**Quod vero sumptis utcunque duobus punctis D, E, aequaliter ab angulo B remotis, transitus per DB fiat {30} tempore aequali tempori reflexionis per BE, hinc colligere possumus.** | **If now we assume any two points D and E, equally distant from the vertex B, we may then infer that the descent along BD takes place in the same time as the ascent along BE.** |

Ducta DF, erit parallela ad BC; constat enim, descensum per AD reflecti per DF: quod si post D mobile feratur per horizontalem DE, impetus in E erit idem cum impetu in D; ergo ex E ascendet in C; ergo gradus velocitatis in D est aequalis gradui in E. | Draw DF parallel to BC; we know that, after descent along AD, the body will ascend along DF; or, if, on reaching D, the body is carried along the horizontal DE, it will reach E with the same momentum [impetus] with which it left D; hence from E the body will ascend as far as C, proving that the velocity at E is the same as that at D. |