Euclid's Elements of Geometry: Chiefly from the Text of Dr. Simson, with Appendix by Thos. Kirkland. the first six booksA. Miller & Company, 1876 - 403 páginas |
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Página 7
... circle CĠH , ( post . 3. ) cutting DF in the point G : and from the center D , at the distance DG , describe the circle GKL , cutting AE in the point L. Then the straight line AL shall be equal to BC BOOK I. PROP . I , II . 7.
... circle CĠH , ( post . 3. ) cutting DF in the point G : and from the center D , at the distance DG , describe the circle GKL , cutting AE in the point L. Then the straight line AL shall be equal to BC BOOK I. PROP . I , II . 7.
Página 11
... BDC is both equal to , and greater than the angle BCD ; which is impossible . Secondly . Let the vertex D of the triangle ADB fall within the riangle ACB . B C F D H Produce AC to E , and AD to F , BOOK I. PROP . VI . VII . 11.
... BDC is both equal to , and greater than the angle BCD ; which is impossible . Secondly . Let the vertex D of the triangle ADB fall within the riangle ACB . B C F D H Produce AC to E , and AD to F , BOOK I. PROP . VI . VII . 11.
Página 13
... . ) therefore the base AD is equal to the base BD . ( 1. 4. ) Wherefore the straight line AB is divided into two equal parts in the point D. Q.E.F. PROPOSITION XI . PROBLEM . To draw a straight line BOOK I. PROP . IX , X. 13.
... . ) therefore the base AD is equal to the base BD . ( 1. 4. ) Wherefore the straight line AB is divided into two equal parts in the point D. Q.E.F. PROPOSITION XI . PROBLEM . To draw a straight line BOOK I. PROP . IX , X. 13.
Página 15
... CBA be not equal to the angle ABD , from the point B draw BE at right angles to CD . ( 1. 11. ) Then the angles CBE , EBD are two right angles . ( def . 10. ) And because the angle CBE is equal to the angles BOOK I. PROP . XII , XIII . 15.
... CBA be not equal to the angle ABD , from the point B draw BE at right angles to CD . ( 1. 11. ) Then the angles CBE , EBD are two right angles . ( def . 10. ) And because the angle CBE is equal to the angles BOOK I. PROP . XII , XIII . 15.
Página 17
... interior opposite angles CBA or BAC . A F B E G Bisect AC in E , ( 1. 10. ) and join BE ; produce BE to F , making EF equal to BE , ( 1. 3. ) and join FC . Because AE is equal to EC , and BE to BOOK I. PROP . XV , XVI . 17.
... interior opposite angles CBA or BAC . A F B E G Bisect AC in E , ( 1. 10. ) and join BE ; produce BE to F , making EF equal to BE , ( 1. 3. ) and join FC . Because AE is equal to EC , and BE to BOOK I. PROP . XV , XVI . 17.
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Términos y frases comunes
A₁ ABCD AC is equal Algebraically angle ABC angle ACB angle BAC angle equal Apply Euc base BC chord circle ABC constr describe a circle diagonals diameter divided draw equal angles equiangular equilateral triangle equimultiples Euclid exterior angle Geometrical given circle given line given point given straight line gnomon greater hypotenuse inscribed intersection isosceles triangle less Let ABC line BC lines be drawn multiple opposite angles parallelogram parallelopiped pentagon perpendicular plane polygon produced Prop proportionals proved Q.E.D. PROPOSITION quadrilateral quadrilateral figure radius ratio rectangle contained rectilineal figure remaining angle right angles right-angled triangle segment semicircle shew shewn similar similar triangles solid angle square on AC tangent THEOREM touch the circle triangle ABC twice the rectangle vertex vertical angle wherefore
Pasajes populares
Página 93 - If a straight line be bisected and produced to any point, the square on the whole line thus produced, and the square on the part of it produced, are together double of the square on half the line bisected, and of the square on the line made up of the half and the part produced. Let the straight line AB be bisected in C, and produced to D ; The squares on AD and DB shall be together double of the squares on AC and CD. CONSTRUCTION. — From the point C draw CE at right angles to AB, and make it equal...
Página 118 - Guido, with a burnt stick in his hand, demonstrating on the smooth paving-stones of the path, that the square on the hypotenuse of a right-angled triangle is equal to the sum of the squares on the other two sides.
Página 145 - If a straight line touch a circle, and from the point of contact a straight line be drawn cutting the circle ; the angles which this line makes with the line touching the circle, shall be equal to the angles which are in the alternate segments of the circle.
Página 88 - If a straight line be divided into two equal parts and also into two unequal parts, the rectangle contained by the unequal parts, together with the square on the line between the points of section, is equal to the square on half the line.
Página 26 - ... upon the same side together equal to two right angles, the two straight lines shall be parallel to one another.
Página 36 - To a given straight line to apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.
Página 144 - The angle in a semicircle is a right angle ; the angle in a segment greater than a semicircle is less than a right angle ; and the angle in a segment less than a semicircle is greater than a right angle.
Página 92 - If a straight line be divided into two equal, and also into two unequal parts, the squares on the two unequal parts are together double of the square on half the line and of the square on the line between the points of section. Let the straight line AB be divided into two equal parts...
Página xv - In every triangle, the square of the side subtending either of the acute angles is less than the squares of the sides containing that angle, by twice the rectangle contained by either of these sides, and the straight line intercepted between the perpendicular let fall upon it from the opposite angle, and the acute angle.
Página 67 - A proposition affirming the possibility of finding such conditions as will render a certain problem indeterminate or capable of innumerable solutions.