The Elements of Plane and Solid GeometryLongmans, Green, and, Company, 1871 - 285 páginas |
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Página viii
... take place ; the proof adopted obviously supposes a simple transference of position , the same face re- maining uppermost throughout . It is true that the language employed is sufficiently general to include every case , but it is ...
... take place ; the proof adopted obviously supposes a simple transference of position , the same face re- maining uppermost throughout . It is true that the language employed is sufficiently general to include every case , but it is ...
Página 10
... take it for granted that the straight line BD must meet the side AC in some point . A And if C and D were two points situated on opposite C B sides of the straight line AB of indefinite length , we should take it for granted that the ...
... take it for granted that the straight line BD must meet the side AC in some point . A And if C and D were two points situated on opposite C B sides of the straight line AB of indefinite length , we should take it for granted that the ...
Página 12
... take away AB from each , therefore BC is greater than AC ~ AB . PROPOSITION 2 . If from the ends of a side of a triangle there be drawn two straight lines to a point within the triangle , these two lines shall be together less than the ...
... take away AB from each , therefore BC is greater than AC ~ AB . PROPOSITION 2 . If from the ends of a side of a triangle there be drawn two straight lines to a point within the triangle , these two lines shall be together less than the ...
Página 25
... take AD equal to AC , and on AC take AE equal to AB and draw DE meeting BC in F. Prove that AF bisects the angle BAC . 5. ABC is an isosceles triangle , of which the side AB is equal to AC , and D is the middle point of BC in AC a ...
... take AD equal to AC , and on AC take AE equal to AB and draw DE meeting BC in F. Prove that AF bisects the angle BAC . 5. ABC is an isosceles triangle , of which the side AB is equal to AC , and D is the middle point of BC in AC a ...
Página 28
... ( Prop . 11 ) . And because CE stands upon AB , therefore AEC + CEB is equal to two right angles ( Prop . 11 ) , therefore AEC + AED is equal to AEC + CEB . Take away the common angle AEC , therefore the angle 28 Geometry .
... ( Prop . 11 ) . And because CE stands upon AB , therefore AEC + CEB is equal to two right angles ( Prop . 11 ) , therefore AEC + AED is equal to AEC + CEB . Take away the common angle AEC , therefore the angle 28 Geometry .
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Términos y frases comunes
ABC and DEF ABCD adjacent angles angle ABC angle ACB angle BAC BC is equal centre circumference coincide common measure construction Corollary diameter dicular dihedral angle distance divided equal angles equal to AC equidistant exterior angle figure four right angles given angle given circle given plane given point given ratio given straight line greater homologous inscribed intersecting straight lines length less Let ABC line of intersection locus magnitudes meet the circle middle point multiple number of sides opposite sides parallelogram pentagon perpen perpendicular plane AC point F produced Prop PROPOSITION PROPOSITION 13 Prove radii radius rectangle regular polygon respectively equal rhombus right angles segments side BC similar triangles Similarly situated square straight line AB straight line BC subtended tangent triangle ABC triangle DEF
Pasajes populares
Página 15 - If two triangles have two sides of the one equal to two sides of the...
Página 101 - Through a given point to draw a straight line parallel to a given straight line. Let A be the given point, and BC the given straight line, it is required to draw a straight line through the point A, parallel to the line BC.
Página 126 - If a straight line be divided into any two parts, the square of the whole line is equal to the squares of the two parts, together with twice the rectangle contained by the parts.
Página 222 - The areas of two circles are to each other as the squares of their radii. For, if S and S' denote the areas, and R and R
Página 188 - If the angle of a triangle be divided into two equal angles, by a straight line which also cuts the base ; the segments of the base shall have the same ratio which the other sides of the triangle have to one another...
Página 204 - The areas of two triangles which have an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles. D c A' D' Hyp. In triangles ABC and A'B'C', ZA = ZA'. To prove AABC = ABxAC. A A'B'C' A'B'xA'C' Proof. Draw the altitudes BD and B'D'.
Página 14 - Upon the same base, and on the same side of it, there cannot be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise those which are terminated in the other extremity.
Página 12 - If, from the ends of the side of a triangle, there be drawn two straight lines to a point within the triangle, these shall be less than, the other two sides of the triangle, but shall contain a greater angle.
Página 161 - Ir there be any number of magnitudes, and as many others, which, taken two and two in order, have the same ratio ; the first shall have to the last of the first magnitudes the same ratio which the first of the others has to the last. NB This is usually cited by the words