hcju/proofwikiqa huggingface dataset:
:500px Let $ABCG$ be a tetrahedron whose base is $\triangle ABC$ and whose apex is $G$. Let $DEFH$ be a tetrahedron whose base is $\triangle DEF$ and whose apex is $H$. Let $ABCG$ and $DEFH$ be equal in volume. It is to be demonstrated that the areas of their bases is reciprocally proportional to their heights. Let the parallelepipeds $BGML$ and $EHQP$ be completed. We have that $ABCG$ is equal to $DEFH$. From: :{{EuclidPropLink|book = XI|prop = 28|title = Parallelepiped cut by Plane through Diagonals of Opposite Planes is Bisected}} and :{{EuclidPropLink|book = XII|prop = 7|title = Prism on Triangular Base divided into Three Equal Tetrahedra}} it follows that: :$BGML = 6 \cdot ABCG$ and: :$EHQP = 6 \cdot DEFH$ Therefore: :$BGML = EHQP$ But from {{EuclidPropLink|book = XI|prop = 34|title = Parallelepipeds are of Equal Volume iff Bases are in Reciprocal Proportion to Heights}}: :$BM : EQ = h \left({EHQP}\right) : h \left({BGML}\right)$ where: :$BM$ and $EQ$ are the bases of $BGML$ and $EHQP$ respectively :$h \left({EHQP}\right)$ denotes the height of $EHQP$ :$h \left({BGML}\right)$ denotes the height of $BGML$. But from {{EuclidPropLink|book = I|prop = 34|title = Opposite Sides and Angles of Parallelogram are Equal}}: :$BM : EQ = \triangle ABC : \triangle DEF$ Therefore by {{EuclidPropLink|book = V|prop = 11|title = Equality of Ratios is Transitive}}: :$\triangle ABC : \triangle DEF = h \left({EHQP}\right) : h \left({BGML}\right)$ But: :$h \left({EHQP}\right) = h \left({DEFH}\right)$ :$h \left({BGML}\right) = h \left({ABCG}\right)$ where: :$h \left({ABCG}\right)$ denotes the height of $ABCG$ :$h \left({DEFH}\right)$ denotes the height of $DEFH$. Therefore: :$\triangle ABC : \triangle DEF = h \left({DEFH}\right) : h \left({ABCG}\right)$ That is the bases of $ABCG$ and $DEFH$ are reciprocally proportional to their heights. {{qed|lemma}} Let $ABCG$ and $DEFH$ be tetrahedra whose bases are reciprocally proportional to their heights: :$\triangle ABC : \triangle DEF = h \left({DEFH}\right) : h \left({ABCG}\right)$ It is to be shown that $ABCG$ is equal to $DEFH$. Let the parallelepipeds $BGML$ and $EHQP$ be completed. We have that: :$\triangle ABC : \triangle DEF = h \left({DEFH}\right) : h \left({ABCG}\right)$ while: :$\triangle ABC : \triangle DEF = BM : EQ$ where $BM$ and $EQ$ are the bases of $BGML$ and $EHQP$ respectively. Therefore by {{EuclidPropLink|book = V|prop = 11|title = Equality of Ratios is Transitive}}: :$BM : EQ = h \left({EHQP}\right) : h \left({BGML}\right)$ From {{EuclidPropLink|book = XI|prop = 34|title = Parallelepipeds are of Equal Volume iff Bases are in Reciprocal Proportion to Heights}}: :$BGML = EHQP$ But from: :{{EuclidPropLink|book = XI|prop = 28|title = Parallelepiped cut by Plane through Diagonals of Opposite Planes is Bisected}} and :{{EuclidPropLink|book = XII|prop = 7|title = Prism on Triangular Base divided into Three Equal Tetrahedra}} it follows that: :$BGML = 6 \cdot ABCG$ and: :$EHQP = 6 \cdot DEFH$ Therefore: :$ABCG = DEFH$ {{qed}} {{Euclid Note|9|XII}}
