Discorsi Propositions | |||||

Discorsi Proposition2/31-th-20 |

THEOREMA XX, PROPOSITIO XXXI. | THEOREM XX, PROPOSITION XXXI |

Si linea recta super horizontalem fuerit utcunque inclinata, planum a dato puncto in horizontali usque ad inclinatam extensum, in quo {252} descensus fit tempore omnium brevissimo, est illud quod bifariam dividit angulum contentum a duabus perpendicularibus a dato puncto extensis, una ad horizontalem lineam, altera ad inclinatam. | If a straight line is inclined at any angle to the horizontal and if, from any assigned point in the horizontal, a plane of quickest descent is to be drawn to the inclined line, that plane will be the one which bisects the angle contained {252} between two lines drawn from the given point, one perpendicular to the horizontal line, the other perpendicular to the inclined line. |

Sit CD linea supra horizontalem ab utcunque inclinata, datoque in horizontali quocunque puncto A, educantur ex eo AC perpendicularis ad AB, AE vero perpendicularis ad CD, et angulum CAE bifariam dividat FA linea: dico, planorum omnium ex quibuslibet punctis lineae CD ad punctum A inclinatorum, extensum {10} per FA esse in quo, tempore omnium brevissimo fiat descensus. Ducatur FG ipsi AE parallela; erunt anguli GFA, FAE coalterni aequales: est autem EAF ipsi FAG aequalis: ergo trianguli latera FG, GA aequalia erunt. Si itaque centro G, intervallo GA, circulus describatur, transibit per F, et horizontalem et inclinatam tanget in punctis A, F; est enim angulus GFC rectus, cum GF ipsi AE sit aequidistans: ex {20} quo constat, lineas omnes usque ad inclinatam ex puncto A productas extra circumferentiam extendi, et, quod consequens est, lationes per ipsas longiori tempore absolvi quam per FA. Quod erat demonstrandum. | Let CD be a line inclined at any angle to the horizontal AB; and from any assigned point A in the horizontal draw AC perpendicular to AB, and AE perpendicular to CD; draw FA so as to bisect the angle CAE.Then, I say, that of all the planes which can be drawn through the point A, cutting the line CD at any points whatsoever AF is the one of quickest descent [in quo tempore omnium brevissimo fiat descensus]. Draw FG parallel to AE; the alternate angles GFA and FAE will be equal; also the angle EAF is equal to the angle FAG. Therefore the sides GF and GA of the triangle FGA are equal. Accordingly if we describe a circle about G as center, with GA as radius, this circle will pass through the point F, and will touch the horizontal at the point A and the inclined line at F; for GFC is a right angle, since GF and AE are parallel. It is clear therefore that all lines drawn from A to the inclined line, with the single exception of FA, will extend beyond the circumference of the circle, (Condition 2/06-th-06) (Condition 202C) thus requiring more time to traverse any of them than is needed for FA. Q. E. D. |

Discorsi Propositions | |||||

Discorsi Proposition2/31-th-20 |