $M$ be also that multiple of $A$.
Also, whatever multiple $K$ is of $D$, let $N$ be also that multiple of $B$.
Now we have that $A : B = C : D$ and of $A, C$ equimultiples $M, G$ have been taken.
We also have that of $B, D$ other arbitrary equimultiples $N, K$ have been taken.
Therefore:
: $M > N \implies G > K$
: $M = N \implies G = K$
: $M < N \implies G < K$
from {{EuclidDefLink|V|5|Equality of Ratios}}.
But $G > K$ and so $M > N$.
But $H \le L$, and:
: $M, H$ are equimultiples $A, E$
: $N, L$ are other, arbitrary equimultiples $B, F$.
Therefore from {{EuclidDefLink|V|7|Greater Ratio}}, $A : B > E : F$.
{{qed}}
{{Euclid Note|13|V}}
Theorem: Relative Sizes of Components of Ratios

{{:Euclid:Proposition/V/14}}
That is, if $a : b = c : d$ then:
:$a > c \implies b > d$
:$a = c \implies b = d$
:$a < c \implies b < d$

Proof:
Let a first magnitude $A$ have the same ratio to a second $B$ as a third $C$ has to a fourth $D$.
:300px
Let $A > C$.
Then from Relative Sizes of Ratios on Unequal Magnitudes $A : B > C : B$.
But $A : B = C : D$.
So from Relative Sizes of Proportional Magnitudes $C : D > C : B$.
But from Relative Sizes of Magnitudes on Unequal Ratios $D > B$.
{{qed|lemma}}
In a similar way it can be shown that:
: if $A = C$ then $B = D$
: if $A < C$ then $B < D$
{{qed}}
{{Euclid Note|14|V}}
Theorem: Ratio Equals its Multiples

{{:Euclid:Proposition/V/15}}
That is:
:$a : b \implies ma = mb$

Proof:
Let $AB$ be the same multiple of $C$ that $DE$ is of $F$.
:300px
So as many magnitudes as there are in $AB$ equal to $C$, so many are there also in $DE$ equal to $F$.
Let $AB$ be divided into the magnitudes $AG, GH, HB$ equal to $C$.
Let $DE$ be divided into the magnitudes $DK, KL, LE$ equal to $F$.
Then the number of magnitudes $AG, GH, GB$ is the same as the number of magnitudes in $DK, KL, LE$.
We have that $AG = GH = HB$ and $DK = KL = LE$.
So from Ratios of Equal Magnitudes it follows that $AG : DK = GH : KL = HB : LE$.
Then from Sum of Components of Equal Ratios $AG : DK = AB : DE$.
But $AG = C$ and $DK = F$.
{{qed}}
{{Euclid Note|15|V}}
Theorem: Proportional Magnitudes are Proportional Alternately

{{:Euclid:Proposition/V/16}}
That is:
:$a : b = c : d \implies a : c = b : d$

Proof:
Let $A, B, C, D$ be four proportional magnitudes, so that as $A$ is to $B$, then so is $C$ to $D$.
We need to show that as $A$ is to $C$, then $B$ is to $D$.
:420px
Let equimultiples $E, F$ be taken of $A, B$.
Let other arbitrary equimultiples $G, H$ be taken of $C, D$.
We have that $E$ is the same multiple of $A$ that $F$ is of $B$.
So from Ratio Equals its Multiples we have that $A : B = E : F$
But $A : B = C : D$.
So from Equality of Ratios is Transitive it follows that $C : D = E : F$.
Similarly, we have that $G, H$ are equimultiples of $C, D$.
So from Ratio Equals its Multiples we have that $C : D = G : H$
So from Equality of Ratios is Transitive it follows that $E : F = G : H$.
But from Relative Sizes of Components of Ratios:
:$E > G \implies F > H$
:$E = G \implies F = H$
:$E < G \implies F < H$
Now:
:$E, F$ are equimultiples of $A, B$
:$G, H$ are equimultiples of $C, D$.
Therefore from {{EuclidDefLink|V|5|Equality of Ratios}}:
:$A : C = B : D$
{{qed}}
{{Euclid Note|16|V}}
Theorem: Magnitudes Proportional Compounded are Proportional Separated

{{:Euclid:Proposition/V/17}}
That is:
:$a : b = c : d \implies \left({a - b}\right) : b = \left({c - d}\right) : d$

