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 - Geometry - 264 pages |
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Results 1-5 of 10
Page 91
... polygon is to polygon in the duplicate ratio of one homologous fide to the o- ther . Let ABCDE , FGHKL , be fimilar polygons , and AB , FG , two homologous fides ; join BE , EC , GL , LH ; then the number of triangles in the polygon ...
... polygon is to polygon in the duplicate ratio of one homologous fide to the o- ther . Let ABCDE , FGHKL , be fimilar polygons , and AB , FG , two homologous fides ; join BE , EC , GL , LH ; then the number of triangles in the polygon ...
Page 92
... polygon to polygon in the duplicate ra- tio of one homologous fide to another . Wherefore , & c . C COR . Hence , if three right lines are proportional , the poly- gon defcribed on the firft is to the fimilar polygon described on the ...
... polygon to polygon in the duplicate ra- tio of one homologous fide to another . Wherefore , & c . C COR . Hence , if three right lines are proportional , the poly- gon defcribed on the firft is to the fimilar polygon described on the ...
Page 130
... polygon ABCDE is to FGHKL as BM fquare is to GN square . For , Join BE , AM , GL , FN ; then , because the polygons ... polygon ABCDE is to the polygon FGHKL in the duplicate ratio of AB to GFf ; but the triangle ABM is to the triangle ...
... polygon ABCDE is to FGHKL as BM fquare is to GN square . For , Join BE , AM , GL , FN ; then , because the polygons ... polygon ABCDE is to the polygon FGHKL in the duplicate ratio of AB to GFf ; but the triangle ABM is to the triangle ...
Page 132
... polygon AXBOCPDR to the polygon EKFLGMHN , fo is the circle ABCD to the figure S ; but the circle ABCD is greater than the polygon in it ; therefore the figure S is greater than the polygon EKFLGMHNf ; but it is lefs ; which is abfurd ...
... polygon AXBOCPDR to the polygon EKFLGMHN , fo is the circle ABCD to the figure S ; but the circle ABCD is greater than the polygon in it ; therefore the figure S is greater than the polygon EKFLGMHNf ; but it is lefs ; which is abfurd ...
Page 138
... polygon AEBFCGDH be infcribed in the circle ABCD , and let the fmall fegments AE , EB , BF , FC , CG , GD , DH , HA , the excefs by which the circle ex- ceeds the polygon , be less than any affigned magnitude ; and , up- on the circle ...
... polygon AEBFCGDH be infcribed in the circle ABCD , and let the fmall fegments AE , EB , BF , FC , CG , GD , DH , HA , the excefs by which the circle ex- ceeds the polygon , be less than any affigned magnitude ; and , up- on the circle ...
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Common terms and phrases
ABCM angle ABC angle BAC arch bafe baſe becauſe bifect Book XI circle ABCD circle EFGH circumference cofine common fection cone contained cylinder defcribe DEFH diameter draw drawn equal angles equal to AC equiangular equilateral equimultiples fame altitude fame multiple fame plain fame proportion fame reafon fecond fegment femicircle fides fimilar folid angle fome fore fquare of AC fubtending given right line greater infcribed join lefs leſs Let ABC magnitudes oppofite parallel parallelogram perpendicular plain angles plain paffing polygon prifms Prop pyramid rectangle right angles right line AB right lined figure Secant Sine ſphere ſquare Tang tangent thefe THEOR theſe triangle ABC triplicate ratio Wherefore whofe ΙΟ
Popular passages
Page 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...
Page 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.
Page 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.
Page 99 - BAC was proved to be equal to ACD : Therefore the whole angle ACE is equal to the two angles ABC, BAC...
Page 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.
Page 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...
Page 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.
Page 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.
Page 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...
Page 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.