 CD : CM. Again, the triangles CAM, CME, having the common vertex M, are to each other as their bases CA, CE ; they are likewise to each other as the polygons A' and B of which they form part ; hence A' : B : :  The Circle and Straight Line - Side 38av John Harris - 1875Uten tilgangsbegrensning - Om denne boken
 | Adrien Marie Legendre - 1822 - 394 sider
...is AM, B' that of the similar circumscribed polygon : A and B are given ; we have to find A' and B'. part : hence A : A' : : CD : CM. Again, the triangles...since AD and ME are parallel, we have CD: CM:: CA: CE; hence A : A' : : A' : B ; hence the polygon A', one of those required, is a mean proportional between... | |
 | Adrien Marie Legendre - 1828 - 346 sider
...having the common vertex A, are to each other as their bases CD, CM ; they are likewise to each oilier as the polygons A and A', of which they form part...since AD and ME are parallel, we have CD : CM : : CA : CE ; hence A : A' : : A' : B ; hence the polygon A' one of those required, is a mean proportional... | |
 | Adrien Marie Legendre - 1830 - 344 sider
...circumscribed polygon : A and B are given ; we have to find A' and B'. First. The triangles ACD, ACM, having the common vertex A, are to each other as their...since AD and ME are parallel, we have CD : CM : : CA : CE ; hence A : A : : A : B ; hence the polygon A' one of those required, is a mean proportional between... | |
 | Adrien Marie Legendre - 1836 - 394 sider
...circumscribed polygon : A and B are given ; we have to find A' and B'. First. The triangles ACD, ACM, having the common vertex A, are to each other as their...since AD and ME are parallel, we have CD : CM : : CA : CE; hence A : A' : : A' : B ; hence the polygon A', one of those required, is a mean proportional... | |
 | Adrien Marie Legendre - 1838 - 382 sider
...bases CD, CM ; they are likewise to each other as the polygons A and A', of which they form part 4 hence A : A' : : CD ' CM. Again, the triangles CAM,...since AD and ME are parallel, we have CD : CM : : CA : CE; hence A : A' : : A' : B ; hence the polygon A', one of those required, is a mean proportional... | |
 | James Bates Thomson - 1844 - 268 sider
...circumscribed polygon : A and B are given : we have to find A' and B'. first. The triangles ACD, ACM, having the common vertex A, are to each other as their...other as the polygons A' and B of which they form part ; therefore A' : B : : CA : CE. But since AD and ME are parallel, we have CD : CM : ; CA : CE ; hence... | |
 | Nathan Scholfield - 1845 - 894 sider
...they are likewise to each other as the polygons A and A', of which they form part: hence A : A'::CD : CM. Again the triangles CAM, CME, having the common...since AD and ME are parallel, we have CD : CM: :CA : CE ; hence A : A': :A' : B ; hence the polygon A', one of those required, is a mean proportional... | |
 | Charles Davies - 1849 - 372 sider
...to First. The triangles ACD, ACM, having thu common vertex A. are to each other as their bases'CD, CM ; they are likewise to each other as the polygons...since AD and ME are parallel, we have CD : CM : : CA : CE; hence A : A' : : A' : B; hence the polygon A',one of those required, is a mean proportional between... | |
 | Adrien Marie Legendre - 1852 - 436 sider
...to each other as the polygons P and P', of which they form part (B. IL, P. 11) : hence, P : P* : : CD : CM. Again, the triangles CAM, CME, having the...; they are likewise to each other as the polygons P' and p of which they form part ; hence, Pr : p :: CA i CR But since AD and ME are parallel, we have,... | |
 | Charles Davies - 1854 - 436 sider
...of which they form part (B. II., P. 11) : hence, P : P1 : : CD : CM. Again, the triangles CAM^CME, having the common vertex M, are to each other as their...; they are likewise to each other as the polygons P' and p of which they form part ; hence, •P' : p :: CA : CE. But since AD and ME are parallel, we... | |
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