If in a triangle two angles be equal to one another, the sides which subtend the equal angles will also be equal to one another.
Let ABC be a triangle
having the angle ABC equal to
the angle ACB;
I say that the side AB is also equal to the side AC.
For, if AB is unequal to AC, one of them is greater.
Let AB be greater; and
from AB the greater
be cut off equal to AC
let DC be joined.
Then, since DB is equal
to AC, and
BC is common,
the two sides DB, BC are equal to the two sides AC, CB respectively;
and the angle DBC is equal to the angle ACB;
therefore the base DC is equal to the base AB,
and the triangle DBC will be equal to the triangle ACB,
the less to the greater;
which is absurd.
Therefore AB is not unequal
it is therefore equal to it.
Therefore etc. Q.E.D.
Book I: Euclid, Book I (ed. Sir Thomas L. Heath 1st Edition, 1908)
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