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 sider |
Inni boken
Resultat 1-5 av 79
Side 3
... line , which is called the circumference , and is such that all straight lines drawn from a certain point within the figure to the circumference , are equal to one another . XVI . And this point is called the center of the circle . XVII ...
... line , which is called the circumference , and is such that all straight lines drawn from a certain point within the figure to the circumference , are equal to one another . XVI . And this point is called the center of the circle . XVII ...
Side 14
... draw a straight line at right angles to a given straight line , from a given point in the same . Let AB be the given straight line , and Ca given point in it . It is required to draw a straight line from the point C at right angles to ...
... draw a straight line at right angles to a given straight line , from a given point in the same . Let AB be the given straight line , and Ca given point in it . It is required to draw a straight line from the point C at right angles to ...
Side 35
... line drawn from A but AD is parallel to BC ; AD is therefore parallel to BC . Wherefore , equal triangles upon , & c . Q. E.D. PROPOSITION XL . THEOREM . Equal triangles upon equal bases in the same straight line , and towards the same ...
... line drawn from A but AD is parallel to BC ; AD is therefore parallel to BC . Wherefore , equal triangles upon , & c . Q. E.D. PROPOSITION XL . THEOREM . Equal triangles upon equal bases in the same straight line , and towards the same ...
Side 53
... line . This problem then becomes the same as Prob . xI , which may be regarded as drawing a line which bisects an ... drawn from the point to the line . From this Prop . it follows that only one perpendicular can be drawn from a given ...
... line . This problem then becomes the same as Prob . xI , which may be regarded as drawing a line which bisects an ... drawn from the point to the line . From this Prop . it follows that only one perpendicular can be drawn from a given ...
Side 71
... line EAF drawn through A , so that EA is equal to AF ; P • H E C through A draw AG parallel to BC , and GH parallel to EF . Then AGHE is a parallelogram , wherefore AE is equal to GH , but EA is equal to AF by hypothesis ; therefore GH ...
... line EAF drawn through A , so that EA is equal to AF ; P • H E C through A draw AG parallel to BC , and GH parallel to EF . Then AGHE is a parallelogram , wherefore AE is equal to GH , but EA is equal to AF by hypothesis ; therefore GH ...
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Vanlige uttrykk og setninger
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
Populære avsnitt
Side 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...
Side 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.
Side 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.
Side 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.
Side 26 - ... upon the same side together equal to two right angles, the two straight lines shall be parallel to one another.
Side 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.
Side 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.
Side 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...
Side 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.
Side 67 - A proposition affirming the possibility of finding such conditions as will render a certain problem indeterminate or capable of innumerable solutions.