Proof:
Let $AB, BE, CD, DF$ be magnitudes which are proportional ''componendo'', so that $AB : BE = CD : DF$.
We need to show that they are proportional ''separando'', that is, that $AE : EB = CF : DF$.
:500px
Let equimultiples $GH, HK, LM, MN$ be taken of $AE, EB, CF, FD$.
Let $KO, NP$ be other arbitrary equimultiples of $EB, FD$.
We have that $GH$ is the same multiple of $AE$ that $HK$ is of $EB$.
So from Multiplication of Numbers is Left Distributive over Addition $GH$ is the same multiple of $AE$ that $GK$ is of $AB$.
But $GH$ is the same multiple of $AE$ that $LM$ is of $CF$.
Therefore $GK$ is the same multiple of $AB$ that $LM$ is of $CF$.
Again, we have that $LM$ is the same multiple of $CF$ that $MN$ is of $FD$.
So from Multiplication of Numbers is Left Distributive over Addition $LM$ is the same multiple of $CF$ that $LN$ is of $CD$.
Therefore $GK, LN$ are equimultiples of $AB, CD$.
Again, we have that $HK$ is the same multiple of $EB$ that $MN$ is of $FD$.
Also, $KO$ is the same multiple of $EB$ that $NP$ is of $FD$.
So from Multiplication of Numbers is Right Distributive over Addition the sum $HO$ is also the same multiple of $EB$ that $MP$ is of $FD$.
We also have that $AB : BE = CD : DF$.
Also, we have that $GK, LN$ are equimultiples of $AB, CD$.
Also, we have that $EB, FD$ are equimultiples of $HO, MP$.
Therefore:
:$GK > HO \implies LN > MP$
:$GK = HO \implies LN = MP$
:$GK < HO \implies LN < MP$
Suppose $GK > HO$.
If we subtract $HK$ from each, we have that $GH > KO$.
But $GK > HO \implies LN > MP$.
Therefore $LN > MP$.
So if $MN$ is subtracted from each, we have that $LM > NP$.
So:
: $GH > KO \implies LM > NP$
Similarly we can prove that:
: $GH = KO \implies LM = NP$
: $GH < KO \implies LM < NP$
Also, we have that:
: $GH, LM$ are equimultiples of $AE, CF$
: $KO, NP$ are equimultiples of $EB, FD$
Therefore $AE : EB = CF : FD$.
{{qed}}
{{Euclid Note|17|V|{{EuclidNoteConverse|book=V|prop=18|title=Magnitudes Proportional Separated are Proportional Compounded}}}}
Theorem: Magnitudes Proportional Separated are Proportional Compounded

{{:Euclid:Proposition/V/18}}
That is:
:$a : b = c : d \implies \paren {a + b} : b = \paren {c + d} : d$

Proof:
Let $AE, EB, CF, FD$ be magnitudes which are proportional ''separando'', that is:
:$AE : EB = CF : FD$
We need to show that they are also proportional ''componendo'', that is:
:$AB : BE = CD : FD$
:300px
Suppose $CD : DF \ne AB : BE$.
Then as $AB : BE$ so will $CD$ be to some magnitude less than $DF$ or greater.
Suppose that it be in that ratio to a less magnitude $DG$.
Then since $AB : BE = CD : DG$ it follows from Magnitudes Proportional Compounded are Proportional Separated that:
:$AE : EB = CG : GD$
But by hypothesis:
:$AE : EB = CF : FD$
So by Equality of Ratios is Transitive we have that:
:$CG : GD = CF : FD$
But $CG > CF$.
Therefore $GD > FD$ from Relative Sizes of Components of Ratios.
But it is also less, which is impossible.
Therefore as $AB$ is to $BE$, so is not $CD$ to a lesser magnitude than $FD$.
Similarly we can show that neither is it in that ratio to a greater.
Hence the result.
{{qed}}
{{Euclid Note|18|V|{{EuclidNoteConverse|book = V|prop = 17|title = Magnitudes Proportional Compounded are Proportional Separated}}}}
Theorem: Proportional Magnitudes have Proportional Remainders

{{:Euclid:Proposition/V/19}}
That is:
:$a : b = c : d \implies \left({a - c}\right) : \left({b - d}\right) = a : b$
where $a : b$ denotes the ratio of $a$ to $b$.

Proof:
As the whole $AB$ is to the whole $CD$, so let the part $AE$ subtracted be to the part $CF$ subtracted.
That is:
:$AB : CD = AE : CF$
We need to show that $EB : FD = AB : CD$.
:200px
We have that :$AB : CD = AE : CF$.
So from Proportional Magnitudes are Proportional Alternately we have that $BA : AE = DC : CF$.
From Magnitudes Proportional Compounded are Proportional Separated we have that $BA : EA = DF : CF$.
From Proportional Magnitudes are Proportional Alternately: $BE : DF = EA : FC$.
But by hypothesis $AE : CF = AB : CD$.
So by Equality of Ratios is Transitive $EB : FD = AB : CD$.
{{qed}}
{{Euclid Note|19|V}}
Theorem: Relative Sizes of Successive Ratios

