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DF be joined,
and on the straight line AB, and at the points A, B on it, let
let the angle GAB be constructed equal to the angle at C,
and
the angle ABG equal to the angle CDF. [I.23]
Therefore
the remaining angle CFD is equal to the angle AGB; [I.32]
therefore,
the triangle FCD is equiangular with the triangle CAB.
Therefore, proportionally,
as FD is to GB,
so is FC to GA,
and
CD to AB. [VI.4]
