The Elements of Plane and Solid GeometryLongmans, Green, and, Company, 1871 - 285 sider |
Inni boken
Resultat 1-5 av 82
Side 6
... respectively , provided they exactly fit into each other , so that within the space in question there is no portion which is not completely filled up , then it is clear that there is a region within this space which belongs as much to ...
... respectively , provided they exactly fit into each other , so that within the space in question there is no portion which is not completely filled up , then it is clear that there is a region within this space which belongs as much to ...
Side 14
... respectively , therefore BA + CA is equal to BD + CD , which is impossible . Next , let the vertex of each triangle be without the other , as in Fig . 5 . Because AB is equal to DB , and DC is equal to AC , therefore AB + DC is equal to ...
... respectively , therefore BA + CA is equal to BD + CD , which is impossible . Next , let the vertex of each triangle be without the other , as in Fig . 5 . Because AB is equal to DB , and DC is equal to AC , therefore AB + DC is equal to ...
Side 17
... respectively , therefore either the angle ABC or the angle ACB may be made to coincide with the angle DEF , and therefore the angle ABC must be equal to the angle ACB , therefore the angles at the base of an isosceles triangle are equal ...
... respectively , therefore either the angle ABC or the angle ACB may be made to coincide with the angle DEF , and therefore the angle ABC must be equal to the angle ACB , therefore the angles at the base of an isosceles triangle are equal ...
Side 18
... respectively , therefore the angle BAC must be equal to the angle EDF . ( Def . 11. ) Therefore the triangles ABC and DEF are equal in all their parts . Next let the triangle DEF be situated as in Fig . 14 . In this case when the points ...
... respectively , therefore the angle BAC must be equal to the angle EDF . ( Def . 11. ) Therefore the triangles ABC and DEF are equal in all their parts . Next let the triangle DEF be situated as in Fig . 14 . In this case when the points ...
Side 19
... respectively with either DE and DF , in the position of Fig . 16 , or with DF and DE , in the position of Fig . 17 ... respectively , to the three sides of the other , the triangles shall be equal in all their parts . Let ABC and DEF be ...
... respectively with either DE and DF , in the position of Fig . 16 , or with DF and DE , in the position of Fig . 17 ... respectively , to the three sides of the other , the triangles shall be equal in all their parts . Let ABC and DEF be ...
Andre utgaver - Vis alle
Vanlige uttrykk og setninger
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
Populære avsnitt
Side 15 - If two triangles have two sides of the one equal to two sides of the...
Side 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.
Side 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.
Side 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
Side 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...
Side 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'.
Side 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.
Side 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.
Side 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