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ProportionalMeans::usage="If you intend to find a square that has the same area as a given rectangle of sides a and b, it is enough to find the proportional mean between a and b, that is, the magnitude x such that = , and take the square of side x. In general, if in a right triangle a perpendicular is drawn from the right angle to the base, the drawn line is the proportional mean of the segments into which the base is divided. Generalizing, if we take a and b, a < b, it is said that the segments x and y are proportional means, between a and b if is fulfilled that, = = if we take b = 2 a, being a the edge of the cube that is intended to duplicate, from the previous double equality it is obtained = a y, 2 = x y, = 2 a x from where it follows that = 2 and, therefore, the cube of ridge x doubles that of edge a. References: Ayerbe Toledano, José María. La cuadratriz de Hipias y los problemas clásicos griegos. Epistemología e Historia de la Ciencia. Vol. 9, No. 1 (2025). Universidad de Sevilla; Personal.";
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Manipulatea=4;A={-a,0};C1={a,0};O1={0,0};B[θ_]:={aCos[θ],0};ζ[θ_]:={aCos[θ],aSin[θ]};p=Grid{,,(ProportionalMean)},NormBt-A,NormC1-Bt,Normζt-Bt,FrameAll,ItemSize->15,ItemStyle->15;ShowParametricPlot[ζ[θ],{θ,0,π},PlotStyle->{Blue,Thick}],GraphicsBlack,PointSize[0.01],Point[A],Text[Style["A",14,Bold],A,{0,-1.2}],Pointζt,TextStyle["D",14,Bold],ζt,{0,-1.2},Point[C1],Text[Style["C",14,Bold],C1,{0,-1.2}],PointBt,TextStyle["B",14,Bold],Bt,{0,-1.2},Point[O1],Text[Style["O",14,Bold],O1,{0,-1.2}],Thick,LineA,ζt,LineC1,ζt,LineBt,ζt,{DarkYellow,Thick,Line[{A,C1}]},AspectRatioAutomatic,AxesTrue,AxesOrigin{0,0},AxesLabel{x,y},PlotRange{{-a-0.1,a+0.1},{-0.1,a}},ImageSize700,Style["Proportional means (cube duplication)",Bold,Large],{{t,0.00002,"Value (t)"},0.000001,a,0.00001},Delimiter,{{p,2,"Segments and their proportional mean"}},ControlPlacementTop
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