Proposition 5.
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. [
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 the whole
AF is equal to the whole
AG,
and in these AB is equal to AC, the remainder BF is equal to the remainder CG.
But
FC was also proved equal to
GB;
therefore the two sides
BF,
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.
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