32. LET E F be drawn parallel to C D. Then, from what has been said upon the second law of motion[92], it is evident, that since the body was moving in the line A B, when it received the impulse in the direction D C; it will have moved after that impulse through the line D E in the same time, as it would have taken up in moving through D F, provided it had received no disturbance in D. But the time of the body’s moving from D to E is supposed to be equal to the time of its moving through A D; therefore the time, which the body would have imployed in moving through D F, had it not been disturbed in D, is equal to the time, wherein it moved through A D: consequently D F is equal in length to A D; for if the body had gone on to move through the line A B without interruption, it would have moved through all parts thereof with the same velocity, and have passed over equal parts of that line in equal portions of time. Now C F being drawn, since A D and D F are equal, the triangular space C D F is equal to the triangular space C A D. Farther, the line E F being parallel to C D, it is proved by EUCLID, that the triangle C E D is equal to the triangle C F D[93]: therefore the triangle C E D is equal to the triangle C A D. 33. AFTER the same manner, if the body receive at E another impulse directed toward the point C, and be turned by that impulse into the line E G; if it move afterwards from E to G in the same space of time, as was taken up by its motion from D to E, or from A to D; then C G being drawn, the triangle C E G is equal to C D E. A third impulse at G directed as the two former to C, whereby the body shall be turned into the line G H, will have also the like effect with the rest. If the body move over G H in the same time, as it took up in moving over E G, the triangle C G H will be equal to the triangle C E G. Lastly, if the body at H be turned by a fresh impulse directed toward C into the line H I, and at I by another impulse directed also to C be turned into the line I K; and if the body move over each of the lines H I, and I K in the same time, as it imployed in moving over each of the preceding lines A D, D E, E G, and G H: then each of the triangles C H I, and C I K will be equal to each of the preceding. Likewise as the time, in which the body moves over A D E, is equal to the time of its moving over E G H, and to the time of its moving over H I K; the space C A D E will be equal to the space C E G H, and to the space C H I K. In the same manner as the time, in which the body moved over A D E G is equal to the time of its moving over G H I K, so the space C A D E G will be equal to the space C G H I K.