In isosceles triangles the angles at the base are equal to one another, and, if the equal straight lines be produced further, the angles under the base will 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 BD, CE be produced
further in a straight line with AB, AC
I say that the angle ABC is equal to the
angle ACB, and 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;
and let the straight lines FC, GB be joined.
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 equal sides subtend,
that is, the angle ACF to the angle AGB,
and the angle AFC to the angle AGB.
And, since the whole AF is equal to the whole AG,
and in these AB is equal to AC,
the remaininder BF is equal to the remainder CG.
But FC was also proved equal to GB;
therefore the two sides BC, FC are equal to the two sides
CG, GB respectively;
and the angle BFC is equal to the angle CGB,
while the base BC is common to them;
therefore the triangle BFC is also equal to the triangle
CGB, and the remaining angles will be equal to the
remaining angles respectively, namely those which the equal
sides subtend;
therefore the angle FBC is equal to the angle GCB,
and the angle BCF to the angle CBG.
Accordingly, since the whole angle ABG was proved
equal to the angle ACF,
and in these the angle CBG is equal to the angle BCF,
the remaining angle ABC is equal to the remaining angle ACB;
and they are at the base of the triangle ABC.
But the angle FBC was also proved equal to the angle GCB;
and they are under the base.
Therefore etc.
Q.E.D