Kitap 1
34
For, since AB is equal to CD, and BC is common, the two sides AB, BC are equal to the two sides DC, CB respectively; and the angle ABC is equal to the angle BCD; therefore the base AC is also equal to DB, and the triangle ABC is equal to the triangle DCB. [I. 4]
34
Therefore the diameter BC bisects the parallelogram ACDB.
34
Q. E. D.
Proposition 35.
35
Enunciation Parallelograms which are on the same base and in the same parallels are equal to one another.
35
Proof. 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.
35
For, since ABCD is a parallelogram, AD is equal to BC. [I. 34]
35
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]
35
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]
35
Let DGE be subtracted from each; therefore the trapezium ABGD which remains is equal to the trapezium EGCF which remains. [C.N. 3]
35
Let the triangle GBC be added to each; therefore the whole parallelogram ABCD is equal to the whole parallelogram EBCF. [C.N. 2]
35
Therefore etc.
35
Q. E. D.
Proposition 36.
