bases equal, the angle which is contained by the two sides of the one shall be equal to the angle which is contained by the two sides, equal to them, of the other.
Let ABC, DEF be two triangles, having the two sides AB, AC equal to the two sides DE, DF, each to each, namely AB to DE, and AC to DF, and also the base BC equal to the base EF: the angle BAC shall be equal to the angle EDF.
For if the triangle ABC be applied to the triangle DEF, so that the point B may be on the point E, and the straight line BC on the straight line EF, the point C will also coincide with the point F, because BC is equal to EF. [Hyp.
Therefore, BC coinciding with EF, BA and AC will coincide with ED and DF.
For if the base BC coincides with the base EF, but the sides BA, CA do not coincide with the sides ED, FD, but have a different situation as EG, FG; then on the same base and on the same side of it there will be two triangles having their sides which are terminated at one extremity of the base equal to one another, and likewise their sides which are terminated at the other extremity.
But this is impossible. [I. 7.
Therefore since the base BC coincides with the base EF, the sides BA, AC must coincide with the sides ED, DF. Therefore also the angle BAC coincides with the angle EDF, and is equal to it. [Axiom 8.
Wherefore, if two triangles &c. q.e.d.
PROPOSITION 9. PROBLEM.
To bisect a given rectilineal angle, that is to divide it into two equal angles.