|Equal triangles which are on the same base and on the same side are also in the same parallels.|
|Let ABC and DBC be equal triangles which are on the same base BC and on the same side of it. Join AD.
I say that AD is parallel to BC.
|If not, draw AE through the point A parallel to the straight line BC, and join EC.||I.31
|Therefore the triangle ABC equals the triangle EBC, for it is on the same base BC with it and in the same parallels.||I.37|
|But ABC equals DBC, therefore DBC also equals EBC, the greater equals the less, which is impossible.||C.N.1|
|Therefore AE is not parallel to BC.
Similarly we can prove that neither is any other straight line except AD, therefore AD is parallel to BC.
|Therefore equal triangles which are on the same base and on the same side are also in the same parallels.|
Next proposition: I.40