If two triangles have the two sides equal to two sides respectively, and have also the base equal to the base, they will also have the angles equal which are contained by the equal straight lines.
DEF be two triangles having the two sides
equal to the two sides DE,
DF respectively, namely
AB to DE
and AC to DF;
and let them have the base BC equal
to the base EF;
I say that the angle BAC is also equal to the angle EDF.
For if the triangle ABC be applied
to the triangle DEF, and if the
point B be placed on the
point E and the straight line
the point C will also coincide with F,
because BC is equal to EF.
Then, BC coinciding with
BA, AC will also coincide with ED, DF;
for, if the base BC coincides with the base EF, and the sides BA, AC do not coincide with ED, DF but fall beside them as EG, GF,
then, given two straight lines constructed on a straight line (from its extremities) and meeting in a point, there will have been constructed on the same straight line (from its extremities), and on the same side of it, two other straight lines meeting in another point and equal to the former two respectively, namely each to that which has the same extremity with it.
But they cannot be so constructed. [I. 7]
Therefore it is not possible that, if the
base BC be applied
to the base EF, the
AC should not coincide
they will therefore coincide,
so that the angle BAC will also coincide with the angle EDF, and will be equal to it.
If therefore etc. Q.E.D.
Book I: Euclid, Book I (ed. Sir Thomas L. Heath 1st Edition, 1908)
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