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.
Let ABC, DEF be two triangles having the two sides AB, AC 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 BC on EF,
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,
so that the angle BAC will also coincide with the angle EDF, and will be equal to it. If therefore etc.
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