The First Six Books: Together with the Eleventh and TwelfthJ. Balfour, 1781 - 520 pages |
From inside the book
Results 6-10 of 79
Page 111
... equiangular pentagon in a given circle . Let ABCDE be the given circle ; it is required to infcribe an equilateral and equiangular pentagon in the circle ABCDE . Defcribe an ifofceles triangle FGH , having each of the a 10. 4 . angles ...
... equiangular pentagon in a given circle . Let ABCDE be the given circle ; it is required to infcribe an equilateral and equiangular pentagon in the circle ABCDE . Defcribe an ifofceles triangle FGH , having each of the a 10. 4 . angles ...
Page 112
... equiangular ; because the cir- cumference AB is equal to the circumference DE : If to each be added BCD , the whole ... equiangular pentagon has been infcribed . Which was to be done . f 27. 3 . PROP . XII . PROB . a ' II . 4 . b 17.3 ...
... equiangular ; because the cir- cumference AB is equal to the circumference DE : If to each be added BCD , the whole ... equiangular pentagon has been infcribed . Which was to be done . f 27. 3 . PROP . XII . PROB . a ' II . 4 . b 17.3 ...
Page 113
... equiangular ; for , fince the angle FKC is equal to the angle FLC , and that the angle HKL is double of the angle FKC , and KLM double of FLC , as was before demonftrated , the angle HKL is equal to KLM : And in like manner it may be ...
... equiangular ; for , fince the angle FKC is equal to the angle FLC , and that the angle HKL is double of the angle FKC , and KLM double of FLC , as was before demonftrated , the angle HKL is equal to KLM : And in like manner it may be ...
Page 114
... equiangular penta- gon ; it is required to infcribe a circle in the pentagon ABCDE . Bifect the angles BCD , CDE by the ftraight lines CF , DF , and from the point F , in which they meet , draw the ftraight lines FB , FA , FE ...
... equiangular penta- gon ; it is required to infcribe a circle in the pentagon ABCDE . Bifect the angles BCD , CDE by the ftraight lines CF , DF , and from the point F , in which they meet , draw the ftraight lines FB , FA , FE ...
Page 115
... equiangular pentagon . Let ABCDE be the given equilateral and equiangular pen . tagon ; it is required to defcribe a circle about it . Bifect the angles BCD , CDE by the ftraight lines CF , FD , a 9. 1 . and from the point F , in which ...
... equiangular pentagon . Let ABCDE be the given equilateral and equiangular pen . tagon ; it is required to defcribe a circle about it . Bifect the angles BCD , CDE by the ftraight lines CF , FD , a 9. 1 . and from the point F , in which ...
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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.