## The First Six Books: Together with the Eleventh and TwelfthJ. Balfour, 1781 - 520 pages |

### From inside the book

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**EFGH**is equal to the fame EBCH : Therefore alfo the parallelogram ABCD is equal to**EFGH**. Wherefore parallelograms , & c . Q. E. D. TR PRO P. XXXVII . с THEOR . RIANGLES upon the fame bafe , and between the fame parallels , are equal to ... Page 147

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**E. F. G. H.**fo G to H : A fhall be to D , as E to H. Because A , B , C are three magnitudes , and E , F , G other three , which , taken two and two , have the fame ratio ; by the foregoing cafe , A is to C , as E to G : But C is to D ... Page 254

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**EFGH**, or greater than it * . First , let it be to a space S less than the circle**EFGH**; and in the circle**EFGH**defcribe the fquare**EFGH**: This fquare is greater than half of the circle**EFGH**; becaufe if , through the points E , F , G ... Page 255

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**EFGH**is half of the fquare defcribed about the circle ; and the circle is less than the fquare described about it ; there- fore the fquare**EFGH**is greater than half of the circle . Di- vide the circumferences EF , FG , GH , HE , each ... Page 256

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**EFGH**: For , if poffible , let it be fo to I , a space greater than the circle**EFGH**: Therefore , inversely , as the fquare of FH to the fquare of BD , fo is the space T to X A B E R K N DF H L M G T S the circle ABCD . But as the space ...### Other editions - View all

### Common terms and phrases

alfo alſo angle ABC angle BAC bafe baſe BC is equal BC is given becauſe the angle becauſe the ratio bifected Book XI cafe centre circle ABCD circumference cone confequently cylinder defcribed demonftrated drawn EFGH equal angles equiangular equimultiples Euclid excefs faid fame manner fame multiple fame ratio fame reafon fecond fegment fide BC fides fimilar firft firſt folid angle fome fore fphere fquare of AC ftraight line AB ftraight line BC given angle given ftraight line given in fpecies given in magnitude given in pofition given magnitude given ratio gnomon greater join lefs likewife oppofite parallel parallelepipeds parallelogram perpendicular plane angles prifms PROP propofition pyramid ratio of BC rectangle contained rectilineal figure right angles ſquare thefe THEOR theſe triangle ABC wherefore

### Popular passages

Page 472 - The circumference of every circle is supposed to be divided into 360 equal parts, called degrees ; and each degree into 60 equal parts, called minutes ; and each minute into 60 equal parts, called seconds ; and these into thirds, &c.

Page 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.

Page 81 - THE straight line drawn at right angles to the diameter of a circle, from the extremity of...

Page 105 - DEF are likewise equal (13. i.) to two right angles ; therefore the angles AKB, AMB are equal to the angles DEG, DEF, of which AKB is equal to DEG ; wherefore the remaining angle AMB is equal to the remaining angle DEF.

Page 167 - AC the same multiple of AD, that AB is of the part which is to be cut off from it : join BC, and draw DE parallel to it : then AE is the part required to be cut off.

Page 10 - When a straight line standing on another straight line makes the adjacent angles equal to one another, each of the angles is called a right angle; and the straight line which stands on the other is called a perpendicular to it.

Page 62 - AB be the given straight line ; it is required to divide it into two parts, so that the rectangle contained by the whole, and one of the parts, shall be equal to the square of the other part.

Page 112 - To describe an equilateral and equiangular pentagon about a given circle. • Let ABCDE be the given circle; it is required to describe an equilateral and equiangular pentagon about the circle ABCDE. Let the angles of a pentagon, inscribed in the circle...

Page 200 - 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...

Page 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.