Book 13
9
But the angle AEC was proved double of the angle ECB; therefore the angle AEC is quadruple of the angle EDC.
9
But the angle AEC was also proved quadruple of the angle BEC; therefore the angle EDC is equal to the angle BEC.
9
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.
9
Therefore, proportionally, as DB is to BE, so is EB to BC. [VI. 4]
9
But EB is equal to CD.
9
Therefore, as BD is to DC, so is DC to CB.
9
And BD is greater than DC; therefore DC is also greater than CB.
9
Therefore the straight line BD has been cut in extreme and mean ratio, and DC is its greater segment. Q. E. D.
PROPOSITION 10.
10
If an equilateral pentagon be inscribed in a circle, the square on the side of the pentagon is equal to the squares on the side of the hexagon and on that of the decagon inscribed in the same circle.
10
Let ABCDE be a circle, and let the equilateral pentagon ABCDE be inscribed in the circle ABCDE.
10
I say that the square on the side of the pentagon ABCDE is equal to the squares on the side of the hexagon and on that of the decagon inscribed in the circle ABCDE.
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]
