Let ABC, DEF be triangles on equal bases BC, EF and in the same parallels BF, AD;
I say that the triangle ABC is equal to the triangle DEF.

For let AD be produced in both directions to G, H;
through B let BG be drawn parallel to CA, [I. 31]
and through F let FH be drawn parallel to DE.

Then each of the figures GBCA, DEFH is a parallelogram;
and GBCA is equal to DEFH;
for they are on equal bases BC, EF and in the same parallels BF, GH. [I. 36]

Moreover the triangle ABC is half of the parallelogram GBCA; for the diameter AB bisects it. [I. 34]

And the triangle ABC is half of the parallelogram DEFH; for the diameter DF bisects it. [I. 34]

[But the halves of equal things are equal to one another.]

Therefore the triangle ABC is equal to the triangle DEF.

Therefore etc. Q.E.D.