Elements of Geometry |
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Page 2
... Fig . 20. ( fig . 20 ) is the sum of the two angles DCB , BCE , and the angle DCB is the difference between the two ... ABC ( fig . 10 ) is a triangle right - angled at A , and the side BC is the hypothenuse . Fig . 11 . Fig . 12 . Fig ...
... Fig . 20. ( fig . 20 ) is the sum of the two angles DCB , BCE , and the angle DCB is the difference between the two ... ABC ( fig . 10 ) is a triangle right - angled at A , and the side BC is the hypothenuse . Fig . 11 . Fig . 12 . Fig ...
Page 7
... ABC , DEF ( fig . 23 ) , Fig . 23 let the angle A be equal to the angle D , the side AB equal to the side DE , and the side AC equal to the side DF ; the two triangles ABC , DEF , will be equal . Indeed , the triangles may be so placed ...
... ABC , DEF ( fig . 23 ) , Fig . 23 let the angle A be equal to the angle D , the side AB equal to the side DE , and the side AC equal to the side DF ; the two triangles ABC , DEF , will be equal . Indeed , the triangles may be so placed ...
Page 8
... ( fig . 23 ) , for example , is the shortest way from B to C ( 3 ) ; BC therefore is less than BA + AC . THEOREM . 41. If , from a point O ( fig . 24 ) , within a triangle ABC , there be drawn straight lines OB , OC , to the extremities ...
... ( fig . 23 ) , for example , is the shortest way from B to C ( 3 ) ; BC therefore is less than BA + AC . THEOREM . 41. If , from a point O ( fig . 24 ) , within a triangle ABC , there be drawn straight lines OB , OC , to the extremities ...
Page 10
... ABC ACB ( fig . 29 ) , the side AC will be equal to the side AB . For , if these sides are not equal , let AB be the greater . Take BDAC , and join DC . The angle DBC is , by hypothesis , equal to ACB , and the two sides DB , BC , are ...
... ABC ACB ( fig . 29 ) , the side AC will be equal to the side AB . For , if these sides are not equal , let AB be the greater . Take BDAC , and join DC . The angle DBC is , by hypothesis , equal to ACB , and the two sides DB , BC , are ...
Page 11
... Fig . 3k , DE , only one perpendicular can be drawn to that line Demonstration . If it be possible , let there be two AB and AC ; produce one of them AB , so that BF AB , and join CF. = = : The triangle CBF is equal to the triangle ABC ...
... Fig . 3k , DE , only one perpendicular can be drawn to that line Demonstration . If it be possible , let there be two AB and AC ; produce one of them AB , so that BF AB , and join CF. = = : The triangle CBF is equal to the triangle ABC ...
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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
Bibliographic information
| Title | Elements of Geometry |
| Author | Adrien Marie Legendre |
| Translated by | John Farrar |
| Publisher | Hilliard, Gray, and Company, 1833 |
| Original from | University of Chicago |
| Digitized | 11 Apr 2012 |
| Length | 235 pages |
| Export Citation | BiBTeX EndNote RefMan |