If the side of the hexagon and that of the decagon inscribed in the same circle be added together, the whole straight line has been cut in extreme and mean ratio, and its greater segment is the side of the hexagon.
若将同一圆内接正六边形与正十边形的边长相加，则整条线段被分成中外比，且其较长线段为正六边形的边长。
圆 O 半径 = 正六边形边长 R；以 R 为弦长依次取点 A、B、C、D、E。AC（六边形边）与 CD（十边形边）相加，把直线 AD 在 C 处分成中外比，且较长段 AC 即六边形边。
正文图形由校订坐标生成；点、线、角、圆可与证明和问答联动。
Let ABC be a circle; of the figures inscribed in the circle ABC let BC be the side of a decagon, CD that of a hexagon, and let them be in a straight line; I say that the whole straight line BD has been cut in extreme and mean ratio, and CD is its greater segment. For let the centre of the circle, the point E, be taken, let EB, EC, ED be joined, and let BE be carried through to A. Since BC is the side of an equilateral decagon, therefore the circumference ACB is five times the circumference BC; therefore the circumference AC is quadruple of CB. But, as the circumference AC is to CB, so is the angle AEC to the angle CEB; [VI. 33] therefore the angle AEC is quadruple of the angle CEB.
设圆ABC，内接正十边形边为BC，正六边形边为CD，且BC与CD共线，则整条线段BD被分成中外比，CD为较长线段。
And, since the angle EBC is equal to the angle ECB, [I. 5] therefore the angle AEC is double of the angle ECB. [I. 32] And, since the straight line EC is equal to CD, for each of them is equal to the side of the hexagon inscribed in the circle ABC, [IV. 15, Por.] the angle CED is also equal to the angle CDE; [I. 5] therefore the angle ECB is double of the angle EDC. [I. 32] But the angle AEC was proved double of the angle ECB; therefore the angle AEC is quadruple of the angle EDC.
取圆心E，连接EB、EC、ED，延长BE至A。因BC为正十边形边，弧ACB是弧BC的五倍，故弧AC是弧CB的四倍，从而角AEC是角CEB的四倍。
But the angle AEC was also proved quadruple of the angle BEC; therefore the angle EDC is equal to the angle BEC. But the angle EBD is common to the two triangles BEC and BED; therefore the remaining angle BED is also equal to the remaining angle ECB; [I. 32] therefore the triangle EBD is equiangular with the triangle EBC. Therefore, proportionally, as DB is to BE, so is EB to BC.
由等腰三角形EBC得角EBC等于角ECB，故角AEC是角ECB的两倍。又因EC等于CD（均为正六边形边），三角形ECD中等角对等边，故角ECB是角EDC的两倍，从而角AEC是角EDC的四倍，故角EDC等于角BEC。
[VI. 4] But EB is equal to CD. Therefore, as BD is to DC, so is DC to CB. And BD is greater than DC; therefore DC is also greater than CB.
三角形BEC与BED有公共角EBD，故剩余角BED等于角ECB，两三角形等角。由比例得DB比BE等于EB比BC，而EB等于CD，故BD比DC等于DC比CB，且BD大于DC，故DC大于CB。