(Edited by Sir Thomas L. Heath, 1908)

[Euclid, ed. Heath, 1908, on

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 [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.

Book I: Euclid, Book I (ed. Sir Thomas L. Heath 1st Edition, 1908)

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