The times of motion over equal planes, unequally inclined, are to each other inversely as the square root of the ratio of the heights of those planes.

1. Let BA and BC be equal planes from the same terminus, B, but unequally inclined; and let horizontal lines AE and CD be drawn to the vertical BD, plane BA having height BE and plane BC height BD.
   
2. And let BI be the mean proportional of these elevations DB and BE, it follows that the ratio of DB to BI is the square root of the ratio of DB to BE.
   
3. I now say that the ratio of the times of descent or movement over planes BA and BC is the same as the inverse ratio of DB to BI, that is, the homologue of the time through BA is the height of the other plane, BC, which [height] is BD, and the homologue of the time through BC is BI.
   
4. It is therefore to be demonstrated that the time through BA is to the time through BC as DB is to BI.
   
5. Draw IS parallel to DC; as already demonstrated, the time of descent through BA is to the time of fall through the vertical BE as BA is to BE, while the time through BE is to the time through BD as BE is to BI; and the time through BD is to the time through BC as BD is to BC, or BI to BS.
   
6. Therefore, by equidistance of ratios, the time through BA will be to the time through BC as BA is to BS, or CB to BS, and also CB is to BS as DB is to BI; therefore the proposition holds.