{{:Euclid:Proposition/V/20}}
That is, let:
:$a : b = d : e$
:$b : c = e : f$
Then:
:$a > c \implies d > f$
:$a = c \implies d = f$
:$a < c \implies d < f$

Proof:
Let there be three magnitudes $A, B, C$, and others $D, E, F$ equal to them in multitude, which taken two and two are in the same ratio:
:$A : B = D : E$
:$B : C = E : F$
and let $A > C$ ''ex aequali''.
We need to show that $D > F$.
Similarly, we need to show that $A = C \implies D = F$ and $A < C \implies D < F$.
:300px
Since $A > C$ we have from Relative Sizes of Ratios on Unequal Magnitudes that $A : B > C : B$.
But $A : B = D : E$ and $C : B = F : E$.
Therefore from Relative Sizes of Proportional Magnitudes $D : E > F : E$.
But from Relative Sizes of Magnitudes on Unequal Ratios it follows that $D > F$.
Similarly we can show that $A = C \implies D = F$ and $A < C \implies D < F$.
{{qed}}
{{Euclid Note|20|V}}
Theorem: Relative Sizes of Elements in Perturbed Proportion

{{:Euclid:Proposition/V/21}}
That is, let:
:$a : b = e : f$
:$b : c = d : e$
Then:
:$a > c \implies d > f$
:$a = c \implies d = f$
:$a < c \implies d < f$

Proof:
Let there be three magnitudes $A, B, C$, and others $D, E, F$ equal to them in multitude, which taken two and two together are in the same ratio.
Let the proportion of them be perturbed, that is:
:$A : B = E : F$
:$B : C = D : E$
Let $A > C$.
Then we need to show that $D > F$.
:350px
We have that $A > C$.
So from Relative Sizes of Ratios on Unequal Magnitudes $A : B > C : B$.
But $A : B = E : F$, and $C : B = E : D$
So from Relative Sizes of Proportional Magnitudes $E : F > E : D$.
But from Relative Sizes of Magnitudes on Unequal Ratios $F < D$ and so $D > F$.
Similarly we can prove that $A = C \implies D = F$ and $A < C \implies D < F$.
{{qed}}
{{Euclid Note|21|V}}
Theorem: Equality of Ratios Ex Aequali

{{:Euclid:Proposition/V/22}}
That is, if:
:$a : b = d : e$
:$b : c = e : f$
then:
:$a : c = d : f$

Proof:
Let there be any number of magnitudes $A, B, C$, and others $D, E, F$ equal to them in multitude, which taken two and two together are in the same ratio, so that:
:$A : B = D : E$
:$B : C = E : F$
Then we need to show that:
:$A : C = D : F$
:550px
Let equimultiples $G, H$ be taken of $A, D$.
Let other arbitrary equimultiples $K, L$ be taken of $B, E$.
Let other arbitrary equimultiples $M, N$ be taken of $C, F$.
We have that $A : B = D : E$.
So from Multiples of Terms in Equal Ratios $G : K = H : L$.
For the same reason, $K : M = L : N$.
We have that there are three magnitudes $G, K, M$ and others $H, L, N$ which taken two and two together are in the same ratio.
So from Relative Sizes of Successive Ratios it follows that:
:$G > M \implies H > N$
:$G = M \implies H = N$
:$G < M \implies H < N$
We also have that $G, H$ are equimultiples of $A, D$ and that $M, N$ are equimultiples of $C, F$.
So from {{EuclidDefLink|V|5|Equality of Ratios}}, $A : C = D : F$.
{{qed}}
{{Euclid Note|22|V}}
Theorem: Equality of Ratios in Perturbed Proportion

{{:Euclid:Proposition/V/23}}
That is, if:
:$a : b = e : f$
:$b : c = d : e$
then:
:$a : c = d : f$

Proof:
Let there be three magnitudes $A, B, C$, and others equal to them in multitude, which taken two and two together are in the same proportion, namely $D, E, F$.
Let the proportion of them be perturbed, that is:
:$A : B = E : F$
:$B : C = D : E$
then we need to show that:
:$A : C = D : F$
:500px
Let equimultiples $A, B, D$ be taken of $G, H, K$.
Let other arbitrary equimultiples $L, M, N$ be taken of $C, E, F$.
From Ratio Equals its Multiples it follows that $A : B = G : H$
For the same reason $E : F = M : N$.
We have that $A : B = E : F$
From Equality of Ratios is Transitive $G : H = M : N$.
Next, we have that $B : C = D : E$.
From Proportional Magnitudes are Proportional Alternately $B : D = C : E$.
From Ratio Equals its Multiples $B : D = H : K$.
We also have that $B : D = C : E$ and $H : K = C : E$.
So from Equality of Ratios is Transitive $H : K = C : E$.
From Ratio Equals its Multiples $C : E = L : M$.
We also have that $C : E = H : K$.
So from Equality of Ratios is Transitive $H : K  = L : M$.
From Proportional Magnitudes are Proportional Alternately $H : L = K : M$.
But we have $G : H = M : N$.
Thus it follows from Relative Sizes of Elements in Perturbed Proportion that:
:$G > L \implies K > N$
:$G = L \implies K = N$
:$G < L \implies K < N$
We have that:
: $G, K$ are equimultiples of $A, D$
: $L, N$ are equimultiples of $C, F$
Therefore $A : C = D : F$.
{{qed}}
{{Euclid Note|23|V}}
Theorem: Sum of Antecedent and Consequent of Proportion

