## Euclid, Elements of Geometry, Book I, Proposition 35 (Edited by Sir Thomas L. Heath, 1908)

### Proposition 35[Euclid, ed. Heath, 1908, on archive.org]

Parallelograms which are one the same base and in the same parallels are equal to one another.

Let ABCD, EBCF be parallelograms on the same base BC and in the same parallels AF, BC;
I say that ABCD is equal to the parallelogram EBCF.

For, since ABCD is a parallelogram,
AD is equal to BC. [I. 34]

For the same reason also
EF is equal to BC,
so that AD is also equal to EF; [C.N. 1]
and DE is common;
therefore the whole AE is equal to the whole DF. [C.N. 2]

But AB is also equal to DC; [I. 34]
therefore the two sides EA, AB are equal to the two sides FD, DC respectively,
and the angle FDC is equal to the angle EAB, the exterior to the interior; [I. 29]
therefore the base EB is equal to the base FC,
and the triangle EAB will be equal to the triangle FDC. [I. 4]

Let DGE be subtracted from each;
therefore the trapazium ABGD which remains is equal to the trapezium EGCF which remains. [C.N. 3]

Let the triangle GBC be added to each;
therefore the thole parallelogram ABCD is equal to the whole parallelogram EBCF. [C.N. 2]

Therefore, etc. Q.E.D.

Next: Proposition 36

Previous: Proposition 34

This proposition in other editions: