And, AB coinciding with DE, AC will fall on DF, because the angle BAC is equal to the angle EDF. [Hypothesis.
Therefore also the point C will coincide with the point F, because AC is equal to DF. [Hypothesis.
But the point B was shewn to coincide with the point E, therefore the base BC will coincide with the base EF;
because, B coinciding with E and C with F, if the base BC does not coincide with the base EF, two straight lines will enclose a space; which is impossible. [Axiom 10.
Therefore the base BC coincides with the base EF, and is equal to it. [Axiom 8.
Therefore the whole triangle ABC coincides with the whole triangle DEF, and is equal to it. [Axiom 8.
And the other angles of the one coincide with the other angles of the other, and are equal to them, namely, the angle ABC to the angle DEF, and the angle ACB to the angle DFE.
Wherefore, if two triangles &c. q.e.d.
PROPOSITION 5. THEOREM.
The angles at the base of an isosceles triangle are equal to one another; and if the equal sides be produced the angles on the other side of the base shall be equal to one another.
Let ABC be an isosceles triangle, having the side AB equal to the side AC, and let the straight lines AB, AC be produced to D and E: the angle ABC shall be equal to the angle ACB, and the angle CBD to the angle BCE.
In BD take any point F,
and from AE the greater cut off AG equal to AF the less, [I.3.