Let the cone be of height $h$. Let the area of the base of the cone be $A$. From Volume of Cylinder, the volume of a cylinder of base $A$ and height $h$ is $A h$. Let the cone be divided by planes parallel to its base each positioned some small distance $d$ apart. Let $d$ be sufficiently small that they can be approximated to cylinders in shape. Let there be $n$ of these small cylinders in total. Let the volumes of each of these small cylinders be: : $v_1, v_2, v_3, \ldots, v_n$ starting from the base of the cone and working up. We have that: :$v_k = d a_k$ where $a_n$ is the  Let the areas of the bases of each of these small cylinders be: : $a_1, a_2, a_3, \ldots, a_n$ starting from the base of the cone and working up. {{improve|clumsy}}
:500px Let the cylinder $AD$ be cut by the plane $GH$ which is parallel to the bases $AB$ and $CD$ of $AD$. Let $GH$ meet the axis of $AD$ at the point $K$. It is to be demonstrated that: :$BG : GD = EK : FK$ where: :$BG$ and $GD$ are the cylinder whose parts together make the cylinder $AD$ :$EK$ and $FK$ are the axes of $BG$ and $GD$. Let $EF$ be produced in both directions to $L$ and $M$. Let there be set out: : any number of axes $EN, NL$ equal to $EK$ and : any number of axes $FO, OM$ equal to $FK$. Let $PW$ be the cylinder on the axis $LM$. Let $PQ$ and $VW$ be the bases of $PW$. Let planes be defined parallel to $AB$ and $CD$ through the points $N$ and $O$. Let the circles $RS$ and $TU$ be drawn in these planes with centers $N, O$ and with circumferences on $PV$ and $QW$. We have that: :$LN = NE = EK$ Therefore from {{EuclidPropLink|book = XII|prop = 11|title = Volume of Cones or Cylinders of Same Height are in Same Ratio as Bases}}: :$QR : RB : BG = PQ : RS : AB$ But the bases $PQ, RS, AB$ are all equal. Therefore the cylinders $QR, RB, BG$ are all equal. So as: :the axes $LN, NE, EK$ are all equal and: :the cylinders $QR, RB, BG$ are all equal and: :the number of axes equals the number of cylinders it follows that: :whatever multiple the axis $KL$ is of axis $EK$, the same multiple cylinder $QG$ will be of cylinder $GB$. For the same reason: :whatever multiple the axis $MK$ is of axis $KF$, the same multiple cylinder $WG$ will be of cylinder $GD$. If the axis is greater than the axis, the cylinder will also be greater than the cylinder. If the axis is less than the axis, the cylinder will also be less than the cylinder. The conditions of {{EuclidDefLink|V|5|Equality of Ratios}} are seen to be fulfilled. Hence the result. {{qed}} {{Euclid Note|13|XII}}
:400px Let $EB$ and $FD$ be cylinders on equal bases, the circles $AB$ and $CD$. It is to be shown that: :$EB : FD = GH : KL$ where $GH$ and $KL$ are the axes of $GH$ and $KL$. Let the axis $KL$ be produced to the point $N$. Let $LN = GH$. Let the cylinder $CM$ be described about the axis $LN$. We have that: :$EB$ and $CM$ have the same height and :$EB$ and $CM$ are on equal bases. From {{EuclidPropLink|book = XII|prop = 11|title = Volume of Cones or Cylinders of Same Height are in Same Ratio as Bases}}: :$EB$ and $CM$ are equal. We have that the cylinder $FM$ has been cut by the plane $CD$ which is parallel to its bases. Therefore by {{EuclidPropLink|book = XII|prop = 13|title = Volumes of Parts of Cylinder cut by Plane Parallel to Opposite Planes are as Parts of Axis}}: :$CM : FD = LN : KL$ But: :$CM = EB$ and: :$LN = GH$ Therefore: :$EB : FD = GH : KL$ But from {{EuclidPropLink|book = XII|prop = 10|title = Volume of Cone is Third of Cylinder on Same Base and of Same Height}}: :$EB : FD = ABG : CDK$ where $ABG$ and $CDK$ are cones whose bases are $AB$ and $CD$ and whose apices are $G$ and $K$. Therefore: :$GH : KL = ABG : CDK$ and: :$GH : KL = EB : FD$ {{qed}} {{Euclid Note|14|XII}}
