## 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 páginas |

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**opposite**fides of it , make the adjacent angles together equal to two right angles , thefe two ftraight lines fhall be in one and the fame ftraight line . At the point B in the straight line AB , let the two ftraight lines BC , BD upon ... Página 33

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**opposite**angle BCA ; which is impoffible ; wherefore BC is not unequal to EF , that is , b 16.1 . it is equal to it ; and AB is equal to DE ; therefore the two AB , BC are equal to the two DE , EF , each to each ; and they contain equal ... Página 71

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**opposite**to one of the equal angles in in each , is common to both ; therefore the other fides are e- qual ; AF therefore is equal to FB . Wherefore , if a straight e 26. 1 . line , & c . Q. E. D. PROP . IV . THEOR . Thin a circle two ... Página 94

... in the alternate fegment of the circle ; and becaufe ABCD is a quadrilateral figure in a circle , the

... in the alternate fegment of the circle ; and becaufe ABCD is a quadrilateral figure in a circle , the

**opposite**angles BAD , BCD are equal to E B F two two right angles ; therefore the angles DBF , DBE 94 THE ELEMENTS. Página 109

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**opposite**to these are equal , viz . FG to GE ; in the fame manner , it may be demon- ftrated that GH , GK are each of them equal to FG or GE ; therefore the four ftraight lines GE , GF , GH , GK , F are equal to one another ; and the ...### Otras ediciones - Ver todas

### Términos y frases comunes

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

### Pasajes populares

Página 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...

Página 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.

Página 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.

Página 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...

Página 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.

Página 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.

Página 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.

Página 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.

Página 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.

Página 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.