Proposition I.35

Parallelograms which are on 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; I.c.n.1

and DE is common;

  • therefore the whole AE is equal to the whole DF. I.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 FDC1 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 2 from each; therefore the trapezium ABGD which remains is equal to the trapezium EGCF which remains. I.c.n.3

Let the triangle GBC be added to each; therefore the whole parallelogram ABCD is equal to the whole parallelogram EBCF. I.c.n.2

Therefore etc.

  • Q. E. D.

References

Footnotes

  1. FDC
    The text has DFC. ↩

  2. DGE
    Euclid speaks of the triangle DGE without any explanation that, in the case which he takes (where AD, EF have no point in common), BE, CD must meet at a point G between the two parallels. He allows this to appear from the figure simply. ↩