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Discorsi Proposition2/30-th-19 |

{251} THEOREMA XIX, PROPOSITIO XXX. | {251} THEOREM XIX, PROPOSITION XXX |

Si ex aliquo puncto lineae horizontalis descendat perpendiculum, ex alio vero puncto in eadem horizontali sumpto ducendum sit planum usque ad perpendiculum, per quod mobile tempore brevissimo usque ad perpendiculum descendat; tale planum erit illud quod de perpendiculo abscindit partem aequalem distantiae puncti accepti in horizontali a termino perpendiculi. | A perpendicular is let fall from any point in a horizontal line; it is required to pass through any other point in this same horizontal line a plane which shall cut the perclendicular and along which a body will descend to the perpendicular in the shortest possible time. Such a plane will cut from the perpendicular a portion equal to the distance of the assumed point in the horizontal from the upper end of the perpendicular. |

Sit perpendiculum BD, ex puncto B horizontalis lineae AC descendens, in qua sit quodlibet punctum C, et in perpendiculo ponatur distantia BE {10} aequalis distantiae BC, et ducatur CE: dico, planorum omnium ex puncto C usque ad perpendiculum inclinatorum, CE esse illud super quo tempore omnium brevissimo fit descensus usque ad perpendiculum. Inclinentur enim, supra et infra, plana CF, CG, et ducatur IK, circulum semidiametro BC descriptum tangens in C, quae erit perpendiculo aequidistans; et ipsi CF parallela sit EK, {20} usque ad tangentem protracta, secans circumferentiam circuli in L: constat, tempus casus per LE esse aequale tempori casus per CE: sed tempus per KE est longius quam per LE: ergo tempus per KE longius est quam per CE. Sed tempus per KE aequatur tempori per CF, cum sint aequales et secundum eandem inclinationem ductae; similiter, cum CG et IE sint aequales et iuxta eandem inclinationem inclinatae, tempora lationum per ipsas erunt aequalia: sed tempus per HE, breviorem ipsa IE, {30} est brevius tempore per IE: ergo tempus quoque per CE (quod aequatur tempori per HE) brevius erit tempore per IE. Patet ergo propositum. | Let AC be any horizontal line and B any point in it from which is dropped the vertical line BD. Choose any point C in the horizontal line and lay off, on the vertical, the distance BE equal to BC; join C and E. Then, I say, that of all inclined planes that can be passed through C, cutting the perpendicular, CE is that one along which the descent to the perpendicular is accomplished in the shortest time. For, draw the plane CF cutting the vertical above E, and the plane CG cutting the vertical below E; and draw IK, a parallel vertical line, touching at C a circle described with BC as radius. Let EK be drawn parallel to CF, and extended to meet the tangent, after cutting the circle at L. (Condition 2/06-th-06) Now it is clear that the time of fall along LE is equal to the time along CE; (Condition 2/02-th-02-cor2) but the time along KE is greater than along LE; therefore the time along KE is greater than along CE. But the time along KE is equal to the time along CF, since they have the same length and the same slope; and, in like manner, it follows that the planes CG and IE, having the same length and the same slope, will be traversed in equal times. (Condition 2/02-th-02-cor2) Also, since HE |

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Discorsi Proposition2/30-th-19 |