Therefore each of the figures FH, HB is a parallelogram;
therefore DH is equal to FG and HK to GB. [I. 34]
Now, since the straight line HE has been drawn parallel to KC, one of the sides of the triangle DKC,
therefore, proportionally, as CE is to ED, so is KH to HD. [VI. 2]
But KH is equal to BG, and HD to GF;
therefore, as CE is to ED, so is BG to GF.
Again, since FD has been drawn parallel to GE, one of the sides of the triangle AGE,
therefore, proportionally, as ED is to DA, so is GF to FA. [VI. 2]
But it was also proved that, as CE is to ED, so is BG to GF;
therefore, as CE is to ED, so is BG to GF,
and, as ED is to DA, so is GF to FA.
Complete text
Book I