The Elements of Euclid,: In which the Propositions are Demonstrated in a New and Shorter Manner Than in Former Translations, and the Arrangement of Many of Them Altered, to which are Annexed Plain and Spherical Trigonometry, Tables of Logarithms from 1 to 10000, and Tables of Sines, Tangents, and Secants, Both Natural and ArtificialJ. Murray, no. 32. Fleetstreet; and C. Elliot, Parliament-square, Edinburgh., 1776 - 264 sider |
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Resultat 1-5 av 31
Side viii
... first definition is challenged by Mr Simpson , which , he fays , ought to be proved ; for this I can fee no reason , or any neceffity of a proof , as the equality of coincident figures . is admitted , ax . 3. Book I. I have taken ...
... first definition is challenged by Mr Simpson , which , he fays , ought to be proved ; for this I can fee no reason , or any neceffity of a proof , as the equality of coincident figures . is admitted , ax . 3. Book I. I have taken ...
Side 27
... first mentioned part . Fa right line be any how cut , the rectangle under the whole Let the right line AB be any how cut in C , the rectangle under AB , BC , is equal to the rectangle under AC , CB , to- gether with the fquare of BC ...
... first mentioned part . Fa right line be any how cut , the rectangle under the whole Let the right line AB be any how cut in C , the rectangle under AB , BC , is equal to the rectangle under AC , CB , to- gether with the fquare of BC ...
Side 30
... first part taken as the fide of the fquare . b . Let the right line AB be any how cut in C , four times the rectangle under AB , BC , together with the fquare of AC , are equal to the fquare of AD ; that is , AB produced to D , so that ...
... first part taken as the fide of the fquare . b . Let the right line AB be any how cut in C , four times the rectangle under AB , BC , together with the fquare of AC , are equal to the fquare of AD ; that is , AB produced to D , so that ...
Side 53
... First , let it pass thro ' the center E , and join BE ; then , because AC is bi- fected in E , and DC added , the rectangle under AD , DC , to- gether with the fquare of CE , are equal to the fquare of DE 2 ; a 6. 2 . but the fquare of ...
... First , let it pass thro ' the center E , and join BE ; then , because AC is bi- fected in E , and DC added , the rectangle under AD , DC , to- gether with the fquare of CE , are equal to the fquare of DE 2 ; a 6. 2 . but the fquare of ...
Side 66
... first to the fecond , and third to the fourth , when there are taken any equimultiples of the first and third , and likewife any equi- multiples of the fecond and fourth ; if the multiple of the firft , be equal to the multiple of the ...
... first to the fecond , and third to the fourth , when there are taken any equimultiples of the first and third , and likewife any equi- multiples of the fecond and fourth ; if the multiple of the firft , be equal to the multiple of the ...
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Vanlige uttrykk og setninger
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Populære avsnitt
Side 93 - Equiangular parallelograms have to one another the ratio which is compounded of the ratios of their sides. Let AC, CF be equiangular parallelograms, having the angle BCD equal to the angle ECG ; the...
Side 78 - ... viz. as A is to B, fo is E to F, and B to C as D to E ; and if the firft A be greater than the third C, then the fourth D will be greater than the fixth F ; if equal, equal ; and, if lefs, lefs.
Side 88 - 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 99 - BAC was proved to be equal to ACD : Therefore the whole angle ACE is equal to the two angles ABC, BAC...
Side 19 - From this it is manifest that if one angle of a triangle be equal to the other two it is a right angle, because the angle adjacent to it is equal to the same two ; (i.
Side 75 - Let AB be the fame multiple of C, that DE is of F : C is to F, as AB to DE. Becaufe AB is the fame multiple of C that DE is of F ; there are as many magnitudes in AB equal to C, as there are in DE equal...
Side 88 - ... reciprocally proportional, are equal to one another. Let AB, BC be equal parallelograms which have the angles at B equal, and let the sides DB, BE be placed in the same straight line ; wherefore also FB, BG are in one straight line (2.
Side 99 - BGC: for the same reason, whatever multiple the circumference EN is of the circumference EF, the same multiple is the angle EHN of the angle EHF: and if the circumference BL be equal to the circumference EN, the angle BGL is also equal to the angle EHN ; (in.
Side 106 - ... but BD, BE, which are in that plane, do each of them meet AB ; therefore each of the angles ABD, ABE is a right angle ; for the same reason, each of the angles CDB, CDE is a right angle: and because AB is equal to DE, and BD...
Side 73 - RATIOS that are the same to the same ratio, are the same to one another. Let A be to B as C is to D ; and as C to D, so let E be to F ; A is to B, as E to F.