Elements of Geometry |
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Results 6-10 of 60
Page 11
... produce one of them AB , so that BF AB , and join CF. = = : The triangle CBF is equal to the triangle ABC . For the angle CBF is a right angle ( 29 ) , as well as CBA , and the side BF BA ; therefore the triangles are equal ( 36 ) , and ...
... produce one of them AB , so that BF AB , and join CF. = = : The triangle CBF is equal to the triangle ABC . For the angle CBF is a right angle ( 29 ) , as well as CBA , and the side BF BA ; therefore the triangles are equal ( 36 ) , and ...
Page 13
... produce it to C ' , mak- ing ACAB ; produce also AB to B ' , making AB ' double of AI . If we designate by A , B , C , the three angles of the triangle ABC , and by A ' , B ' , C ' , the three angles of the triangle A B'C ' , we say ...
... produce it to C ' , mak- ing ACAB ; produce also AB to B ' , making AB ' double of AI . If we designate by A , B , C , the three angles of the triangle ABC , and by A ' , B ' , C ' , the three angles of the triangle A B'C ' , we say ...
Page 16
... produce the side AB toward D , the exterior angle CBD will be equal to the sum of the two opposite interior angles A and C ; for , by adding to each the angle ABC , the sums are each equal to two right angles . THEOREM . 64. The sum of ...
... produce the side AB toward D , the exterior angle CBD will be equal to the sum of the two opposite interior angles A and C ; for , by adding to each the angle ABC , the sums are each equal to two right angles . THEOREM . 64. The sum of ...
Page 17
... produced ever so far . Demonstration . For , if they should meet in any point 0 , there would be two perpendiculars OF , OG , let fall from the same point O upon the same straight line FG , which is impossi- ble ( 50 ) . THEOREM . 70 ...
... produced ever so far . Demonstration . For , if they should meet in any point 0 , there would be two perpendiculars OF , OG , let fall from the same point O upon the same straight line FG , which is impossi- ble ( 50 ) . THEOREM . 70 ...
Page 19
... produced will meet AB in a determinate point ; therefore , for a still stronger 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 ...
... produced will meet AB in a determinate point ; therefore , for a still stronger 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 ...
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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