hcju/proofwikiqa huggingface dataset:
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}}
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}}
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}}
Let a first magnitude $AB$ have to a second $C$ the same ratio as a third $DE$ has to a fourth $F$. Let also a fifth $BG$ have to the second $C$ the same ratio as a sixth $EF$ has to a fourth $F$. :300px We need to show that $AG : C = DH : F$. We have that $BG : C = EH : F$ and so $C : BG = F : EH$. We also have that $AB : C = DE : F$. So from Equality of Ratios Ex Aequali $AB : BG = DE : EH$. From Magnitudes Proportional Separated are Proportional Compounded it follows that $AG : GB = DH : HE$. But $BG : C = EH : F$. So from Equality of Ratios Ex Aequali, $AG : C = DH : F$. {{qed}} {{Euclid Note|24|V}}
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 $AB + F > CD + E$. :250px 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}}
: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}}
Let $\triangle ABC$ be a triangle, and $DE$ be drawn parallel to the side $BC$. We need to show that $BD : DA = CE : EA$. :200px Let $BE$ and $CD$ be joined. From Triangles with Equal Base and Same Height have Equal Area the area of $\triangle BDE$ is the same as the area of $\triangle CDE$. From Ratios of Equal Magnitudes: : $\triangle BDE : \triangle ADE = \triangle CDE : \triangle ADE$ From Areas of Triangles and Parallelograms Proportional to Base: : $\triangle BDE : \triangle ADE = BD : DA$ By the same reasoning: : $\triangle CDE : \triangle ADE = CE : EA$ From Equality of Ratios is Transitive: : $BD : DA = CE : EA$ {{qed|lemma}} Now let the sides $AB, AC$ of $\triangle ABC$ be cut proportionally, so that $BD : DA = CE : EA$. Join $DE$.  We need to show that $DE \parallel BC$. We use the same construction as above. From Areas of Triangles and Parallelograms Proportional to Base we have that: :$BD : DA = \triangle BDE : \triangle ADE$ :$CE : EA = \triangle CDE : \triangle ADE$ From Equality of Ratios is Transitive: : $\triangle BDE : \triangle ADE = \triangle CDE : \triangle ADE$ So from Magnitudes with Same Ratios are Equal, the area of $\triangle BDE$ is the same as the area of $\triangle CDE$. But these triangles are on the same base $DE$. So from Equal Sized Triangles on Same Base have Same Height, it follows that $DE$ and $BC$ are parallel. {{qed}} {{Euclid Note|2|VI}}
=== $(1)$ implies $(2)$ === :300px Let $CE$ be drawn through $C$ parallel to $DA$. Let $BA$ be produced so as to meet $CE$ at $E$. From Parallelism implies Equal Alternate Angles we have that: :$\angle ACE = \angle CAD$ But by hypothesis: :$\angle CAD = \angle BAD$ and so: :$\angle DAB = \angle ACE$ From Parallelism implies Equal Corresponding Angles: :$\angle BAD = \angle AEC$ But from above $\angle ACE = \angle BAD$, so: :$\angle ACE = \angle AEC$ So from Triangle with Two Equal Angles is Isosceles: :$AC = AE$ Since $AD \parallel EC$, from Parallel Line in Triangle Cuts Sides Proportionally: :$BD : DC = BA : AE$ But $AE = AC$, so: :$BD : DC = AB : AC$ {{Qed|lemma}} === $(2)$ implies $(1)$ === Now suppose $BD : DC = AB : AC$. Join $AD$. Using the same construction, from Parallel Line in Triangle Cuts Sides Proportionally: :$BD : DC = AB : AE$ From Equality of Ratios is Transitive: :$BA : AC = BA : AE$ So $AC = AE$ from Magnitudes with Same Ratios are Equal. So from Isosceles Triangle has Two Equal Angles: :$\angle AEC = \angle ACE$ But from Parallelism implies Equal Corresponding Angles: :$\angle AEC = \angle BAD$ Also, from Parallelism implies Equal Alternate Angles: :$\angle ACE = \angle CAD$ Therefore $\angle BAD = \angle CAD$ and so $AD$ has bisected $\angle BAC$. {{qed}} {{Euclid Note|3|VI}}
