| On the same straight line there cannot be constructed two similar and unequal segments of circles on the same side. | ||
| For, if possible, on the same straight line AB let two similar and unequal segments of circles ACB and ADB be constructed on the same side. Draw ACD through, and join CB and DB. | ||
| Then, since the segment ACB is similar to the segment ADB, and similar segments of circles are those which admit equal angles, the angle ACB equals the angle ADB, the exterior to the interior, which is impossible. | III.Def.11
I.16 | |
| Therefore on the same straight line there cannot be constructed two similar and unequal segments of circles on the same side. | ||
| Q.E.D. | ||
Book III Introduction - Proposition III.22 - Proposition III.24.