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Euclid: Elementa
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Proposition 35. 
THEOR. 25. PROPOS. 35. 
第三十五題 
Parallelograms which are on the same base and in the same parallels are equal to one another. 
PARALLELOGRAMMA super eadem basi, & in eisdem parallelis constituta, inter se sunt æqualia. 
兩平行方形。若同在平行線內。又同底。則兩形必等。 
Let ABCD, EBCF be parallelograms on the same base BC and in the same parallels AF, BC;  I say that ABCD is equal to the parallelogram EBCF. 
INTER duas parallelas A B, C D, super basi C D, existant duo parallelogramma C D E A, C D B F. (Dicuntur autem parallelogramma in eisdem esse parallelis, quando duo latera opposita partes sunt parallelarum, ut in exemplo proposito cernitur)  Dico ipsa parallelogramma inter se esse æqualia, non quoad angulos & latera, sed quoad aream, seu capacitatem. 
解曰。甲乙、丙丁、兩平行線內。有丙丁戊甲、與丙丁乙己、兩平行方形。同丙丁底。題言此兩形等。等者。不謂腰等、角等。謂所函之地等。  後言形等者、多倣此。 
For, since ABCD is a parallelogram, AD is equal to BC. [I. 34]  For the same reason also EF is equal to BC,  so that AD is also equal to EF; [C.N. 1]  and DE is common; therefore the whole AE is equal to the whole DF. [C.N. 2]  But AB is also equal to DC; [I. 34]  therefore the two sides EA, AB are equal to the two sides FD, DC respectively,  and the angle FDC is equal to the angle EAB, the exterior to the interior; [I. 29]  therefore the base EB is equal to the base FC, and the triangle EAB will be equal to the triangle FDC. [I. 4]  Let DGE be subtracted from each; therefore the trapezium ABGD which remains is equal to the trapezium EGCF which remains. [C.N. 3]  Let the triangle GBC be added to each;  therefore the whole parallelogram ABCD is equal to the whole parallelogram EBCF. [C.N. 2] 
Cadat enim primo punctum F, inter A, & E. Quoniam igitur in parallelogrammo C D E A, recta A E, æqualis est rectæ C D,  oppositæ & eidem C D, æqualis est F B, in parallelogrammo C D B F, opposita;  Erunt A E, F B, inter se æquales.  Dempta igitur communi F E, remanebit A F, ipsi E B, æqualis:  Est autem & A C, ipsi E D, oppositæ æqualis in parallelogrammo C D E A;    & angulus B E D, angulo F A C, externus interno.  Quare triangulum F A C, triangulo B E D, æquale erit.    Addito igitur communi trapezio C D E F,  fiet totum parallelogrammum C D E A, toti parallelogrammo C D B F, æquale. 
先論曰。設己在甲戊之內。其丙丁戊甲、與丙丁乙己。皆平行方形。丙丁同底。  則甲戊、與丙丁。己乙、與丙丁。各相對之兩邊各等。本篇三四。  而甲戊、與己乙、亦等。公論一。  試於甲戊、己乙、兩線。各減己戊。卽甲己、與戊乙、亦等。公論三。  而甲丙、與戊丁、元等。本篇三四。    乙戊丁外角。與己甲丙內角乂等。本篇廿九。  則乙戊丁、與己甲丙、兩角形必等矣。本篇四。    次於兩角形。每加一丙丁戊己無法四邊形。  則丙丁戊甲、與丙丁乙己、兩平行方形等也。公論二。 
Therefore etc.  Q. E. D. 
  Quod est propsitum.
CADAT secundo punctum F, in punctum E. Dico rursus, parallelogramma C D E A, C D B E, æquala esse. Erunt enim, ut prius, rectæ AE, EB, æquales, nec non & anguli B E D, E A C; atque adeo triangula, E A C, B E D, æqualia. Addito igitur communi triangulo C D E, fient parallelogramma C D E A, C D B E, æqualia.
CADAT tertio punctum F, ultra E, ita ut recta CF, secet rectam D E, in G. Quoniam igitur, ut prius, rectæ A E, F B, sunt æquales; si communis addatur E F; erit tota A F, toti E B, æqualis, nec non & anguli B E D, F A C, æquales erunt; atque adeo triangulum F A C, triangulo B E D, æquale. Ablato ergo communi triangulo E G F, remanebit trapezium A E G C, trapezio FGDB, æquale. Quocirca addito communi triangulo C D G, fiet totum parallelogrammum C D E A, toti parallelogrammo C D B F, æquale. Parallelogramma igitur super eadem basi, & in eisdem parallelis constituta, inter se sunt æqualia. Quod erat demonstrandum.

SCHOLION
CONVERTEMVS facile hanc propositionem, hoc modo.
PARALLELOGRAMMA æqualia super eandem basin, ad easdemque partes censtituta, erunt inter easdem parallelas.
SINT duo parallelogramma æqualia A B C D, C D E F, super eandem basin C D, & ad easdem partes. Dico rectam A B, productam in directum iacere ipsi E F, & propterea ipsa parallelogramma inter easdem esse parallelas. Alias enim A B, producta vel cadet infra E F, vel supra. Cadat primo infra, qualis est A H. Erit igitur parallelogrammum C D G H, æquale parallelogrammo A B C D; Ponitur autem eidem parallelogrammo A B C D, æquale parallelogrammum C D E F. Quare parallelogramma C D E F, C D G H, æqualia erunt, totum & pars, quod est absurdum. Non ergo cadet A B infra E F.
CADAT secundo A B, producta supra E F. Cadet igitur F E, protracta infra AB. Quare, ut prius, erunt parallelogramma A B C D, C D H G, æqualia, totum & pars, quod est absurdum. Idem absurdum consequeretur, si C F, D E, producerentur usque ad A B, protractam. Eademque demonstratio conueniet omnibus casibus, qui occurrere possunt, hoc est, siue punctum E, sit ultra B, siue non, ut perspicuum est. Non ergo cadet A B, supra E F; sed nec infra, ut demonstratum est; ergo producta in directum iacet ipsi E F; ac proinde parallelogramma A B C D, C D E F, in eisdem sunt parallelis. 
  次論曰。設己、戊、同點。依前甲戊、與戊乙等。乙戊丁、與戊甲丙、兩角形等。本論四。而每加一戊丁丙角形。則丙丁戊甲、與丙丁乙戊、兩平行方形必等。公論二。

後論曰。設己點在戊之外。而丙己、與戊丁、兩線交於庚。依前甲戊、與己乙、兩線等。而每加一戊己線。卽戊乙、與甲己、兩線亦等。公論二。因顯己甲丙、與乙戊丁、兩角形亦等。本篇四。次每減一己戊庚角形。則所存戊庚丙甲、與乙己庚丁、兩無法四邊形、亦等。公論三。次於兩無法形每加一庚丁丙角形則丙丁戊甲、與丙丁乙己、兩平行方形必等。公論二。 

 
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