Let $\triangle ABC, \triangle DCE$ be equiangular triangles such that: :$\angle ABC = \angle DCE$ :$\angle BAC = \angle CDE$ :$\angle ACB = \angle CED$ :300px Let $BC$ be placed in a straight line with $CE$. From Two Angles of Triangle Less than Two Right Angles $\angle ABC + \angle ACB$ is less than two right angles. As $\angle ACB = \angle DEC$, it follows that $\angle ABC + \angle DEC$ is also less than two right angles. So from the Parallel Postulate, $BA$ and $ED$, when produced, will meet. Let this happen at $F$. We have that $\angle ABC = \angle DCE$. So from Equal Corresponding Angles implies Parallel Lines: : $BF \parallel CD$ Again, we have that $\angle ACB = \angle CED$. Again from Equal Corresponding Angles implies Parallel Lines: : $AC \parallel FE$ Therefore by definition $\Box FACD$ is a parallelogram. Therefore from Opposite Sides and Angles of Parallelogram are Equal $FA = DC$ and $AC = FD$. Since $AC \parallel FE$, it follows from Parallel Line in Triangle Cuts Sides Proportionally that $BA : AF = BC : CE$. But $AF = CD$ so $BA : AF = BC : CE$. From Proportional Magnitudes are Proportional Alternately, $AB : BC = DC : CE$. Since $CD \parallel BF$, from Parallel Line in Triangle Cuts Sides Proportionally $BC : CE = FD : DE$. But $FD = AC$ so $BC : CE = AC : DE$. So from Proportional Magnitudes are Proportional Alternately, $BC : CA = CE : ED$. It then follows from Equality of Ratios Ex Aequali that $BA : AC = CD : DE$. {{qed}} {{Euclid Note|4|VI}}
Let $\triangle ABC, \triangle DEF$ be triangles whose sides are proportional, so that: :$ AB : BC = DE : EF$ :$ BC : CA = EF : FD$ :$ BA : AC = ED : DF$ We need to show that : $\angle ABC = \angle DEF$ : $\angle BCA = \angle EFD$ : $\angle BAC = \angle EDF$ :400px On the straight line $EF$, and at the points $E, F$ on it, construct $\angle FEG = \angle ABC$ and $\angle EFG = \angle ACB$. From Sum of Angles of Triangle Equals Two Right Angles, the remaining angle at $A$ equals the remaining angle at $G$. Therefore $\triangle ABC$ is equiangular with $\triangle GEF$. From Equiangular Triangles are Similar, the sides about the equal angles are proportional, and those are corresponding sides which subtend the equal angles. So: : $AB : BD = GE : EF$ But by hypothesis: : $AB : BC = DE : EF$ So from Equality of Ratios is Transitive : $DE : EF = GE : EF$ So each of $DE, GE$ has the same ratio to $EF$. So from Magnitudes with Same Ratios are Equal: : $DE = GE$ For the same reason: : $DF = GF$ So we have that $DE = EG$, $EF$ is common and $DF = FG$. So from Triangle Side-Side-Side Equality: : $\triangle DEF = \triangle GEF$ That is: : $\angle DEF = \angle GEF, \angle DFE = \angle GFE, \angle EDF = \angle EGF$ As $\angle GEF = \angle ABC$ it follows that: : $\angle ABC = \angle DEF$ For the same reason $\angle ACB = \angle DFE$ and $\angle BAC = \angle EDF$. Hence the result. {{Qed}} {{Euclid Note|5|VI}}
