Let ABC be an isosceles triangle having the side AB equal to the side AC;
and let the straight lines BD, CE, be produced further in a straight line with AB, AC. [Post. 2]

I say that the angle ABC is equal to the angle ACB, and the angle CBD to the angle BCE.

Let a point F be taken at random on BD;
from AE the greater let AG be cut off equal to AF the less; [I. 3]
and let the straight lines FC, GB be joined. [Post. 1]

Then, since AF is equal to AG and AB to AC,
the two sides FA, AC are equal to the two sides GA, AB respectively;
and they contain a common angle, the angle FAG.

Therefore the base FC is equal to the base GB,
and the triangle AFC is equal to the triangle AGB, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend,
that is, the angle ACF to the angle ABG,
and the angle AFC to the angle AGB. [I. 4]

And, since [I. 4]

Again, let sides subtending equal angles be equal, as AB to DE;
I say again that the remaining sides will be equal to the remaining sides, namely AC to DF and BC to EF, and further the remaining angle BAC is equal to the remaining angle EDF.

For if BC is unequal to EF, one of them is greater.

Let BC be greater, if possible, and let BH be made equal to EF; let AH be joined.

Then, since BH is equal to EF, and AB to DE,
the two sides AB, BH are equal to the two sides DE, EF respectively, and they contain equal angles;
therefore the base AH is equal to the base DF,
and the triangle ABH is equal to the triangle DEF,
and the remaining angles will be equal to the remaining angles, namely those which equal sides subtend; [I. 4]
therefore the angle BHA is equal to the angle EFD.

But the angle EFD is equal to the angle BCA;
therefore, in the triangle AHC, the exterior angle BHA is equal to the interior and opposite angle BCA:
which is impossible. [I. 16]
Therefore BC is not unequal to EF,
and is therefore equal to it.

But AB is also equal to DE;
therefore the two sides AB, BC are equal to the two sides DE, EF respectively, and they contain equal angles;
therefore the base AC is equal to the base DF,
the triangle ABC equal to the triangle DEF,
and the remaining angle BAC equal to the remaining angle EDF. [I. 4]

Therefore etc. Q.E.D.