第3卷命题 23 · 同侧不能作相似不等圆弓形

For, if possible, on the same straight line AB let two similar and unequal segments of circles ACB, ADB be constructed on the same side; let ACD be drawn through, and let CB, DB be joined.

假设可能，在直线AB同侧作出两个相似且不相等的圆弓形ACB和ADB。

Then, since the segment ACB is similar to the segment ADB, and similar segments of circles are those which admit equal angles, [III. Def. 11] the angle ACB is equal to the angle ADB, the exterior to the interior: which is impossible.

连接ACD，并连接CB和DB。

由于弓形ACB与ADB相似，根据定义，它们所含的角相等，故角ACB等于角ADB。

但角ACB是三角形ADB的外角，应大于内对角ADB，矛盾。因此假设不成立。