Let $\triangle ABC, \triangle DEF$ be triangles having $\angle BAC = \angle EDF$ and two sides proportional, so that $BA : AC = ED : DF$. We need to show that $\triangle ABC$ is equiangular with $\triangle DEF$, such that $\angle ABC = \angle DEF$ and $\angle ACB = \angle DFE$. :350px Let $\angle FDG$ be constructed equal to either $\angle BAC$ or $\angle EDF$, and $\angle DFG = \angle ACB$. From Sum of Angles of Triangle Equals Two Right Angles: : $\angle ABC = \angle DGF$ Therefore $\triangle ABC$ is equiangular with $\triangle DGF$. From Equiangular Triangles are Similar: : $BA : AC = GD : DF$ But by hypothesis: : $BA : AC = ED : DF$ So from Equality of Ratios is Transitive: : $ED : DF = GD : DF$ From Magnitudes with Same Ratios are Equal: : $ED = DG$ But $DF$ is common to $\triangle DEF$ and $\triangle DGF$, and $\angle EDF = \angle GDF$. So from Triangle Side-Angle-Side Equality: : $\triangle DEF = \triangle DGF$ But $\angle DFG = \angle ACB$ and so: : $\angle ABC = \angle DFE$ But by hypothesis: : $\angle BAC = \angle EDF$ Therefore by Sum of Angles of Triangle Equals Two Right Angles: : $\angle ABC = \angle DEF$ Hence the result. {{qed}} {{Euclid Note|6|VI}}
Let $\triangle ABC, \triangle DEF$ be triangles such that: : $\angle BAC = \angle EDF$ : two sides adjacent to $\angle ABC$ and $\angle DEF$ proportional, so that $AB : BC = DE : EF$ : $\angle ACB$ and $\angle DFE$ either both acute or both not acute. We need to show that $\angle ABC = \angle DEF$ and $\angle BCA = \angle EFD$. First assume that $\angle ACB$ and $\angle DFE$ are both acute. :350px Suppose $\angle ABC \ne \angle DEF$. Then one of them is greater, WLOG $\angle ABC$. On $AB$ and at point $B$, let $\angle ABG$ be constructed equal to $\angle DEF$. Since $\angle BAG = \angle EDF$ and $\angle ABG = \angle DEF$, from Sum of Angles of Triangle Equals Two Right Angles: : $\angle AGB = \angle DFE$ So $\triangle ABG$ is equiangular with $\triangle DEF$. So from Equiangular Triangles are Similar: : $AB : BG = DE : EF$ But by hypothesis: : $DE : EF = AB : BC$ Thus from Equality of Ratios is Transitive: : $AB : BG = AB : BC$ So from Magnitudes with Same Ratios are Equal: : $BC = BG$ So from Isosceles Triangle has Two Equal Angles: : $\angle BCG = \angle BGC$ By hypothesis $\angle BCG$ is less than a right angle. So $\angle BGC$ is less than a right angle. So from Two Angles on Straight Line make Two Right Angles $\angle AGB$ is greater than a right angle. But this was proved equal to $\angle DFE$. But by hypothesis $\angle DFE$ is less than a right angle. From this contradiction we conclude that: : $\angle ABC = \angle DEF$ We also have that $\angle BAC = \angle DEF$. So from Sum of Angles of Triangle Equals Two Right Angles: : $\angle ACB = \angle DFE$ So, as we wanted to show, $\triangle ABC$ is equiangular with $\triangle DEF$. Now, suppose $\angle ACB$ and $\angle DFE$ are both not acute: :350px From the same construction we show similarly that $BC = BG$ and so from Isosceles Triangle has Two Equal Angles, $\angle BCG = \angle BGC$. But $\angle BCG$ is not less than a right angle. So neither is $\angle BGC$ less than a right angle. So in $\triangle BGC$, $\angle BCG + \angle BGC$ are not less than two right angles. From Two Angles of Triangle Less than Two Right Angles, this is impossible. So from this contradiction we conclude that: : $\angle ABC = \angle DEF$ Hence, similarly to above, $\triangle ABC$ is equiangular with $\triangle DEF$. {{qed}} {{Euclid Note|7|VI}}
Let $\triangle ABC$ be a right-angled triangle such that $\angle BAC$ is a right angle. Let $AD$ be drawn perpendicular to $BC$. We need to show that $\triangle ABD$ is similar to $\triangle CAD$ which is similar to $\triangle CBA$ which is in turn similar to $\triangle ABD$. :250px We have that $\angle BAC = \angle ADB$ as both are right angles. Then $\angle ABC$ is common to both $\triangle ABC$ and $\triangle ABD$. So from Sum of Angles of Triangle Equals Two Right Angles: : $\angle ACB = \angle BAD$ So $\triangle ABC$ is equiangular with $\triangle ABD$. From Equiangular Triangles are Similar it follows that $\triangle ABC$ is similar to $\triangle ABD$. Similarly we see that $\triangle ABC$ is equiangular with $\triangle ACD$. Therefore each of $\triangle ABD$ and $\triangle ACD$ are similar to $\triangle ABC$. Finally, $\triangle ABD$ and $\triangle ACD$ are equiangular with each other and so similar to each other. {{qed}}
