greater than the base EF.
Of the two sides DE, DF, let DE be the side which is not greater than the other. At the point D in the straight line DE, make the angle EDG equal to the angle BAC, [I. 23.
and make DG equal to AC or DF, [I. 3.
and join EG, GF.
Because AB is equal to DE, [Hypothesis.
and AC to DG; [Construction.
the two sides BA, AC are equal to the two sides ED, DG, each to each;
and the angle BAC is equal to the angle EDG; [Constr.
therefore the base BC is equal to the base EG. [I. 4.
And because DG is equal to DF, [Construction.
the angle DGF is equal to the angle DFG. [I. 5.
But the angle DGF is greater than the angle EGF. [Ax. 9.
Therefore the angle DFG is greater than the angle EGF. Much more then is the angle EFG greater than the angle EGF. [Axiom 9.
And because the angle EFG of the triangle EFG is greater than its angle EGF, and that the greater angle is subtended by the greater side, [I. 19.
therefore the side EG is greater than the side EF.
But EG was shewn to be equal to BC;
therefore BC is greater than EF.
Wherefore, if two triangles &c. q.e.d.
PROPOSITION 25. THEOREM.
If two triangles have two sides of the one equal to two sides of the other, each to each, but the base of the one
