## 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 sider |

### Fra bogen

Resultater 6-10 af 100

Side 22

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**BC**, CD , each to each ; and the angle ACD is equal to the angle BCD ; therefore the bafe AD is equal to the bafe DB , and the ftraight**line**B is divided into two equal parts in the point D. Which was to be done , C A A : ע B See N. a 3. . Side 36

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**line**, & c . Q. E. D. 13. I. a 29. I. b 27.1 . I. PROP . XXX . THEOR .. TRAIGHT**lines**which are parallel to the fame ...**BC**the given straight**line**; it is required to draw a ftraight**line**. through the point A , parallel to the ftraight ... Side 37

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**line BC**. Which was to be done . PROP . XXXII . THEOR . IF a fide of any triangle be produced , the exterior angle is equal to the two interior and oppofite angles ; and the three interior angles of every triangle are equal to two right ... Side 39

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**line BC**meets the two ftraight lines b 4. I. AC , BD , and makes the alternate angles ACB , CBD equal to one another , AC is parallel to BD ; and it was fhown to be c 27. I , equal to it . Therefore ftraight lines , & c . Q. E. D. PRO P ... Side 43

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**BC**. In the fame manner , it can be demonstrated that no other**line**but AD is parallel to**BC**; AD is therefore parallel to it . Wherefore equal triangles upon , & c . Q. E. D. B PROP . XL . THEOR . Estraight**line**, and towards the fame ...### Andre udgaver - Se alle

### Almindelige termer og sætninger

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

### Populære passager

Side 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...

Side 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.

Side 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.

Side 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...

Side 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.

Side 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.

Side 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.

Side 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.

Side 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.

Side 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.