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 38
Side 11
... THEOR . Fone fide of a triangle be produced , the outward angle will be greater than either of the inward oppofite angles . Let ABC be a triangle , and one of its fides BC be produced to D , the outward angle ACD will be greater than ...
... THEOR . Fone fide of a triangle be produced , the outward angle will be greater than either of the inward oppofite angles . Let ABC be a triangle , and one of its fides BC be produced to D , the outward angle ACD will be greater than ...
Side 20
... THEOR . HE oppofite fides and oppofite angles of every parallelo- gram are equal ; and the diameter divides it into two equal parts . Let ABCD be a parallelogram , the oppofite fides AB , CD , AC , BD , are equal ; the angle CAB equal ...
... THEOR . HE oppofite fides and oppofite angles of every parallelo- gram are equal ; and the diameter divides it into two equal parts . Let ABCD be a parallelogram , the oppofite fides AB , CD , AC , BD , are equal ; the angle CAB equal ...
Side 21
... THEOR . QUAL triangles , conftitute upon the fame bafe , on the fame fide , are between the fame parallels . Let the equal triangles ABC , DBC , be conftitute upon the fame bafe , BC , on the fame fide ; the right line AD , that joins ...
... THEOR . QUAL triangles , conftitute upon the fame bafe , on the fame fide , are between the fame parallels . Let the equal triangles ABC , DBC , be conftitute upon the fame bafe , BC , on the fame fide ; the right line AD , that joins ...
Side 29
... THEOR . F a right line be any how cut , the fquare of the whole line , and one of the parts , is equal to twice the rectangle contained by the whole line , and said part , together with the square of the o- ther part . Let the right ...
... THEOR . F a right line be any how cut , the fquare of the whole line , and one of the parts , is equal to twice the rectangle contained by the whole line , and said part , together with the square of the o- ther part . Let the right ...
Side 30
... THEOR . a right line be cut into two parts , four times the rectangle under the whole line , and one of the parts , together with the Square of the other part , are equal to the fquare of the whole line , and the first part taken as the ...
... THEOR . a right line be cut into two parts , four times the rectangle under the whole line , and one of the parts , together with the Square of the other part , are equal to the fquare of the whole line , and the first part taken as 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.