Euclid's Elements of Geometry: The First Six, the Eleventh and Twelfth BooksJ. Rivington, 1765 - 464 sider |
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Side 41
... ABCD , E B C F be conftituted upon the fame bafe BC , and between the fame parallels AF , BC : Ifay , the parallelogram A B C D is equal to the parallelogram E BCF . A DE F C For because A B C D is a parallelogram [ by prop . 34. ] AD ...
... ABCD , E B C F be conftituted upon the fame bafe BC , and between the fame parallels AF , BC : Ifay , the parallelogram A B C D is equal to the parallelogram E BCF . A DE F C For because A B C D is a parallelogram [ by prop . 34. ] AD ...
Side 42
... A B C D , E F G H are equal to one another . For join BE , CH . A E F H G But Then because B C is equal to FG , and ... ABCD ; for it has the fame base B C , and is conftituted between the fame parallels , BC , A H. By the like way of ...
... A B C D , E F G H are equal to one another . For join BE , CH . A E F H G But Then because B C is equal to FG , and ... ABCD ; for it has the fame base B C , and is conftituted between the fame parallels , BC , A H. By the like way of ...
Side 45
... A B C D , and the triangle EBC have the fame base BC , and be between the fame parallels BC , AE : I fay , the parallelogram ABCD will be double to the triangle B E C. For draw A C. E Then the triangle ABC [ by prop . 37. ] is equal to ...
... A B C D , and the triangle EBC have the fame base BC , and be between the fame parallels BC , AE : I fay , the parallelogram ABCD will be double to the triangle B E C. For draw A C. E Then the triangle ABC [ by prop . 37. ] is equal to ...
Side 46
... A B C D , whofe diameter is A c ; and let the parallelograms E H , F G be about the fame ; then BK , KD are called the complements of them : I fay , the complement в K is equal to the com- plement K D. For because A B C D is a ...
... A B C D , whofe diameter is A c ; and let the parallelograms E H , F G be about the fame ; then BK , KD are called the complements of them : I fay , the complement в K is equal to the com- plement K D. For because A B C D is a ...
Side 48
... A B C D , and the given right - lined angle be E : It is required to constitute a parallelogram equal to the given right - lined figure AB CD , having one angle equal to the angle E. Draw D B , and conftitute [ by prop . 42. ] the ...
... A B C D , and the given right - lined angle be E : It is required to constitute a parallelogram equal to the given right - lined figure AB CD , having one angle equal to the angle E. Draw D B , and conftitute [ by prop . 42. ] the ...
Andre utgaver - Vis alle
Euclid's Elements of Geometry: The First Six, the Eleventh and Twelfth Books Euclid,David Gregory Ingen forhåndsvisning tilgjengelig - 2023 |
Euclid's Elements of Geometry: The First Six, the Eleventh and Twelfth Books Euclid,David Gregory Ingen forhåndsvisning tilgjengelig - 2023 |
Euclid's Elements of Geometry: The First Six, the Eleventh and Twelfth Books Euclid,David Gregory Ingen forhåndsvisning tilgjengelig - 2016 |
Vanlige uttrykk og setninger
A B C D alfo alſo angle ABC becauſe the angle bifected centre circle A B C circumference cone confequent cylinder defcribed demonftrated diameter equal angles equiangular equimultiples Euclid EUCLID's ELEMENTS fame altitude fame multiple fame ratio fame reafon fecond fegment femidiameter fhall fides A B fimilar fince firft firſt fixth folid angle folid parallelepipedon fome fphere ftand given circle given right line given triangle greater infcribed interfect join leffer lefs leſs parallel parallelogram perpendicular polygon prifm PROP propofition proportional pyramid rectangle contained regular polygon remaining angle right angles right line A B right lined figure right-lined SCHOLIUM ſquare thefe THEOR theſe thofe thoſe trapezium triangle ABC twice the fquare vertex the point Wherefore whofe bafe whoſe baſe
Populære avsnitt
Side 247 - If two triangles have one angle of the one equal to one angle of the other and the sides about these equal angles proportional, the triangles are similar.
Side 30 - If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, viz. either the sides adjacent to the equal...
Side 248 - But it was proved that the angle AGB is equal to the angle at F ; therefore the angle at F is greater than a right angle : But by the hypothesis, it is less than a right angle ; which is absurd.
Side 18 - When a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right, and the straight line standing on the other is called a perpendicular to that on which it stands.
Side 32 - Let the straight line EF, which falls upon the two straight lines AB, CD, make the alternate angles AEF, EFD equal to one another; AB is parallel to CD.
Side 56 - Therefore all the angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has sides.
Side 391 - KL: but the cylinder CM is equal to the cylinder EB, and the axis LN to the axis GH; therefore as the cylinder EB to...
Side 110 - If any two points be taken in the circumference of a circle, the straight line which joins them shall fall within the circle.
Side 130 - When you have proved that the three angles of every triangle are equal to two right angles...
Side 183 - FK : in the same manner it may be demonstrated, that FL, FM, FG are each of them equal to FH, or FK : therefore the five straight lines FG, FH, FK, FL, FM are equal to one another : wherefore the circle described from the centre F, at the distance of...