## 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 |

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Resultater 1-5 af 96

Side 6

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**cafe**in which the folid Angles are contained by no more than three plain Angles ; nor of this**Cafe**is there any Demonftration in the Elements we now have , though it be quite neceffary there fhould be one . Now , upon the 10th ... Side 19

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**cafe**in which the vertex of each of the tri- angles is without the other triangle , becaufe AC is equal to AD , the angle ACD is equal to the angle ADC : But the angle ACD is greater than the angle BCD ; therefore the angle A ADC is ... Side 20

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**cafe**in which the ver- tex of one triangle is upon a fide of the other needs no de- monftration . a B Therefore upon the fame bafe , and on the fame fide of it , there cannot be two triangles that have their fides which are terminated ... Side 33

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**cafe**, the other fides fhall be equal , AC to DF , and BC to FF ; and alfo the third angle BAC to the third EDF . B a D HC E For , if BC be not equal to EF , let BC be the greater of them , and make BH equal to EF , and join AH ; and ... Side 85

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**cafe**, that the angle GEC is double of the angle GDC , and that GEB a part of the first is double of GDB a part of the other ; therefore the re- G maining angle BEC is double of the remaining angle BDC . Therefore the angle at the ...### 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.