:400px Let $ABCD$ and $EFGH$ be two concentric circles whose centers are at $K$. Let $ABCD$ be the larger. It is required that a regular polygon with an even number of sides be inscribed in $ABCD$ such that it does not touch $EFGH$. Let the straight line $BKD$ be drawn through the center $K$. From the point $G$ let $GA$ be drawn perpendicular to $BD$ and carried through to $ABCD$ at $C$. From {{EuclidPropLink|book = III|prop = 16|qual=Porism|title = Line at Right Angles to Diameter of Circle}}: :$AC$ is tangent to $EFGH$. Let the arc $BAD$ of $ABCD$ be bisected repeatedly. By {{EuclidPropLink|book = X|prop = 1|title = Existence of Fraction of Number Smaller than Given}}: :this can be done until an arc remains less than $AD$. Let such remaining arc be $LD$. Let $LM$ be drawn perpendicular to $BD$ and carried through to $ABCD$ at $N$. Let $LD$ and $DN$ be joined. From {{EuclidPropLink|book = III|prop = 3|title = Conditions for Diameter to be Perpendicular Bisector}} :$LD = DN$ We have that $LN \parallel AC$. We also have that $AC$ is tangent to $EFGH$. Therefore $LN$ does not touch $EFGH$. Therefore $LD$ and $DN$ are far from touching $EFGH$. We can then fit straight lines equal to $LD$ into $ABCD$ as chords going all the way round the circle. Thus a regular polygon with an even number of sides has been inscribed in $ABCD$ such that it does not touch $EFGH$. {{qed}} {{Euclid Note|16|XII}}
:800px We have that the polyhedron as constructed in {{EuclidPropLink|book = XII|prop = 17|title = Construction of Polyhedron in Outer of Concentric Spheres}} is divided into pyramids which are similar in number and arrangement. But from {{EuclidPropLink|book = XII|prop = 8|qual=Porism|title = Volumes of Similar Tetrahedra are in Triplicate Ratio of Corresponding Sides}}: :similar pyramids are to one another in the triplicate ratio of their corresponding sides. Therefore the pyramid whose base is $KBPS$ and whose apex is $A$ has to the similarly arranged pyramid in the other sphere the triplicate ratio of their corresponding sides: that is, $AB$ to the radius of the other sphere. The same applies to all the other pyramids. It follows from {{EuclidPropLink|book = V|prop = 12|title = Sum of Components of Equal Ratios}}: :the sum total of all the pyramids that form the whole polyhedron in the one sphere has to the sum total of all the pyramids that form the whole polyhedron in the other sphere the triplicate ratio of the radii of the spheres. That is, the triplicate ratio of the diameters of the spheres. {{qed}} {{Euclid Note|17|XII}}
:600px Let the spheres $ABC$ and $DEF$ be described. Let their diameters be $BC$ and $EF$. It is to be demonstrated that: :$ABC : DEF = BC^3 : EF^3$ Suppose it is not the case that: :$ABC : DEF = BC^3 : EF^3$ Then: :$ABC : O = BC^3 : EF^3$ where $O$ is either less than or greater than $DEF$. Let $O$ be equal to a sphere $GHK$ which is less than $DEF$. Let $DEF$ be described about $GHK$ with the same center as $GHK$. From {{EuclidPropLink|book = XII|prop = 17|title = Construction of Polyhedron in Outer of Concentric Spheres}}: :let a polyhedron $X$ be