Elements of Geometry and Trigonometry |
From inside the book
Results 1-5 of 46
Page 20
In the same manner it may be shown that the angle E is equal to B , and the angle F to C : hence the two triangles are equal ( Prop . VI . Sch . ) . Scholium . It may be observed that the equal angles lie op- posite the equal sides ...
In the same manner it may be shown that the angle E is equal to B , and the angle F to C : hence the two triangles are equal ( Prop . VI . Sch . ) . Scholium . It may be observed that the equal angles lie op- posite the equal sides ...
Page 40
In the same way it may be shown that like powers or roots of proportional quantities are proportionals . PROPOSITION XIII . THEOREM . If there be two sets of proportional quantities , the products of the corresponding terms will be ...
In the same way it may be shown that like powers or roots of proportional quantities are proportionals . PROPOSITION XIII . THEOREM . If there be two sets of proportional quantities , the products of the corresponding terms will be ...
Page 46
... through the three given points A , B , C. We have now shown that one circumference can always be made to pass through three given points , not in the same straight line : we say farther , that but one can be described through them .
... through the three given points A , B , C. We have now shown that one circumference can always be made to pass through three given points , not in the same straight line : we say farther , that but one can be described through them .
Page 48
If the two parallels DE , IL , are tangents , the one at H , the other at K , draw the parallel secant AB ; and , from what has just been shown , we shall have MH - HP , MK = KP ; and hence the whole arc HMK = HPK .
If the two parallels DE , IL , are tangents , the one at H , the other at K , draw the parallel secant AB ; and , from what has just been shown , we shall have MH - HP , MK = KP ; and hence the whole arc HMK = HPK .
Page 51
... the angle ACB will be equal to DCE . For , if these angles are not equal , suppose ACB to be the greater , and let ACI be taken equal to DCE . From what has just been shown , we shall have AI - DE : but , by hypothesis , AB is ...
... the angle ACB will be equal to DCE . For , if these angles are not equal , suppose ACB to be the greater , and let ACI be taken equal to DCE . From what has just been shown , we shall have AI - DE : but , by hypothesis , AB is ...
What people are saying - Write a review
We haven't found any reviews in the usual places.
Other editions - View all
Common terms and phrases
ABCD adjacent altitude base become Book called centre chord circle circumference circumscribed common cone consequently construction contained corresponding cosine Cotang cylinder described diameter difference distance divided draw drawn equal equation equivalent evident expressed extremities fall figure follows formed formulas four frustum give given gles greater half hence homologous included inscribed intersection less likewise logarithm manner means measured meet middle multiplied opposite parallel parallelogram parallelopipedon pass perpendicular plane polygon prism PROBLEM Prop proportional PROPOSITION pyramid quantities radii radius ratio reason rectangle regular remaining right angles Scholium segment shown sides similar sine solid solid angle sphere spherical triangle square straight line Suppose surface taken tang tangent THEOREM third triangle triangle ABC vertex whole
Bibliographic information
| Title | Elements of Geometry and Trigonometry |
| Author | Adrien Marie Legendre |
| Editor | Charles Davies |
| Translated by | David Brewster |
| Edition | 4 |
| Publisher | A.S. Barnes and Company, 1834 |
| Original from | the University of California |
| Digitized | 14 Oct 2010 |
| Length | 359 pages |
| Export Citation | BiBTeX EndNote RefMan |