We have that $ED$ is parallel to one of the sides of $\triangle ABC$. So from Parallel Line in Triangle Cuts Sides Proportionally: : $CD : DA = BE : EA$ 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: : $AB = n \cdot AE$ {{qed}} {{Euclid Note|9|VI|It is in fact only a particular case of {{EuclidPropLink|book=VI|prop=10|title=Construction of Similarly Cut Straight Line}}.<br/>The proof given here is that given by {{AuthorRef|Robert Simson}}, with a refinement by {{AuthorRef|Johann Wilhelm von Camerer|Camerer}}, as {{AuthorRef|Euclid}}'s original was unnecessarily specific.}}
We have that each of $FH$ and $HB$ is a parallelogram. So from Opposite Sides and Angles of Parallelogram are Equal: : $DH = FG$ and $HK = GB$ We have that $HE \parallel KC$, and $KC$ is one of the sides of $\triangle DKC$. So from Parallel Line in Triangle Cuts Sides Proportionally: : $CE : ED = KH : HD$ But $KH = BG$ and $HD = GF$. So: : $CE : ED = BG : GF$ Again, we have $FD \parallel GE$, and $GE$ is one of the sides of $\triangle AGE$. So from Parallel Line in Triangle Cuts Sides Proportionally: : $ED : DA = GF : FA$ So $AB$ has been cut at $F$ and $G$ similarly to $AC$ at $D$ and $E$. {{qed}} {{Euclid Note|10|VI}}
First note that every Fibonacci number $F_n$ is itself a '''Zeckendorf representation''' of itself, where the sequence $\left\langle{c_i}\right\rangle$ contains $1$ term. === Existence === We will use strong induction on $n$. We have that: :$F_2 = 1$ :$F_3 = 2$ :$F_4 = 3$ so each of the integers $1$, $2$, and $3$ have a '''Zeckendorf representation'''. ==== Basis of the Induction ==== $4$ has the '''Zeckendorf representation''': :$4 = 1 + 3 = F_2 + F_4$ This is the basis for the induction. ==== Induction Hypothesis ==== Let the induction hypothesis be that each $n \le k$ has a '''Zeckendorf representation'''. The induction step remains to be proved: that $k + 1$ has a '''Zeckendorf representation'''. ==== Induction Step ==== If $k + 1$ is a Fibonacci number, then it is itself a '''Zeckendorf representation'''. In that case, the induction step holds. Suppose that $k + 1$ is not a Fibonacci number. Then: :$\exists j \in \Z: F_j < k + 1 < F_{j + 1}$ Consider $a = k + 1 - F_j$. Since $a \le k$, it must have a '''Zeckendorf representation''' by the induction hypothesis. In addition: {{begin-eqn}} {{eqn | l = F_j + a       | r = k + 1       | c =  }} {{eqn | o = <       | r = F_{j + 1}       | c =  }} {{eqn | r = F_j + F_{j - 1}       | c =  }} {{eqn | l = a       | o = <       | r = F_{j - 1}       | c =  }} {{end-eqn}} Thus the '''Zeckendorf representation''' of $a$ does not contain $F_{j - 1}$. Thus $k + 1$ has the '''Zeckendorf representation''' of $a + F_j$. It follows by strong induction that for every positive integer there exists a '''Zeckendorf representation'''. {{qed|lemma}} === Uniqueness === Let $n \in \Z_{>0}$. {{AimForCont}} $S$ and $T$ are distinct '''Zeckendorf representation''' for $n$. Consider $S' = S \setminus T$ and $T' = T \setminus S$, that is, each of the sets without their common elements. Since $S$ and $T$ had equal sums: :$(1): \quad \displaystyle \sum S' = \sum