Euclid's Elements: Or, Second Lessons in Geometry,in the Order of Simson's and Playfair's Editions ... |
From inside the book
Results 6-10 of 17
Page 59
... tangent AD , equal to C ; through the centre O , draw DF , B cutting the circle in E , F. Then the rectangle DEX DF equals the square of AD ( a ) , or its equal C ; and the difference of DE and DF is EF , which is equal to AB ...
... tangent AD , equal to C ; through the centre O , draw DF , B cutting the circle in E , F. Then the rectangle DEX DF equals the square of AD ( a ) , or its equal C ; and the difference of DE and DF is EF , which is equal to AB ...
Page 61
... tangent GAH ( a ) ; at A , the point of contact , make the angle HAC equal to the angle E , also the angle GAB equal ... tangents through the points A , B , с ( c ) , to meet in the points M , L , N. There fore LM , LN , MN , meet ...
... tangent GAH ( a ) ; at A , the point of contact , make the angle HAC equal to the angle E , also the angle GAB equal ... tangents through the points A , B , с ( c ) , to meet in the points M , L , N. There fore LM , LN , MN , meet ...
Page 63
... tangents ( b ) to the circle , through the points A , B , C , D , at right angles to the diameters , meeting in the points F , G , H , K. Now , because of the equal alternate angles ( c ) GBD , BDK , the side GH is parallel to FK : for ...
... tangents ( b ) to the circle , through the points A , B , C , D , at right angles to the diameters , meeting in the points F , G , H , K. Now , because of the equal alternate angles ( c ) GBD , BDK , the side GH is parallel to FK : for ...
Page 65
... tangents GH , HK , KL , LM , MG ( 6 ) . If these tangents be equal , and their angles B equal , the required pentagon is described about the circle . Find the centre F , and draw the radii FB , FC , FD ; join FK , FL . G A E M F D C L K ...
... tangents GH , HK , KL , LM , MG ( 6 ) . If these tangents be equal , and their angles B equal , the required pentagon is described about the circle . Find the centre F , and draw the radii FB , FC , FD ; join FK , FL . G A E M F D C L K ...
Page 68
... tangents be drawn , touch- ing the circle at right angles to the diameters , the angles of the in- scribed hexagon will be in the sides of the one described . Def . 2 , 4 , b . 4 . Tangent literally means they touch . See def . 2 , b ...
... tangents be drawn , touch- ing the circle at right angles to the diameters , the angles of the in- scribed hexagon will be in the sides of the one described . Def . 2 , 4 , b . 4 . Tangent literally means they touch . See def . 2 , b ...
Other editions - View all
Euclid's Elements, Or Second Lessons in Geometry, in the Order of Simson's ... D. M'Curdy No preview available - 2017 |
Euclid's Elements, Or Second Lessons in Geometry, in the Order of Simson's ... D. M'Curdy No preview available - 2017 |
Common terms and phrases
ABCD alternate angles angle ACD angles ABC angles equal antecedents Argument base BC bisected centre Chart chord circle ABC circumference Constr Denison Olmsted diameter draw drawn equal angles equal arcs equal radii equal sides equals the squares equiangular equilateral equilateral polygon equimultiples exterior angle fore Geometry given circle given rectilineal given straight line given triangle gles gnomon greater inscribed isosceles isosceles triangle join less meet multiple opposite angles parallelogram parallelopipeds pentagon perimeter perpendicular plane polygon produced propositions Q. E. D. Recite radius ratio rectangle rectangle contained rectilineal figure School secant segment semicircle similar sine square of AC tangent third touches the circle triangle ABC unequal Wherefore
Popular passages
Page 90 - If two triangles have one angle of the one equal to one angle of the other, and the sides about the equal angles proportionals, the triangles shall be equiangular, and shall have those angles equal which are opposite to the homologous sides.
Page 117 - In the same way it may be proved that a : b : : sin. A : sin. B, and these two proportions may be written a : 6 : c : : sin. A : sin. B : sin. C. THEOREM III. t8. In any plane triangle, the sum of any two sides is to their difference as the tangent of half the sum of the opposite angles is to the tangent of half their difference. By Theorem II. we have a : b : : sin. A : sin. B.
Page 92 - IN a right-angled triangle, if a perpendicular be drawn from the right angle to the base, the triangles on each side of it are similar to the whole triangle, and to one another.
Page 79 - THEOREM. lf the first has to the second the same ratio which the third has to the fourth, but the third to the fourth, a greater ratio than the fifth has to the sixth ; the first shall also have to the second a greater ratio than the fifth, has to the sixth.
Page 87 - If a straight line be drawn parallel to one of the sides of a triangle, it shall cut the other sides, or those sides produced, proportionally...
Page 26 - Triangles upon equal bases, and between the same parallels, are equal to one another.
Page 94 - Equal parallelograms which have one angle of the one equal to one angle of the other, have their sides about the equal angles reciprocally proportional ; and parallelograms that have one angle of the one equal to one angle of the other, and their sides about the equal angles reciprocally proportional, are equal to one another.
Page 12 - THE angles at the base of an isosceles triangle are equal to one another : and, if the equal sides be produced, the angles upon the other side of the base shall be equal.
Page 133 - If a straight line stand at right angles to each of two straight lines at the point of their intersection, it shall also be at right angles to the plane which passes through them, that is, to the plane in which they are.
Page 13 - AB be the greater, and from it cut (3. 1.) off DB equal to AC the less, and join DC ; therefore, because A in the triangles DBC, ACB, DB is equal to AC, and BC common to both, the two sides DB, BC are equal to the two AC, CB. each to each ; and the angle DBC is equal to the angle ACB; therefore the base DC is equal to the base AB, and the triangle DBC is< equal to the triangle (4. 1.) ACB, the less to 'the greater; which is absurd.