H
E
E
L
E
M
E
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T
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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.
