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 |
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Resultat 1-5 av 90
Side 8
... equal to BC . Because the point B is the center of the circle CGH , therefore BC is equal to BG ; ( def . 15. ) and because D is the center of the circle GKL , therefore DL is equal to DG , and DA , DB parts of them are equal ; ( 1. 1 ...
... equal to BC . Because the point B is the center of the circle CGH , therefore BC is equal to BG ; ( def . 15. ) and because D is the center of the circle GKL , therefore DL is equal to DG , and DA , DB parts of them are equal ; ( 1. 1 ...
Side 9
... equal sides are opposite shall be equal , each to each , viz . the angle ABC to the angle DEF , and the angle ACB to the angle DFE . D B C E F For , if the triangle ABC be applied to the triangle DEF , so that the point A may be on D ...
... equal sides are opposite shall be equal , each to each , viz . the angle ABC to the angle DEF , and the angle ACB to the angle DFE . D B C E F For , if the triangle ABC be applied to the triangle DEF , so that the point A may be on D ...
Side 10
... equal to the triangle AGB , also the remaining angles of the one are equal to the remaining angles of the other , each to each , to which the equal sides are opposite ; viz . the angle ACF to the angle ABG , and the angle AFC to the ...
... equal to the triangle AGB , also the remaining angles of the one are equal to the remaining angles of the other , each to each , to which the equal sides are opposite ; viz . the angle ACF to the angle ABG , and the angle AFC to the ...
Side 11
... equal to CA the less , ( 1. 3. ) and join DC . Then , in the triangles DBC , ABC , because DB is equal to AC , and BC is common to both triangles , the two sides DB , BC are equal to the two sides AC , CB , each to each ; and the angle ...
... equal to CA the less , ( 1. 3. ) and join DC . Then , in the triangles DBC , ABC , because DB is equal to AC , and BC is common to both triangles , the two sides DB , BC are equal to the two sides AC , CB , each to each ; and the angle ...
Side 12
... equal to BD in the triangle BCD , therefore the angle BDC is equal to the angle BCD , ( 1. 5. ) but the angle BDC has been proved greater than BCD , wherefore the angle BDC is both equal to , and greater than the angle BCD ; ⚫ which is ...
... equal to BD in the triangle BCD , therefore the angle BDC is equal to the angle BCD , ( 1. 5. ) but the angle BDC has been proved greater than BCD , wherefore the angle BDC is both equal to , and greater than the angle BCD ; ⚫ which is ...
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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.