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.
be an isosceles triangle having the side AB
equal to the side AC
; and let the straight lines BD
be produced further in a straight line with AB
. [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
be joined. [Post. 1
Then, since AF
is equal to AG
, 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.
was also proved equal to GB
therefore the two sides BF
are equal to the two sides CG
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.
Q. E. D.