First, let A B (in the first figure) be any strait line taken in any solid body; and let A D be any arch drawn upon any centre C and radius CA. Let the point B be understood to describe towards the same parts the arch B E, like and equal to the arch A D. Now in the same time in which the point A transmits the arch A D, the point B, which by reason of its simple motion is supposed to be carried with a velocity equal to that of A, will transmit the arch B E; and at the end of the same time the whole A B will be in D E; and therefore A B and D E are equal. And seeing the arches A D and B E are like and equal, their subtending strait lines AD and BE will also be equal; and therefore the four-sided figure A B D E will be a parallelogram. Wherefore A B is carried parallel to itself. And the same may be proved by the same method, if any other strait line be taken in the same moved body in which the strait line A B was taken. So that all strait lines, taken in a body moved with simple circular motion, will be carried parallel to themselves. Coroll. I. It is manifest that the same will also happen in any body which hath simple motion, though not circular. For all the points of any strait line whatsoever will describe lines, though not circular, yet equal; so that though the crooked lines A D and B E were not arches of circles, but of parabolas, ellipses, or of any other figures, yet both they, and their subtenses, and the strait lines which join them, would be equal and parallel. Coroll. II. It is also manifest, that the radii of the equal circles A D and B E, or the axis of a sphere, will be so carried, as to be always parallel to the places in which they formerly were. For the strait line B F drawn to the centre of the arch B E being equal to the radius A C, will also be equal to the strait line F E or C D; and the angle B F E will be equal to the angle A C D. Now the intersection of the strait lines C A and B E being at G, the angle C G E (seeing B E and A D are parallel) will be equal to the angle D A C. But the angle E B F is equal to the same angle D A C; and therefore the angles C G E and E B F are also equal. Wherefore A C and B F are parallel; which was to be demonstrated. [Sidenote: If circular motion be made about a resting centre, and in that circle there be an epicycle whose revolution is made the contrary way, in such manner that in equal times it make equal angles, every strait line taken in that epicycle will be so carried, that it will always be parallel to the places in which it formerly was.]