Book 13
10
For let the centre of the circle, the point F, be taken, let AF be joined and carried through to the point G, let FB be joined, let FH be drawn from F perpendicular to AB and be carried through to K, let AK, KB be joined, let FL be again drawn from F perpendicular to AK, and be carried through to M, and let KN be joined.
10
Since the circumference ABCG is equal to the circumference AEDG, and in them ABC is equal to AED, therefore the remainder, the circumference CG, is equal to the remainder GD.
10
But CD belongs to a pentagon; therefore CG belongs to a decagon.
10
And, since FA is equal to FB, and FH is perpendicular, therefore the angle AFK is also equal to the angle KFB. [I. 5, I. 26]
10
Hence the circumference AK is also equal to KB; [III. 26] therefore the circumference AB is double of the circumference BK; therefore the straight line AK is a side of a decagon.
10
For the same reason AK is also double of KM.
10
Now, since the circumference AB is double of the circumference BK, while the circumference CD is equal to the circumference AB, therefore the circumference CD is also double of the circumference BK.
10
But the circumference CD is also double of CG; therefore the circumference CG is equal to the circumference BK.
10
But BK is double of KM, since KA is so also; therefore CG is also double of KM.
10
But, further, the circumference CB is also double of the circumference BK, for the circumference CB is equal to BA.