inscribed inside $DEF$ which does not touch $GHK$. Let a polyhedron $Y$ be inscribed inside $ABC$ which is similar to the one inscribed inside $DEF$. From {{EuclidPropLink|book = XII|prop = 17|qual=Porism|title = Construction of Polyhedron in Outer of Concentric Spheres}}: :$X : Y = BC^3 : EF^3$ But we have that: :$ABC : GHK = BC^3 : EF^3$ Therefore: :$ABC : GHK = X : Y$ So from {{EuclidPropLink|book = V|prop = 16|title = Proportional Magnitudes are Proportional Alternately}}: :$ABC : X = GHK : Y$ But $ABC > X$, as $ABC$ completely encloses $X$. Therefore $GHK > Y$. But we have that $GHK < Y$, because $Y$ completely encloses $GHK$. From this impossibility it follows that it is not the case that: :$ABC : O = BC^3 : EF^3$ where $O < DEF$. Similarly it can be shoen that it is not the case that: :$DEF : O = EF^3 : BC^3$ where $O < ABC$. Now it is to be shown that it is not the case that: :$ABC : O = BC^3 : EF^3$ where $O$ is greater than $DEF$. Suppose that: :$ABC : LMN = BC^3 : EF^3$ where $LMN$ is a sphere which is greater than $DEF$. Then: :$LMN : ABC = EF^3 : BC^3$ We have that: :$LMN > DEF$ From {{EuclidPropLink|book = XII|prop = 2|qual=Lemma|title = Areas of Circles are as Squares on Diameters}}: :$LMN : ABC = DEF : Z$ where $Z$ is some sphere which is less than $ABC$. Therefore: :$DEF : Z  = EF^3 : BC^3$ where $Z$ is less than $ABC$. This has been shown to be impossible. Therefore it is not the case that: :$ABC : LMN : BC^3 : EF^3$ where $LMN > DEF$. It follows that: :$ABC : DEF = BC^3 : EF^3$ {{qed}} {{Euclid Note|18|XII}}
:350px Let the line segment $AB$ be cut in extreme and mean ratio at the point $C$. Let $AC$ be the greater segment. Let the straight line $AD$ be produced in a straight line with $CA$. Let $AD = \dfrac {AB} 2$. It is to be demonstrated that: :$CD^2 = 5 \cdot AD^2$ Let the squares $AE$ and $DF$ be drawn on $AB$ and $DC$. Let the figure in $DF$ be drawn. Let $FC$ be produced to $G$. We have that $AB$ has been cut in extreme and mean ratio at $C$. Therefore from: :{{EuclidPropLink|book = VI|prop = 17|title = Rectangles Contained by Three Proportional Straight Lines}} and: :{{EuclidDefLink|VI|3|Extreme and Mean Ratio}} it follows that: :$AB \cdot BC = AC^2$ From the construction: :$CE = AB \cdot BC$ :$FH = AC^2$ Therefore: :$CE = FH$ We have that: :$BA = 2 \cdot AD$ while: :$BA = KA$ :$AD = AH$ Therefore: :$KA = 2 \cdot AH$ But from {{EuclidPropLink|book = VI|prop = 1|title = Areas of Triangles and Parallelograms Proportional to Base}} :$KA : AH = CK : CH$ Therefore: :$CK = 2 \cdot CH$ But: :$LH + HC = 2 \cdot CH$ Therefore: :$KC = LH + HC$ But: :$CE = HF$ Therefore $AE$ equals the gnomon $MNO$. We have that: :$BA = 2 \cdot AD$ Therefore: :$BA^2 = 4 \cdot A^2$ That is: :$AE = 4 \cdot DH$ But: :$AE = MNO$ Therefore: :$MNO = 4 \cdot AP$ Therefore: :$DF = 5 \cdot AP$ But: :$DF$ is the square on $DC$ and: :$AP$ is the square on $DA$. Hence the result. {{qed}} {{Euclid Note|1|XIII|{{EuclidNoteConverse|prop=2|title=Converse of Area of Square on Greater Segment of Straight Line cut in Extreme and Mean Ratio}}}}
