Let ABCDE be a circle, and let the equilateral pentagon ABCDE be
inscribed in the circle ABCDE.
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.
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, [p. 458] 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.
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.
But CD belongs to a pentagon; therefore CG belongs to a decagon.
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]
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.
For the same reason AK is also double of KM.
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.
But the circumference CD is also double of CG; therefore the circumference
CG is equal to the circumference BK.
But BK is double of KM, since KA is so also; therefore CG is also
double of KM. [p. 459]
But, further, the circumference CB is also double of the circumference
BK, for the circumference CB is equal to BA.
Therefore the whole circumference GB is also double of BM; hence
the angle GFB is also double of the angle BFM. [VI. 33]
But the angle GFB is also double of the angle FAB, for the angle
FAB is equal to the angle ABF.
Therefore the angle BFN is also equal to the angle FAB.
But the angle ABF is common to the two triangles ABF and BFN; therefore
the remaining angle AFB is equal to the remaining angle BNF; [I.
32] therefore the triangle ABF is equiangular with the triangle
BFN.
Therefore, proportionally, as the straight line AB is to BF, so
is FB to BN; [VI. 4] therefore the rectangle AB, BN is equal to
the square on BF. [VI. 17]
Again, since AL is equal to LK, while LN is common and at right
angles, therefore the base KN is equal to the base AN; [I. 4] therefore
the angle LKN is also equal to the angle LAN.
But the angle LAN is equal to the angle KBN; therefore the angle
LKN is also equal to the angle KBN.
And the angle at A is common to the two triangles AKB and AKN.
Therefore the remaining angle AKB is equal to the remaining angle
KNA; [I. 32] therefore the triangle KBA is equiangular with the
triangle KNA.
Therefore, proportionally, as the straight line BA is to AK, so
is KA to AN; [VI. 4] therefore the rectangle BA, AN is equal to
the square on AK. [VI. 17]
But the rectangle AB, BN was also proved equal to the square on
BF; [p. 460] therefore the rectangle AB, BN together with the rectangle
BA, AN, that is, the square on BA [II. 2], is equal to the square
on BF together with the square on AK.
And BA is a side of the pentagon, BF of the hexagon [IV. 15, Por.],
and AK of the decagon.
Therefore etc.
Q. E. D.
