Book 1
34
Therefore ABC, DCB are two triangles having the two angles ABC, BCA equal to the two angles DCB, CBD respectively, and one side equal to one side, namely that adjoining the equal angles and common to both of them, BC; therefore they will also have the remaining sides equal to the remaining sides respectively, and the remaining angle to the remaining angle; [I. 26] therefore the side AB is equal to CD, and AC to BD, and further the angle BAC is equal to the angle CDB.
34
And, since the angle ABC is equal to the angle BCD, and the angle CBD to the angle ACB, the whole angle ABD is equal to the whole angle ACD. [C.N. 2] And the angle BAC was also proved equal to the angle CDB.
34
Therefore in parallelogrammic areas the opposite sides and angles are equal to one another.
34
I say, next, that the diameter also bisects the areas.
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]
