Let ACDB be a parallelogram, of which BC is a diameter; the opposite sides and angles of the figure shall be equal to one another, and the diameter BC shall bisect it.
Because AB is parallel to CD, and BC meets them, the alternate angles ABC, BCD are equal to one another. [I. 29.
And because AC is parallel to BD, and BC meets them, the alternate angles ACB, CBD are equal to one another. [I. 29.
Therefore the two triangles ABC, BCD have two angles ABC, BCA in the one, equal to two angles DCB, CBD in the other, each to each, and one side BC is common to the two triangles, which is adjacent to their equal angles;
therefore their other sides are equal, each to each, and the third angle of the one to the third angle of the other, namely, the side AB equal to the side CD, and the side AC equal to the side BD, and the angle BAC equal to the angle CDB. [I. 26.
And because 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. [Ax. 2.
And the angle BAC has been shewn to be equal to the angle CDB.
Therefore the opposite sides and angles of a parallelogram are equal to one another.
Also the diameter bisects the parallelogram.
For AB being equal to CD, and BC common,
the two sides AB, BC are equal to the two sides DC, CB each to each;
and the angle ABC has been shewn to be equal to the angle BCD;
therefore the triangle ABC is equal to the triangle BCD, [1. 4.
and the diameter BC divides the parallelogram ACDB into two equal parts.
Wherefore, the opposite sides &c. q.e.d.
