hcju/proofwikiqa huggingface dataset:
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 Equality 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 Equality: : $\triangle FCK = \triangle FLC$ 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: :P $\angle FKC = \angle FLC$ : $\angle HKL = 2 \angle FKC$ : $\angle KLM = 2 \angle FLC$ 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}}
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 Equality $\triangle BCF = \triangle DCF$ 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 Equality: : $\triangle FHC = \triangle FKC$ 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}}
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}}
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}}
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}}
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}}
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}}
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}}
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 Multiplication of Numbers Distributes over Addition, $AE$ 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}}
{{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 Axioms: $\R \text D$: Distributivity }} {{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}}
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}}}}
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}}}}
: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}}}}
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}}}}
Let $A : B = C : D$ and $C : D = E : F$. :550px 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}}
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}}
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}}
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}}
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}}
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 $D$. :400px 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 $A, B$, and $G, H$ are equimultiples of $C, D$. Therefore from {{EuclidDefLink|V|5|Equality of Ratios}}: :$A : C = B : D$ {{qed}} {{Euclid Note|16|V}}
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}}}}
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}}}}
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}}
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}}