:350px Let the square on $AB$ be five times the square on $AC$. Let $CD = 2 \cdot AC$. It is to be demonstrated that when $CD$ is cut in extreme and mean ratio, the greater segment is $CB$. Let the squares $AF$ and $CG$ be described on $AB$ and $CF$ respectively. Let the figure $AF$ be drawn. Let $BE$ be produced from $FB$. We have that: :$BA^2 = 5 \cdot AC^2$ That is: :$AF = 5 \cdot AH$ Therefore the gnomon $MNO$ is $4$ times $AH$. We have that: :$DC = 2 \cdot CA$ Therefore: :$DC^2 = 4 \cdot CA^2$ that is: :$CG = 4 \cdot AH$ But: :$MNO = 4 \cdot AH$ Therefore: :$MNO = CG$ We have that: :$DC = CK$ and: :$AC = CH$ Therefore from {{EuclidPropLink|book = VI|prop = 1|title = Areas of Triangles and Parallelograms Proportional to Base}}: :$KB = 2 \cdot BH$ But we also have: :$LH + HB = 2 \cdot BH$ Therefore: :$KB = LH + HB$ But: :$MNO = CG$ Therefore: :$HF = BG$ and as $CD = DG$: :$BG = CD \cdot DB$ Also: :$HF = CB^2$ Therefore: :$CD \cdot DB = CB^2$ Therefore: :$DC : CB = CB : BD$ But: :$DC > CB$ therefore: :$CB > BD$ Therefore, when $CD$ is cut in extreme and mean ratio, the greater segment is $CB$. {{qed}} {{Euclid Note|2|XIII|{{EuclidNoteConverse|prop=1|title=Area of Square on Greater Segment of Straight Line cut in Extreme and Mean Ratio}}}}
:350px Let the line $AB$ be cut in extreme and mean ratio at the point $C$. Let $AC$ be the greater segment. Let $AC$ be bisected at $D$. It is to be demonstrated that $BD^2 = 5 \cdot DC^2$. Let the squares $AE$ be described on $AB$. Let the figure be drawn as above. We have that: : $AC = 2 \cdot DC$ Therefore: : $AC^2 = 4 \cdot DC^2$ That is: :$RS = 4 \cdot FG$ We have by definition of extreme and mean ratio that: :$AB \cdot BC = AC^2$ and: :$CE = AB \cdot BC$ Therefore: :$CE = RS$ But: :$RS = 4 \cdot FG$ Therefore: :$CE = 4 \cdot FG$ We have that: :$AD = DC$ and so: :$HK = KF$ Hence: :$GF = HL$ Therefore: :$GK = KL$ that is: :$MN = NE$ Hence: :$MF = CG$ Therefore: :$CG = FE$ and so: :$CG + CN = FE + CN$ That is, the gnomon $OPQ$ equals the rectangle $CE$. But $CE = 4 \cdot GF$ Therefore: :$OPQ = 4 \cdot GF$ Therefore: :$OPQ + FG = 5 \cdot GF$ But: :$OPQ + FG = DN$ while: :$DN = DB^2$ and: :$GF = DC^2$ Therefore: :$BD^2 = 5 \cdot DC^2$ {{qed}} {{Euclid Note|3|XIII}}
:300px Let the line $AB$ be cut in extreme and mean ratio at the point $C$. Let $AC$ be the greater segment. It is to be demonstrated that: :$AB^2 + BC^2 = 3 \cdot CA^2$ Let the square $ADEB$ be described on $AB$. Let the figure be drawn as above. We have that $AB$ be cut in extreme and mean ratio at $C$ such that $AC > CB$. Therefore from: :{{EuclidPropLink|book = VI|prop = 17|title = Rectangles Contained by Three Proportional Straight Lines}} and: :{{EuclidDefLink|VI|3|Extreme and Mean Ratio}} it follows that: :$AB \cdot BC = AC^2$ We have that: :$AK = AB \cdot BC$ and: :$HG = AC^2$ Therefore: :$AK = HG$ We have that: :$AF = FE$ So: :$AF + CK = FE + CK$ Therefore: :$AK = CE$ and so: :$AK + CE = 2 \cdot CE$ But: :$AK + CE$ are the gnomon $LMN$ and the square $CK$. Therefore: :$LMN + CK = 2 \cdot AK$ But we have that: :$AK = HG$ Therefore: :$LMN + CK + HG = 3 \cdot HG$ But: :$LMN + CK + HG = AE + CK$ while: :$HG = AC^2$ Therefore: :$AB^2 + BC^2 = 3 \cdot CA^2$ {{qed}} {{Euclid Note|4|XIII}}