T'$ {{WLOG}}, assume that $S'$ is empty. Then in order for $T'$ to have the same sum using only positive integers, $T'$ must also be empty. But as $S \ne T$, it follows that neither $S'$ nor $T'$ is empty. Let the largest element of $S'$ be $F_S$. Let the largest element of $T'$ be $F_T$. Since $S'$ and $T'$ contain no common elements, $F_S \ne F_T$. {{WLOG}}, let $F_S < F_T$. From Sum of Non-Consecutive Fibonacci Numbers: :$\displaystyle \sum S' < F_{S + 1}$ and so: :$\displaystyle \sum S' < F_T$ But from $(1)$: :$\displaystyle \sum S' = \sum T'$ From this contradiction, it cannot be the case that $S'$ and $T'$ are non-empty. So $S' = T' = \O$ and so $S = T$. Thus the '''Zeckendorf representation''' is unique as desired. {{qed}} {{Namedfor|Edouard Zeckendorf|cat = Zeckendorf}}
The proof proceeds by induction on $j$ for $j \ge 2$. For all $j \in \N_{> 0}$, let $P \left({j}\right)$ be the proposition: :$\displaystyle \sum_{F_i \mathop \in S} F_i < F_{j + 1}$ Let the term '''allowable set''' be used to mean a non-empty set of distinct non-consecutive Fibonacci numbers not containing $F_0$ or $F_1$. === Basis for the Induction === The only possible '''allowable set''' whose largest member is $F_2 = 1$ is the set: :$\left\{{F_2}\right\} = \left\{{1}\right\}$ This has sum $1$, which is strictly less than $F_3 = 2$. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true. So this is our induction hypothesis: :$\displaystyle \sum_{F_i \mathop \in S} F_i < F_{k + 1}$ where $S$ is an '''allowable set''' whose largest element is $F_k$. Then we need to show: :$\displaystyle \sum_{F_i \mathop \in S} F_i < F_{k + 2}$ where $S$ is an '''allowable set''' whose largest element is $F_{k + 1}$. === Induction Step === This is our induction step: Let $S$ be an '''allowable set''' whose greatest element is $F_{k + 1}$. Let $S' := S \setminus \left\{ {F_{k + 1} }\right\}$. That is, $S'$ is $S$ without its largest element $F_{k + 1}$. We have that $S$ cannot contain consecutive Fibonacci numbers. Hence the largest element of $S'$ is at most $F_{k - 1}$. So by the induction hypothesis: :$\displaystyle \sum S' < F_k$ Then: {{begin-eqn}} {{eqn | l = \sum S       | r = \sum S' + F_{k + 1}       | c =  }} {{eqn | o = <       | r = F_k + F_{k + 1}       | c =  }} {{eqn | o = <       | r = F_{k + 2}       | c = Definition of Fibonacci numbers }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall j \in \N_{> 0}: \sum_{F_i \mathop \in S} F_i < F_{j + 1}$ {{qed}} Category:Fibonacci Numbers b6wxmmurlhhrlq2e4wdoimlsc7gpd3b
:300px We have that: : $BC \parallel DE$ So from Parallel Line in Triangle Cuts Sides Proportionally: : $AB : BD = AC : CE$ But $BD = AC$ and so: : $AB : AC = AC : CE$ as required. {{qed}} {{Euclid Note|11|VI|It is a special case of {{EuclidPropLink|book=VI|prop=12|title=Construction of Fourth Proportional Straight Line}}.}}
:450px From Parallel Line in Triangle Cuts Sides Proportionally: : $DG : GE = DH : HF$ But: : $DG = A, GE = B, DH = C$ Hence the result. {{qed}} {{Euclid Note|12|VI}}
:250px From Relative Sizes of Angles in Segments, $\angle ADC$ is a right angle. So from the porism to Perpendicular in Right-Angled Triangle makes two Similar Triangles, $DB$ is the mean proportional between $AB$ and $BC$. {{qed}} {{Euclid Note|13|VI|Also see {{EuclidPropLink|book=II|prop=2|title=Construction of Square equal to Given Polygon}} for what amounts to an application of this technique.<br/>Note that nowhere in {{BookLink|The Elements|Euclid}} is the term '''mean proportional''' specifically defined.}}