The Elements of Plane and Solid GeometryLongmans, Green, and, Company, 1871 - 285 sider |
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Side xviii
... REGULAR POLYGONS IV . PROBLEMS OF CONSTRUCTION CONNECTED WITH RATIO AND PROPORTION BOOK VII . 175 200 218 227 On Planes , and Lines in Space . I. MISCELLANEOUS PROPOSITIONS 242 II . ON THE PERPENDICULARS AND OBLIQUES TO PLANES 255 III ...
... REGULAR POLYGONS IV . PROBLEMS OF CONSTRUCTION CONNECTED WITH RATIO AND PROPORTION BOOK VII . 175 200 218 227 On Planes , and Lines in Space . I. MISCELLANEOUS PROPOSITIONS 242 II . ON THE PERPENDICULARS AND OBLIQUES TO PLANES 255 III ...
Side 219
... regular polygon . Let ABCDE be any regular pentagon , then Ist . It is always possible to de- scribe a circle about ABCDE . Bisect each of the angles at A and B by the straight lines AO , and BO , meeting at O , and join OE . Because ...
... regular polygon . Let ABCDE be any regular pentagon , then Ist . It is always possible to de- scribe a circle about ABCDE . Bisect each of the angles at A and B by the straight lines AO , and BO , meeting at O , and join OE . Because ...
Side 221
... regular polygon coincide . Corollary 2. - If there be two regular polygons , the common centre of the inscribed and circumscribing circle of the one is homologous to the common centre of the inscribed and circumscribing circle of the ...
... regular polygon coincide . Corollary 2. - If there be two regular polygons , the common centre of the inscribed and circumscribing circle of the one is homologous to the common centre of the inscribed and circumscribing circle of the ...
Side 224
... regular polygon be inscribed in each circle , the number of sides in each polygon being the same . Because the polygons are regular polygons , with the same number of sides in each , therefore they are similar ; therefore the ratio of ...
... regular polygon be inscribed in each circle , the number of sides in each polygon being the same . Because the polygons are regular polygons , with the same number of sides in each , therefore they are similar ; therefore the ratio of ...
Side 225
... regular polygon inscribed in the centre . Let OE , OF , & c . , be perpendiculars drawn from O to the sides AB , BC , & c . , in succession , then OE , OF , & c . , are all equal ( Bk . II . Prop . 7 ) . Fig . 38 . A E B F C Because the ...
... regular polygon inscribed in the centre . Let OE , OF , & c . , be perpendiculars drawn from O to the sides AB , BC , & c . , in succession , then OE , OF , & c . , are all equal ( Bk . II . Prop . 7 ) . Fig . 38 . A E B F C Because the ...
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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