Kitap 1
7
Proof. For, if possible, given two straight lines AC, CB constructed on the straight line AB and meeting at the point C, let two other straight lines AD, DB be constructed on the same straight line AB, on the same side of it, meeting in another point D and equal to the former two respectively, namely each to that which has the same extremity with it, so that CA is equal to DA which has the same extremity A with it, and CB to DB which has the same extremity B with it; and let CD be joined.
7
Then, since AC is equal to AD, the angle ACD is also equal to the angle ADC; [I. 5] therefore the angle ADC is greater than the angle DCB; therefore the angle CDB is much greater than the angle DCB.
7
Again, since CB is equal to DB, the angle CDB is also equal to the angle DCB. But it was also proved much greater than it: which is impossible.
7
Therefore etc.
7
Q. E. D.
Proposition 8.
8
Enunciation 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.
8
Proof. 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;
8
I say that the angle BAC is also equal to the angle EDF.
8
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, the point C will also coincide with F, because BC is equal to EF.