:400px Let the line $AB$ be cut in extreme and mean ratio at the point $C$. Let $AC$ be the greater segment. Let $AD = AC$. It is to be demonstrated that $DB$ is cut in extreme and mean ratio at the point $A$ where $BA > AD$. Let the square $AE$ be described on $AB$. Let the figure be drawn as above. We have that $AB$ is cut in extreme and mean ratio at $C$. Therefore from: :{{EuclidPropLink|book = VI|prop = 17|title = Rectangles Contained by Three Proportional Straight Lines}} and: :{{EuclidDefLink|VI|3|Extreme and Mean Ratio}} it follows that: :$AB \cdot BC = AC^2$ We have that: :$CE = AB \cdot BC$ and: :$CH = AC^2$ THerefore: :$CH = HC$ But: :$HE = CE$ and: :$DH = HC$ Therefore: :$DH = HE$ Therefore: :$DK = AE$ Thus: :$DK = BD \cdot DA$ That is: :$BD \cdot DA = AB^2$ Therefore: :$DB : BA = BA : AD$ and: :$DB > BA$ Therefore by {{EuclidPropLink|book = V|prop = 14|title = Relative Sizes of Components of Ratios}}: :$BA > AD$ Hence the result as stated. {{qed}} {{Euclid Note|5|XIII}}
:400px Let $AB$ be a rational straight line. Let $AB$ be cut in extreme and mean ratio at the point $C$. Let $AC$ be the greater segment. It is to be demonstrated that each of $AC$ and $CB$ is the irrational straight line known as apotome. Let $BA$ be produced. Let $AD = \dfrac {BA} 2$. We have that $AB$ is cut in extreme and mean ratio at $C$. From {{EuclidPropLink|book = XIII|prop = 1|title = Area of Square on Greater Segment of Straight Line cut in Extreme and Mean Ratio}}: :$CD^2 = 5 \cdot DA^2$ Therefore $CD^2 : DA^2$ is the ratio that a number has to a number. Therefore by {{EuclidPropLink|book = X|prop = 6|title = Magnitudes with Rational Ratio are Commensurable}}: :$CD^2$ is commensurable with $DA^2$. But by {{EuclidDefLink|X|4|Rational Area}}: :$DA^2$ is rational. Therefore by {{EuclidDefLink|X|4|Rational Area}}: :$CD^2$ is also rational. Therefore $CD$ is rational. But $CD^2 : DA^2$ is not the ratio that a square number has to a square number. Therefore by {{EuclidPropLink|book = X|prop = 9|title = Commensurability of Squares}}: :$CD$ and $DA$ are incommensurable in length. Thus $CD$ and $DA$ are commensurable in square only. Therefore by definition, $AC$ is an apotome. {{qed|lemma}} We have that $AB$ is cut in extreme and mean ratio at $C$. Therefore from: :{{EuclidPropLink|book = VI|prop = 17|title = Rectangles Contained by Three Proportional Straight Lines}} and: :{{EuclidDefLink|VI|3|Extreme and Mean Ratio}} it follows that: :$AB \cdot BC = AC^2$ The square on the apotome $AC$, applied to the rational straight line $AB$, produces $BC$ as breadth. Therefore from {{EuclidPropLink|book = X|prop = 97|title = Square on Apotome applied to Rational Straight Line}}: :$CB$ is a first apotome. {{qed}} {{Euclid Note|6|XIII}}
:250px Let $ABCDE$ be an equilateral pentagon. Let $A, B, C$ be three vertices of $ABCDE$ taken in order such that $\angle A = \angle B = \angle C$. Let $AC, BE, FD$ be joined. We have that $CB$ and $BA$ are equal to $BA$ and $AE$ respectively. We also have that $\angle CBA = \angle BAE$. Therefore from {{EuclidPropLink|book = I|prop = 4|title = Triangle Side-Angle-Side Equality}}: :$AC = BE$ and: :$\triangle ABC = \triangle ABE$ Thus: :$\angle BCA = \angle BEA$ :$\angle ABE = \angle CAB$ So from {{EuclidPropLink|book = I|prop = 6|title = Triangle with Two Equal Angles is Isosceles}}: :$AF = CF$ But we have that: :$AC = BE$ Therefore: :$FC = FE$ But: :$CD = DE$ Therefore: :$FC = FE$ and :$CD = ED$ while $FD$ is common. So from {{EuclidPropLink|book = I|prop = 8|title = Triangle Side-Side-Side Equality}}: :$\angle FCD = \angle FED$ But we have already