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alſo angle ABC angle BAC baſe BC BC is equal BC is given becauſe the angle becauſe the ratio biſe&ted Book XI caſe cauſe circle ABCD circumference conſequently conſtruction cylinder demonſtrated deſcribed drawn EFGH equal angles equiangular equimultiples exceſs fides fimilar firſt fore given angle given in magnitude given in poſition given in ſpecies given magnitude given ratio given ſtraight line gnomon greater join laſt leſs likewiſe muſt oppoſite P R O parallel parallelepipeds parallelogram paſs thro perpendicular plane angles priſm Prop proportionals Propoſition pyramid reŽtangle contained right angles ſaid ſame manner ſame multiple ſame ratio ſame reaſon ſecond ſegment ſhall ſhewn ſides ſolid angle ſome ſpace ſphere ſquare of AC ſtraight line BC ſum T H E O theſe thoſe triangle ABC uſe wherefore
Page 157 - 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. Next, let each of the angles at C, F be not less than a right angle ; the triangle ABC is also, in this case, equiangular to the triangle DEF.
Page 1 - 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, fourths, &c.
Page 45 - If a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts, is equal to the rectangle contained by the two parts, together with the square of the aforesaid part.
Page 73 - THE straight line drawn at right angles to the diameter of a circle, from the extremity of...
Page 163 - 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 - ELF. Join BC, EF ; and because the circles ABC, DEF are equal, the straight lines drawn from their centres are equal: therefore the two sides BG, GC, are equal to the two EH, HF ; and the angle at G is equal to the angle at H ; therefore the base BC is equal (4.
Page 323 - Equiangular parallelograms have to one another the ratio which is compounded of the ratios of their sides.
Page 80 - EA : and because AD is equal to DC, and DE common to the triangles ADE, CDE, the two sides AD, DE are equal to the two CD, DE, each to each ; and the angle ADE is equal to the angle CDE, for each of them is a right angle ; therefore the base AE is equal (4.