Euclid's Elements: Or, Second Lessons in Geometry,in the Order of Simson's and Playfair's Editions ...Collins, Brother & Company, 1846 - 138 páginas |
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Página 18
... exceeds the interior C ( d ) ; much more does ABC exceed C. Therefore , in every side , & c . B Q. E. D. Recite ( a ) , p . 3 ; ( c ) , p . 5 ; ( b ) , pos . 1 ; ( d ) , p . 16 . 19 Th . The greater angle of every triangle is sub ...
... exceeds the interior C ( d ) ; much more does ABC exceed C. Therefore , in every side , & c . B Q. E. D. Recite ( a ) , p . 3 ; ( c ) , p . 5 ; ( b ) , pos . 1 ; ( d ) , p . 16 . 19 Th . The greater angle of every triangle is sub ...
Página 22
... exceeds its interior opposite angle EFG ( a ) , which were said to be equal . Therefore AB and CD , neither meet nor ... exceed BGH and GHD ( a ) : but AGH and BGH are equal to two right angles ( b ) ; therefore BGH and GHD are less than ...
... exceeds its interior opposite angle EFG ( a ) , which were said to be equal . Therefore AB and CD , neither meet nor ... exceed BGH and GHD ( a ) : but AGH and BGH are equal to two right angles ( b ) ; therefore BGH and GHD are less than ...
Página 42
... exceeds the side DE ( ƒ ) ; and so the part DF exceeds A E B the whole DE , which cannot be admitted ( g ) . Wherefore AB falls not without the circle ; and it cannot fall upon the circumference because it is a straight line ( h ) : it ...
... exceeds the side DE ( ƒ ) ; and so the part DF exceeds A E B the whole DE , which cannot be admitted ( g ) . Wherefore AB falls not without the circle ; and it cannot fall upon the circumference because it is a straight line ( h ) : it ...
Página 48
... exceeds that on KF ; therefore , of the two squares which remain , that on EH is less than that on EK ; and so , EH is less than EK : therefore BC is nearer the centre than FG ( ƒ ) . Wherefore , the diameter is the greatest chord , & c ...
... exceeds that on KF ; therefore , of the two squares which remain , that on EH is less than that on EK ; and so , EH is less than EK : therefore BC is nearer the centre than FG ( ƒ ) . Wherefore , the diameter is the greatest chord , & c ...
Página 51
... exceeds the interior ADB , ( c ) : therefore , since one an- gle cannot be equal to , and greater than another , the segments ACB , ADB are not similar . Therefore , upon one side of the same chord , & c . Q. E. D. Recite ( a ) p . 10 ...
... exceeds the interior ADB , ( c ) : therefore , since one an- gle cannot be equal to , and greater than another , the segments ACB , ADB are not similar . Therefore , upon one side of the same chord , & c . Q. E. D. Recite ( a ) p . 10 ...
Otras ediciones - Ver todas
Euclid's Elements, Or Second Lessons in Geometry, in the Order of Simson's ... D. M'Curdy Sin vista previa disponible - 2017 |
Euclid's Elements, Or Second Lessons in Geometry, in the Order of Simson's ... D. M'Curdy Sin vista previa disponible - 2017 |
Términos y frases comunes
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 equi equiangular equilateral equilateral polygon equimultiples exterior angle fore Geometry given circle given rectilineal given straight line gles gnomon greater half 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 rectilineal figure School segment semicircle similar similar triangles sine square of AC tangent touches the circle triangle ABC unequal Wherefore
Pasajes populares
Página 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.
Página 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.
Página 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.
Página 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.
Página 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...
Página 26 - Triangles upon equal bases, and between the same parallels, are equal to one another.
Página 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.
Página 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.
Página 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.
Página 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.