Let ABC be a triangle having the angle ABC equal to the angle ACB.

I say that the side AB also equals the side AC.

If line AB does not equal line AC, then one of them is greater.

Let line AB be greater.

Cut off line BD from line AB the greater equal to line AC the less, and join line CD.

Since line BD equals line AC, and line BC is common, therefore the two sides line BD and line BC equal the two sides line AC and line BC respectively, and the angle DBC equals the angle ACB.

Therefore the base CD equals the base AB, and the triangle DBC equals the triangle ACB, the less equals the greater, which is absurd.

Therefore if in a triangle two angles equal one another, then the sides opposite the equal angles also equal one another.

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