Elements of Geometry |
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Page v
... give to it all the exact- ness and precision of which it is susceptible . Perhaps I might have attained this object by calling a straight line that which can have only one position between two given points . For , from this essential ...
... give to it all the exact- ness and precision of which it is susceptible . Perhaps I might have attained this object by calling a straight line that which can have only one position between two given points . For , from this essential ...
Page xii
... gives also A : B :: C : D A : C :: B : D ; and , since the ratios A : C , B : D , are equal , we obtain B + A : D + C :: A : C , or : : B : D , B - AD - C :: A : C , or : : B : D , :: a result which may be thus enunciated ; In any ...
... gives also A : B :: C : D A : C :: B : D ; and , since the ratios A : C , B : D , are equal , we obtain B + A : D + C :: A : C , or : : B : D , B - AD - C :: A : C , or : : B : D , :: a result which may be thus enunciated ; In any ...
Page 14
... gives A = A ' + B ' ; therefore A + B + C A + B ' + C ' . Moreover , since , by hypothesis , we have AC < AB , and , consequently , C'B ' < AC ' , it will be seen , that , in the triangle AC'B ' , the angle at A , designated by A ' , is ...
... gives A = A ' + B ' ; therefore A + B + C A + B ' + C ' . Moreover , since , by hypothesis , we have AC < AB , and , consequently , C'B ' < AC ' , it will be seen , that , in the triangle AC'B ' , the angle at A , designated by A ' , is ...
Page 49
... gives a quadruple square ( fig . 103 ) , a triple Fig . 103 . line a square nine times as great , and so on . THEOREM . 174. The area of any parallelogram is equal to the product of its base by its altitude . Demonstration . The ...
... gives a quadruple square ( fig . 103 ) , a triple Fig . 103 . line a square nine times as great , and so on . THEOREM . 174. The area of any parallelogram is equal to the product of its base by its altitude . Demonstration . The ...
Page 51
... gives BC = EF ; but , on account of the parallels , IG = BC , and DG EF , therefore HIGD is equal to the square described upon BC . These two parts being taken from the whole square , there remain the two rectangles BCGI , EFIH , which ...
... gives BC = EF ; but , on account of the parallels , IG = BC , and DG EF , therefore HIGD is equal to the square described upon BC . These two parts being taken from the whole square , there remain the two rectangles BCGI , EFIH , which ...
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Common terms and phrases
ABC fig adjacent angles altitude angle ACB angle BAC base ABCD bisect centre chord circ circular sector circumference circumscribed common cone consequently construction convex surface Corollary cube cylinder Demonstration diagonals diameter draw drawn equal angles equiangular equilateral equivalent faces figure formed four right angles frustum GEOM given point gles greater hence homologous sides hypothenuse inclination intersection isosceles triangle JOHN CRERAR LIBRARY join less Let ABC let fall line AC mean proportional measure the half meet multiplied number of sides oblique lines opposite parallelogram parallelopiped perimeter perpendicular plane MN polyedron prism produced proposition radii radius ratio rectangle regular polygon right angles Scholium sector segment semicircle semicircumference side BC similar solid angle sphere spherical polygons spherical triangle square described straight line tangent THEOREM three angles triangle ABC triangular prism triangular pyramids vertex vertices whence