Euclid's Elements of Geometry, Bøker 1-6;Bok 11Henry Martyn Taylor The University Press, 1895 - 657 sider |
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Resultat 1-5 av 89
Side x
... shew that , if the method of superposition be used , we need not take as a postulate " all right angles are equal to one another , " but that we may deduce this theorem from other postulates which have been already assumed . Another new ...
... shew that , if the method of superposition be used , we need not take as a postulate " all right angles are equal to one another , " but that we may deduce this theorem from other postulates which have been already assumed . Another new ...
Side 21
... shew how Euclid with this self- imposed restriction solved the problem , which without such a restric- tion could have been solved more readily . After the problems in the first three propositions have been solved , we may assume that ...
... shew how Euclid with this self- imposed restriction solved the problem , which without such a restric- tion could have been solved more readily . After the problems in the first three propositions have been solved , we may assume that ...
Side 27
... Shew that , if the angles ABC and ACB at the base of an isosceles triangle be bisected by the straight lines BD and CD , DBC will be an isosceles triangle . 2. BAC is a triangle having the angle B double of the angle A. If BD bisect the ...
... Shew that , if the angles ABC and ACB at the base of an isosceles triangle be bisected by the straight lines BD and CD , DBC will be an isosceles triangle . 2. BAC is a triangle having the angle B double of the angle A. If BD bisect the ...
Side 31
... shew that the sides CB , CD are equal , and that the diagonal AC bisects the angle BCD . 7. ACB , ADB are two triangles on the same side of AB , such that AC is equal to BD , and AD is equal to BC , and AD and BC intersect at 0 : shew ...
... shew that the sides CB , CD are equal , and that the diagonal AC bisects the angle BCD . 7. ACB , ADB are two triangles on the same side of AB , such that AC is equal to BD , and AD is equal to BC , and AD and BC intersect at 0 : shew ...
Side 49
... Shew by joining the angular point A of a triangle to any point in the opposite side BC between B and C that the angles ABC , BCA are together less than two right angles . 3. Not more than two equal straight lines can be drawn from a ...
... Shew by joining the angular point A of a triangle to any point in the opposite side BC between B and C that the angles ABC , BCA are together less than two right angles . 3. Not more than two equal straight lines can be drawn from a ...
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Vanlige uttrykk og setninger
ABCD ADDITIONAL PROPOSITION AE is equal angle ABC angle ACB angle BAC angular points bisected bisectors centre of similitude chord circle ABC circumscribed circle coincide CONSTRUCTION Coroll cut the circle describe a circle diagonals diameter equal angles equiangular equilateral triangle Euclid EXERCISES figure fixed point given circle given point given straight line given triangle greater harmonic range hypotenuse inscribed circle intersect isosceles triangle Let ABC locus magnitudes meet middle points opposite sides pairs parallel parallelepiped parallelogram perpendicular plane angles polygon PROOF Prop quadrilateral radical axis radius rectangle contained regular polygon required to prove respectively rhombus right angles right-angled triangle shew side BC Similarly solid angle sphere square on AC straight line drawn straight line joining tangent tetrahedron theorem triangle ABC triangles are equal trihedral angle twice the rectangle vertex vertices Wherefore
Populære avsnitt
Side 70 - If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, viz.
Side 218 - The angle which an arc of a circle subtends at the centre is double that which it subtends at any point on the remaining part of the circumference.
Side 279 - If a straight line be bisected and produced to any point, the rectangle contained by the whole line thus produced and the part of it produced, together •with the square...
Side 148 - If a straight line be divided into any two parts, the squares on the whole line, and on one of the parts, are equal to twice the rectangle contained by the whole and that part, together with the square on the other part.
Side 137 - If there be two straight lines, one of which is divided into any number of parts, the rectangle contained by the two straight lines is equal to the rectangles contained by the undivided line, and the several parts of the divided line.
Side 372 - To find a mean proportional between two given straight lines. Let AB, BC be the two given straight lines ; it is required to find a mean proportional between them. Place AB, BC in a straight line, and upon AC describe the semicircle ADC, and from the point B draw (9.
Side 78 - ... the same side together equal to two right angles ; the two straight lines shall be parallel to one another.
Side 303 - To inscribe, an equilateral and equiangular pentagon in a given circle. Let ABCDE be the given circle. It is required to inscribe an equilateral...
Side 420 - PROPOSITION 5. The locus of a point, the ratio of whose distances from two given points is constant, is a circle*.
Side 300 - To inscribe a circle in a given square. Let ABCD be the given square ; it is required to inscribe a circle in ABCD.