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 FED 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.

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