Elements of Geometry |
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Page 19
... reason , the straight line FD , comprehended in the an- gle EFZ , will meet AB . 2. Let us suppose that the sum of the interior angles AEF + CFE is greater than two right angles ; if we produce AE to- ward B , and CF toward D , the sum ...
... reason , the straight line FD , comprehended in the an- gle EFZ , will meet AB . 2. Let us suppose that the sum of the interior angles AEF + CFE is greater than two right angles ; if we produce AE to- ward B , and CF toward D , the sum ...
Page 21
... reason is this ; if we produce FE toward H , the angle DEH would have its sides parallel to those of the angle BAC , but the two angles would not be equal . In this case , the angle DEH and the angle BAC would together make two right ...
... reason is this ; if we produce FE toward H , the angle DEH would have its sides parallel to those of the angle BAC , but the two angles would not be equal . In this case , the angle DEH and the angle BAC would together make two right ...
Page 22
... reason , AB is parallel to CD ; therefore the quadrilateral ABCD is a parallelogram . THEOREM . 84. If two opposite sides AB , CD ( fig . 44 ) , of a quadrilateral are equal and parallel , the two other sides will also be equal and ...
... reason , AB is parallel to CD ; therefore the quadrilateral ABCD is a parallelogram . THEOREM . 84. If two opposite sides AB , CD ( fig . 44 ) , of a quadrilateral are equal and parallel , the two other sides will also be equal and ...
Page 28
... reason ; it will , then , be in both of these lines at the same time . But two lines can cut each other in only one point ( 32 ) ; there is , therefore , only one circle , whose circum- ference can pass through three given points . 108 ...
... reason ; it will , then , be in both of these lines at the same time . But two lines can cut each other in only one point ( 32 ) ; there is , therefore , only one circle , whose circum- ference can pass through three given points . 108 ...
Page 29
... reason , CF > CI . But CF CG , since the chords AB , DE , are equal . Therefore CG > CI , and of two unequal chords , the less is at the greater distance from the centre . 110 . THEOREM . The perpendicular BD ( fig . 54 ) , at the ...
... reason , CF > CI . But CF CG , since the chords AB , DE , are equal . Therefore CG > CI , and of two unequal chords , the less is at the greater distance from the centre . 110 . THEOREM . The perpendicular BD ( fig . 54 ) , at the ...
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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