proved that: :$\angle BCA = \angle AEB$ Therefore: :$\angle BCD = \angle AED$ But by hypothesis: :$\angle BCD = \angle A = \angle B$ Therefore: :$\angle AED = \angle A = \angle B$ Similarly it can be proved that: :$\angle CDE = \angle A = \angle B = \angle C$ Therefore $ABCDE$ is equiangular. {{qed|lemma}} Next, suppose that $A, C, D$ are the three vertices of $ABCDE$ such that $\angle A = \angle C = \angle D$. Let $BD$ be joined. We have that $BA = BC$ and $AE = CD$. Also, they contain equal angles. Therefore from {{EuclidPropLink|book = I|prop = 4|title = Triangle Side-Angle-Side Equality}}: :$BE = BD$ and: :$\triangle ABE = \triangle BCD$ Therefore: :$\angle AEB = \angle CDB$ But $BE = BD$. Therefore from {{EuclidPropLink|book = I|prop = 5|title = Isosceles Triangle has Two Equal Angles}}: :$\angle BED = \angle BDE$ Therefore: :$\angle AED = \angle DDE$ But by hypothesis: :$\angle CDE = \angle A = \angle C$ Therefore: :$\angle AED = \angle A = \angle C$ For the same reason: :$\angle ABC = \angle A = \angle C = \angle D$ Therefore $ABCDE$ is equiangular. {{qed}} {{Euclid Note|7|XIII}}
:250px Let $ABCDE$ be a regular pentagon. Let the straight lines $AC$ and $BE$ subtend two vertices $A$ and $B$ taken in order. Let $AC$ and $BE$ cut each other at $H$. It is to be demonstrated that $AC$ and $BE$ are cut in extreme and mean ratio at the point $H$, and that the greater segment equals the side of $ABCDE$. Let the circle $ABCDE$ circumscribe the pentagon $ABCDE$. We have that: :$EA = AB$ and: :$AB = BC$ and they contain equal angles. Therefore from {{EuclidPropLink|book = I|prop = 4|title = Triangle Side-Angle-Side Equality}}: :$AC = BE$ and: :$\triangle ABC = \triangle ABE$ Therefore: :$\triangle BAC = \triangle ABE$ From {{EuclidPropLink|book = I|prop = 31|title = Sum of Angles of Triangle equals Two Right Angles}}: :$\angle AHE = 2 \cdot \angle BAH$ We have that the arc $EDC$ is twice the arc $DC$. Therefore from: :{{EuclidPropLink|book = III|prop = 28|title = Straight Lines Cut Off Equal Arcs in Equal Circles}} and: :{{EuclidPropLink|book = VI|prop = 33|title = Angles in Circles have Same Ratio as Arcs}} it follows that: :$\angle EAC = 2 \cdot \angle BAC$ Therefore: :$\angle HAE = \angle AHE$ Hence from {{EuclidPropLink|book = I|prop = 6|title = Triangle with Two Equal Angles is Isosceles}}: :$HE = EA$ That is: :$HE = AB$ We have that: :$BA = AE$ and: :$\angle ABE = \angle AEB$ But it has been proved that: :$\angle ABE = \angle BAH$ Therefore: :$\angle BEA = \angle BAH$ We also have that $\angle ABE$ is common to $\triangle ABE$ and $\triangle ABH$. Therefore from {{EuclidPropLink|book = I|prop = 32|title = Triangle with Two Equal Angles is Isosceles}}: :$\angle BAE = \angle AHB$ Therefore $\triangle ABE$ has the same corresponding angles with $\triangle ABH$. Therefore from {{EuclidPropLink|book = VI|prop = 4|title = Equiangular Triangles are Similar}}: :$EB : BA = AB : BH$ But: :$BA = EH$ Therefore: :$BE : EH = EH : HB$ and: :$BE > EH$ Therefore: :$EH > HB$ Therefore $BE$ is cut in extreme and mean ratio at the point $H$, and that the greater segment equals the side of $ABCDE$. Similarly it is shown that $AC$ is also cut in extreme and mean ratio at the point $H$, and that the greater segment equals the side of $ABCDE$. {{qed}} {{Euclid Note|8|XIII}}