Euclid, Elements I 34©
translated by Henry Mendell (Cal. State U., L.A.)
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Prop. 34: The opposite sides and angles of parallelogram regions are equal to one another, and the diameters bisect them.
(diagram 1) Let there be a parallelogram region, AGDB, and a diameter of it, BG. I say that the opposite sides and angles of parallelogram AGDB are equal to one another and that diameter BG bisects them.
(diagram 2) For since AB is equal to GD and straight-line BG falls into them, (diagram ) the alternate angles by ABG, BGD are equal to one another. (diagram 4) Again, since AG is parallel to BD and BG falls into them, the alternate angles by AGB, GBD are equal to one another. (diagram 5)In fact, two triangles, ABG, BGD, having two angles, those by ABG, BGA equal to two, those by BGD, GBD, respectively, and one side equal to one side, what’s common at the equal angles, BG. (diagram 6) Therefore, they will have the remaining sides equal to the remaining sides, respectively, angle to the remaining angle. Therefore, side AB is equal to GD, and AG to BD, and, furthermore, the angle by BAG to that by GDB. (diagram 7) Since the angle by ABG is equal to that by BGD, and that by GBD to that by AGB, a whole, therefore, that by ABD, is equal to a whole, that by AGD. But the angle BAG was also shown equal to that by GDB. Therefore, the opposite sides and angles of parallelogram regions are equal to one another.
(diagram 5) I say, in fact, that the diameter also bisects them. For since AB is equal to GD and BG is common, in fact, two, AB, BG are equal to two, GD, BG, respectively. And the angle by ABG is equal to the that by BGD. (diagram 6) Therefore, a base, AG, is equal to DB. Therefore, triangle ABG is equal to triangle BGD. Therefore, diameter BG bisects parallelogram ABGD, just what it was required to show.