The Elements of Euclid, Viz: The Errors, by which Theon, Or Others, Have Long Ago Vitiated These Books, are Corrected; and Some of Euclid's Demonstrations are Restored. Also the Book of Euclid's Data, in Like Manner Corrected. the first six books, together with the eleventh and twelfthJ. Nourse, London, and J. Balfour, Edinburgh, 1775 - 520 Seiten |
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Seite 11
... polygons , by more than four straight lines . XXIV . Of three fided figures , an equilateral triangle is that which has three equal fides . XXV . An ifofceles triangle , is that which has only two fides equal . Book 1 . A AA XXVI . A ...
... polygons , by more than four straight lines . XXIV . Of three fided figures , an equilateral triangle is that which has three equal fides . XXV . An ifofceles triangle , is that which has only two fides equal . Book 1 . A AA XXVI . A ...
Seite 176
... polygon ABCDE has to the polygon FGHKL the duplicate ratio of that which the fide AB has to the fide FG . Join BE , EC , GL , LH : And because the polygon ABCDE is fimilar fimilar to the polygon FGHKL , the angle BAE is 176 THE ELEMENTS.
... polygon ABCDE has to the polygon FGHKL the duplicate ratio of that which the fide AB has to the fide FG . Join BE , EC , GL , LH : And because the polygon ABCDE is fimilar fimilar to the polygon FGHKL , the angle BAE is 176 THE ELEMENTS.
Seite 177
... polygons have to one another , the antecedents being ABE , ÉBC , ECD , and the confequents FGL , LGH , LHK : And the polygon ABCDE has to the polygon FGHKL the du- plicate ratio of that which the fide AB has to the homologous fide FG ...
... polygons have to one another , the antecedents being ABE , ÉBC , ECD , and the confequents FGL , LGH , LHK : And the polygon ABCDE has to the polygon FGHKL the du- plicate ratio of that which the fide AB has to the homologous fide FG ...
Seite 178
... polygon ABCDE to the polygon FGHKL But the triangle ABE has to the triangle FGL , the duplicate ratio of that which the fide AB has to the homologous fide FG .. Therefore alfo the polygon ABCDE has to the poly- gon FGHKL the duplicate ...
... polygon ABCDE to the polygon FGHKL But the triangle ABE has to the triangle FGL , the duplicate ratio of that which the fide AB has to the homologous fide FG .. Therefore alfo the polygon ABCDE has to the poly- gon FGHKL the duplicate ...
Seite 214
... polygon BCDEF : F E D Therefore all the angles at the bafes of the triangles are toge ther ther greater than all the angles of the polygon : 214 THE ELEMENTS.
... polygon BCDEF : F E D Therefore all the angles at the bafes of the triangles are toge ther ther greater than all the angles of the polygon : 214 THE ELEMENTS.
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alfo alſo angle ABC angle BAC bafe baſe BC is equal BC is given becauſe the angle bifected Book XI cafe circle ABCD circumference cone confequently cylinder defcribed demonftrated diameter drawn equal angles equiangular equimultiples Euclid excefs faid fame manner fame multiple fame ratio fame reafon fecond fegment fhall fhewn fide BC fimilar firft firſt folid angle fome fore fphere fquare of AC ftraight line AB given angle given ftraight line given in fpecies given in magnitude given in pofition given magnitude given ratio gnomon greater join lefs likewife line BC oppofite parallel parallelepipeds parallelogram perpendicular polygon prifm propofition proportionals pyramid Q. E. D. PROP rectangle contained rectilineal figure right angles thefe THEOR theſe triangle ABC vertex wherefore
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Seite 32 - 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...
Seite 165 - D ; wherefore the remaining angle at C is equal to the remaining angle at F ; Therefore the triangle ABC is equiangular to the triangle DEF.
Seite 170 - 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.
Seite 10 - When several angles are at one point B, any ' one of them is expressed by three letters, of which ' the letter that is at the vertex of the angle, that is, at ' the point in which the straight lines that contain the ' angle meet one another, is put between the other two ' letters, and one of these two is...
Seite 55 - If a straight line be divided into two equal parts, and also into two unequal parts; the rectangle contained by the unequal parts, together with the square of the line between the points of section, is equal to the square of half the line.
Seite 32 - ... then shall the other sides be equal, each to each; and also the third angle of the one to the third angle of the other. Let ABC, DEF be two triangles which have the angles ABC, BCA equal to the angles DEF, EFD, viz.
Seite 45 - To describe a parallelogram that shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.
Seite 211 - AB shall be at right angles to the plane CK. Let any plane DE pass through AB, and let CE be the common section of the planes DE, CK ; take any point F in CE, from which draw FG in the plane DE at right D angles to CE ; and because AB is , perpendicular to the plane CK, therefore it is also perpendicular to every straight line in that plane meeting it (3.
Seite 38 - F, which is the common vertex of the triangles ; that is, together with four right angles. 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.
Seite 304 - Thus, if B be the extremity of the line AB, or the common extremity of the two lines AB, KB, this extremity is called a point, and has no length : For if it have any, this length must either be part of the length of the line AB, or of the line KB.