Equal triangles which are on the same base and on the same side are also in the same parallels.
Let ABC, DBC
be equal triangles which are on the same base
BC and on the same
side of it;
[I say that they are also in the same parallels.] And [For] let AD be joined;
I say that AD is parallel to BC.
For, if not, let AE be drawn through
the point A
parallel to the straight line BC,
and let EC be joined.
Therefore the triangle ABC is equal
to the triangle EBC;
for it is on the same base BC with it and in the same parallels. [I. 37]
But ABC is equal to
therefore DBC is also equal to EBC, [C.N. 1]
the greater to the less: which is impossible.
Therefore AE is not parallel to BC.
Similarly we can prove that neither is any other straight line except
therefore AD is parallel to BC.
Therefore etc. Q.E.D.
Book I: Euclid, Book I (ed. Sir Thomas L. Heath 1st Edition, 1908)
Next: Proposition 40
Previous: Proposition 38
This proposition in other editions: