Elements of Geometry and Trigonometry |
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Page 10
... hence , a line has length only , without breadth or thickness . 6. If we regard a limited portion of a line , that which separates such portion from the part , at either extremity , is called a Point . But this mark of division forms no ...
... hence , a line has length only , without breadth or thickness . 6. If we regard a limited portion of a line , that which separates such portion from the part , at either extremity , is called a Point . But this mark of division forms no ...
Page 25
... Hence , the point D , falling at the same time in the two straight lines BA and CA , must fall at their intersection A : hence , the two triangles EDF , BAC , coincide with each other , and consequently , are equal ( a . 14 ) . Cor ...
... Hence , the point D , falling at the same time in the two straight lines BA and CA , must fall at their intersection A : hence , the two triangles EDF , BAC , coincide with each other , and consequently , are equal ( a . 14 ) . Cor ...
Page 30
... hence they are equal ( P. 5 ) . But the part cannot be equal to the whole ( A. 8 ) ; hence , there is no inequality between the sides BA and AC ; therefore , the triangle BAC is isosceles . PROPOSITION XIII . THEOREM . The greater side ...
... hence they are equal ( P. 5 ) . But the part cannot be equal to the whole ( A. 8 ) ; hence , there is no inequality between the sides BA and AC ; therefore , the triangle BAC is isosceles . PROPOSITION XIII . THEOREM . The greater side ...
Page 32
... hence , they would be equal ( A. 10 ) , and a part would be equal to the whole , which is impossible . PROPOSITION XV . THEOREM . If from a point without a straight line , a perpendicular be let fall on the line , and oblique lines be ...
... hence , they would be equal ( A. 10 ) , and a part would be equal to the whole , which is impossible . PROPOSITION XV . THEOREM . If from a point without a straight line , a perpendicular be let fall on the line , and oblique lines be ...
Page 37
... hence , H F we shall have two lines GB , GL , drawn through the same point G and parallel to CD , which is impossible ( A. 13 ) : hence , HGB + GHD is equal to two right angles . In the same manner it may be proved that HGA + GHC is ...
... hence , H F we shall have two lines GB , GL , drawn through the same point G and parallel to CD , which is impossible ( A. 13 ) : hence , HGB + GHD is equal to two right angles . In the same manner it may be proved that HGA + GHC is ...
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Elements of Geometry and Trigonometry: From the Works of A. M. Legendre Adrien Marie Legendre,Charles Davies No preview available - 2016 |
Common terms and phrases
adjacent angles altitude angle ACB angle BAD bisect centre chord circ circumference circumscribed common comp cone consequently convex surface cosē Cosine Cosine D Cotang cylinder diagonal diameter distance divided draw drawn equations equivalent feet figure find the area frustum given angle given line gles greater hence homologous homologous sides hypothenuse included angle inscribed circle intersect less Let ABC let fall logarithm magnitudes measured by half middle point number of sides opposite parallel parallelogram parallelopipedon pendicular perimeter perpendicular plane MN polyedral angle polyedron PROBLEM PROPOSITION pyramid quadrant radii radius ratio rectangle regular polygon right angles right-angled triangle Scholium secant segment side BC similar sinē sine slant height solidity sphere spherical polygon spherical triangle square described straight line Tang tangent THEOREM triangle ABC triangular prism triedral angles vertex vertices ΙΟ
Popular passages
Page 24 - If two triangles have two sides and the included angle of the one, equal to two sides and the included angle of the other, each to each, the two triangles will be equal.
Page 38 - That, if a straight line falling on two straight lines make the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles.
Page 227 - A spherical triangle is a portion of the surface of a sphere, bounded by three arcs of great circles.
Page 271 - The circumference of every circle is supposed to be divided into 360 equal parts, called degrees...
Page 43 - Hence, the interior angles plus four right angles, is equal to twice as many right angles as the polygon has sides, and consequently, equal to the sum of the interior angles plus the sum of the exterior angles.
Page 215 - The surface of a sphere is equal to the product of its diameter by the circumference of a great circle.
Page 107 - If two triangles have two angles of the one equal to two angles of the other, each to each, and also one side of the one equal to the corresponding side of the other, the triangles are congruent.
Page 93 - The area of a parallelogram is equal to the product of its base and altitude.
Page 231 - The angles of spherical triangles may be compared together, by means of the arcs of great circles described from their vertices as poles and included between their sides : hence it is easy to make an angle of this kind equal to a given angle.
Page 232 - F, be respectively poles of the sides BC, AC, AB. For, the point A being the pole of the arc EF, the distance AE is a 'quadrant ; the point C being the pole of the arc DE, the distance CE is likewise a quadrant : hence the point E is...