{{:Euclid:Proposition/V/25}}
That is, if $a : b = c : d$ and $a$ is the greatest and $d$ is the least, then:
:$a + d > b + c$

Proof:
Let the four magnitudes $AB, CD, E, F$ be proportional, so that $AB : CD = E : F$.
Let $AB$ be the greatest and $F$ the least.
We need to show that $AB + F > CD + E$.
:260px
Let $AG = E, CH = F$.
We have that $AB : CD = E : F$, $AG = E, F = CH$.
So $AB : CD = AG : CH$.
So from Proportional Magnitudes have Proportional Remainders $GB : HD = AB : CD$.
But $AB > CD$ and so $GB > HD$.
Since $AG = E$ and $CH = F$, it follows that $AG + F = CH + E$.
We have that $GB > HD$.
So add $AG + F$ to $GB$ and $CH + E$ to $HD$.
It follows that $AB + F > CD + E$.
{{qed}}
{{Euclid Note|25|V}}
Theorem: Areas of Triangles and Parallelograms Proportional to Base

{{:Euclid:Proposition/VI/1}}
Let $ABC$ and $ACD$ be triangles.
Let $EC, CF$ be parallelograms under the same height.
Then:
:$AC : CD = \triangle ABC : \triangle ACD = \Box EC : \Box CF$
where:
:$AC : CD$ denotes the ratio of the length of $AC$ to that of $CD$
:$\triangle ABC : \triangle ACD$ denotes the ratio of the area of $\triangle ABC$ to that of $\triangle ACD$
:$\Box EC : \Box CF$ denotes the ratio of the area of parallelogram $EC$ to that of parallelogram $CF$.

Proof:
:300px
Let $BD$ be produced in both directions to the points $H, L$ and let any number of straight lines, for example, $BG, GH$ be made equal to the base $BC$, and any number of straight lines, for example, $DK, KL$ be made equal to the base $CD$.
Let $AG, AH, AK, AL$ be joined.
Since $CB = BG = GH$ it follows from Triangles with Equal Base and Same Height have Equal Area that $\triangle ABC = \triangle AGB = \triangle AHG$.
Therefore, whatever multiple the base $HC$ is of the base $BC$, that multiple also is $\triangle AHC$ of $\triangle ABC$.
For the same reason, whatever multiple the base $LC$ is of the base $CD$, that multiple also is $\triangle ALC$ of $\triangle ACD$.
If the base $HC$ is equal to the base $CL$, from Triangles with Equal Base and Same Height have Equal Area, $\triangle AHC = \triangle ACL$.
If the base $HC$ is greater than the base $CL$, $\triangle AHC > \triangle ACL$.
If the base $HC$ is less than the base $CL$, $\triangle AHC < \triangle ACL$.
So we have four magnitudes, two bases $BC, CD$ and two triangles $ABC, ACD$.
We also have that equimultiples have been taken of the base $BC$ and $\triangle ABC$, namely the base $HC$ and $\triangle AHC$.
Also we have equimultiples of the base $CD$ and $\triangle ADC$, namely the base $LC$ and $\triangle ALC$.
It has been proved that:
:$HC > CL \implies \triangle AHC > ALC$
:$HC = CL \implies \triangle AHC = ALC$
:$HC < CL \implies \triangle AHC < ALC$
Therefore from {{EuclidDefLink|V|5|Equality of Ratios}} $BC : CD = \triangle ABC : \triangle ACD$.
Next, from Parallelogram on Same Base as Triangle has Twice its Area, the parallelogram $EC$ is twice the area of $\triangle ABC$.
Similarly, the parallelogram $FC$ is twice the area of $\triangle ACD$.
So from Ratio Equals its Multiples, $\triangle ABC : \triangle ACD = \Box FC : \Box RC$.
The result follows from Equality of Ratios is Transitive.
{{qed}}
{{Euclid Note|1|VI}}