Now, since the angle DCE is equal to the angle ABC,
BF is parallel to CD. [I. 28]
Again, since the angle ACB is equal to the angle DEC,
AC is parallel to FE. [I. 28]
Therefore FACD is a parallelogram;
therefore FA is equal to DC, and AC to FD. [I. 34]
And, since AC has been drawn parallel to FE, one side of the triangle FBE,
therefore, as BA is to AF, so is BC to CE. [VI. 2]
But AF is equal to CD;
therefore, as BA is to CD, so is BC to CE,
and alternately, as AB is to BC, so is DC to CE. [V. 16]
Again, since CD is parallel to BF,
therefore, as BC is to CE, so is FD to DE. [VI. 2]
But FD is equal to AC;
therefore, as BC is to CE, so is AC to DE,
and alternately, as BC is to CA, so is CE to ED. [V. 16]
Since then it was proved that, as AB is to BC, so is DC to CE,
and, as BC is to CA, so is CE to ED;
therefore, ex aequali, as BA is to AC, so is CD to DE. [V. 22]