AB is the greater: it is required to cut off from AB the greater, a part equal to C the less.
From the point A draw the straight line AD equal to C;
and from the centre A, at the distance AD, describe the circle DEF meeting AB at E. [Postulate 3.
AE shall be equal to C.
Because the point A is the centre of the circle DEF, AE is equal to AD. [Definition 15.
But C is equal to AD. [Construction.
Therefore AE and C are each of them equal to AD.
Therefore AE is equal to C. [Axiom 1.
Wherefore from AB the greater of two given straight lines a part AE has been cut off equal to C the less. q.e.f.
PROPOSITION 4. THEOREM.
If two triangles have two sides of the one equal to two sides of the other, each to each, and have also the angles contained by those sides equal to one another, they shall also have their bases or third sides equal; and the two triangles shall be equal, and their other angles shall be equal, each to each, namely those to which the equal sides are opposite.
Let ABC, DEF be two triangles which have the two sides AB, AC equal to the two sides DE, DF, each to each, namely, AB to DE, and AC to DF, and the angle BAC equal to the angle EDF: the base BC shall be equal to the base EF, and the triangle ABC to the triangle DEF, and the other angles shall be equal, each to each, to which the equal sides are opposite, namely, the angle ABC to the angle DEF, and the angle ACB to the angle DFE.