2/09-th-09
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2/09-th-09 |
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| THEOREMA IX, PROPOSITIO IX. | THEOREM IX, PROPOSITION IX |
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| Si a puncto in linea horizonti parallela duo plana utcunque inclinentur, et a linea secentur, quae cum ipsis angulos faciat permutatim aequales {10} angulis ab iisdem planis et horizontali contentis, lationes in partibus a dicta linea sectis, temporibus aequalibus absolventur. | If from any point on a horizontal line two planes, inclined at any angle, are drawn, and if they are cut by a line which makes with them angles alternately equal to the angles between these planes and the horizontal, then the times required to traverse those portions of the plane cut off by the aforesaid line are equal. |
| Ex puncto C horizontalis lineae X duo plana utcumque inflectantur CD, CE, et in quolibet puncto lineae CD constituatur angulus CDF, angulo XCE aequalis; secet autem linea DF planum CE in F, adeo ut anguli CDF, CFD angulis XCE, LCD permutatim sumptis sint aequales: dico, tempora descensuum per CD, CF esse aequalia. {20} Quod autem (posito angulo CDF aequali angulo XCE) angulus CFD sit aequalis angulo DCL, manifestum est. Dempto enim angulo communi DCF, ex tribus angulis trianguli CDF, aequalibus duobus rectis, quibus aequantur anguli omnes ad lineam LX in puncto C constitutis, remanent in triangulo duo CDF, CFD, duobus XCE, LCD aequales; positus autem est CDF ipsi XCE aequalis; ergo reliquus CFD reliquo DCL. Ponatur planum CE aequale plano CD, et ex punctis D, E perpendiculares agantur DA, EB ad horizontalem XL, ex C vero ad DF ducatur {30} perpendicularis CG; et quia angulus CDG angulo ECB est aequalis, et recti sunt DGC, CBE, erunt trianguli CDG, CBE aequianguli, et ut DC ad CG, ita CE ad EB: est autem DC aequalis CE: ergo CG aequalis erit BE: cumque triangulorum DAC, CGF anguli C, A angulis F, G sint aequales, erit ut CD ad DA, ita FC ad CG, et, permutando, ut DC ad CF, ita DA ad CG seu BE. Ratio itaque elevationum planorum aequalium CD, CE est {228} eadem cum ratione longitudinum DC, CF; ergo, ex corollario primo praecedentis propositionis sextae, tempora descensuum in ipsis erunt aequalia: quod erat probandum. | Through the point C on the horizontal line X, draw two planes CD and CE inclined at any angle whatever: at any point in the line CD lay off the angle CDF equal to the angle XCE; let the line DF cut CE at F so that the angles CDF and CFD are alternately equal to XCE and LCD; then, I say, the times of descent over CD and CF are equal. Now since the angle CDF is equal to the angle XCE by construction, it is evident that the angle CFD must be equal to the angle DCL. For if the common angle DCF be subtracted from the three angles of the triangle CDF, together equal to two right angles, (to which are also equal all the angles which can be described about the point C on the lower side of the line LX) there remain in the triangle two angles, CDF and CFD, equal to the two angles XCE and LCD; but, by hypothesis, the angles CDF and XCE are equal; hence the remaining angle CFD is equal to the remainder DCL. Take CE equal to CD; from the points D and E draw DA and EB perpendicular to the horizontal line XL; and from the point C draw CG perpendicular to DF. Now since the angle CDG is equal to the angle ECB and since DGC and CBE are right angles, it follows that the triangles CDG and CBE are equiangular; consequently DC:CG = CE:EB. But DC is equal to CE, and therefore CG is equal to EB. Since also the angles at C and at A, in the triangle DAC, are equal to the angles at F and G in the triangle CGF, we have CD:DA = FC:CG and, permutando, DC:CF = DA:CG = DA:BE. (Condition 2/06-th-06-cor3) Thus the ratio of the heights of the equal planes CD and CE is the same as the ratio of the lengths DC and CF. Therefore, by {228} Corollary I of Prop. VI, the times of descent along these planes will be equal. Q. E. D. |
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| Aliter idem: | An alternative proof is the following: |
| ducta FS perpendiculari ad horizontalem AS. Quia triangulum CSF simile est triangulo DGC, erit ut SF ad FC, ita GC ad CD; et quia triangulum CFG simile est triangulo DCA, erit ut FC ad CG, ita CD ad DA; ergo, ex aequali, ut SF ad CG, ita CG ad DA: media est {10} igitur CG inter SF, DA, et ut DA ad SF, ita quadratum DA ad quadratum CG. Rursus, cum triangulum ACD simile sit triangulo CGF, erit ut DA ad DC, ita GC ad CF, et, permutando, ut DA ad CG, ita DC ad CF, et ut quadratum DA ad quadratum CG, ita quadratum DC ad quadratum CF; sed ostensum est, quadratum DA ad quadratum CG esse ut linea DA ad lineam FS; ergo, ut quadratum DC ad quadratum CF, ita linea DA ad FS; ergo, ex praecedenti septima, cum planorum CD, CF {20} elevationes DA, FS duplam habeant rationem eorumdem planorum, tempora lationum per ipsa erunt aequalia. | Draw FS perpendicular to the horizontal line AS. Then, since the triangle CSF is similar to the triangle DGC, we have SF:FC = GC:CD; and since the triangle CFG is similar to the triangle DCA, we have FC:CG = CD:DA. Hence, ex aequali, SF:CG = CG:DA. Therefore CG is a mean proportional between SF and DA, while DA:SF =DA^2:CG^2. Again since the triangle ACD is similar to the triangle CGF, we have DA:DC = GC:CF and, permutando, DA:CG = DC:CF: also DA^2:CG^2 = DC^2:CF^2. But it has been shown that DA^2:CG^2 = DA:SF.Therefore DC^2:CF^2 = DA:FS. (Condition 2/07-th-07) Hence from the above Prop. VII, since the heights DA and FS of the planes CD and CF are to each other as the squares of the lengths of the planes, it follows that the times of descent along these planes will be equal. |
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2/09-th-09 |
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