Let triangle ABC be an isosceles triangle having the line AB equal to the line AC,

and let the straight lines line BD and line CE be produced further in a straight line with line AB and line AC.

I say that the angle ABC equals the angle ACB, and the angle CBD equals the angle BCE.

Take an arbitrary point F on line BD. Cut off line AG from line AE the greater equal to line AF the less, and join line FC and line GB.

Since line AF equals line AG, and line AB equals line AC, therefore, line FA and line AC equal line AG and line AB, respectively, and they contain a common angle, the angle FAG.

Therefore, line FC equals line GB, triangle AFC equals triangle AGB, and the remaining angles equal the remaining angles respectively, namely those opposite the equal sides, that is, the angle ACF equals the angle ABG, and the angle AFC equals the angle AGB.

Since line AF equals line AG, and in these line AB equals line AC, therefore, the remainder line BF equals the remainder line CG.

But line FC was also proved equal to line GB, therefore, line BF and line FC equal line CG and line GB respectively, and the angle BFC equals the angle CGB, while line BC is common to them. Therefore, triangle BFC also equals triangle CGB, and the remaining angles equal the remaining angles respectively, namely those opposite the equal sides. Therefore, the angle FBC equals the angle GCB, and the angle BCF equals the angle CBG.

Accordingly, since the whole angle ABG was proved equal to the angle ACF, and in these the angle CBG equals the angle BCF, the remaining angle ABC equals 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, in isosceles triangles, the angles at the base equal one another, and, if the equal straight lines are produced further, then the angles under the base equal one another.

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