Kitap 13
10
Therefore the whole circumference GB is also double of BM; hence the angle GFB is also double of the angle BFM. [VI. 33]
10
But the angle GFB is also double of the angle FAB, for the angle FAB is equal to the angle ABF.
10
Therefore the angle BFN is also equal to the angle FAB.
10
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.
10
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]
10
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.
10
But the angle LAN is equal to the angle KBN; therefore the angle LKN is also equal to the angle KBN.
10
And the angle at A is common to the two triangles AKB and AKN.
10
Therefore the remaining angle AKB is equal to the remaining angle KNA; [I. 32] therefore the triangle KBA is equiangular with the triangle KNA.
10
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]
10
But the rectangle AB, BN was also proved equal to the square on BF; 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.
