What people are saying - Write a review
We haven't found any reviews in the usual places.
Other editions - View all
AC is equal angle ABC angle BAC base BC BC is given bisected Book circle ABCD circumserence cone cylinder demonstrated described Desinition diameter draw drawn equal angles equiangular equimultiples Euclid excess faid fame altitude fame manner fame multiple fame ratio fame reason fame straight line fore given angle given in magnitude given in position given in species given magnitude given ratio given straight line gnomon greater join less Let ABC opposite parallel parallelogram pass thro perpendicular polygon prisms proportionals Proposition pyramid Q^E. D. PROP radius rectangle contained rectilineal sigure right angles segment shewn side BC similar sine sirst solid angle solid parallelepipeds square of AC straight line AB straight line BC tangent THEOR tiple triangle ABC vertex 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 469 - 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.