Elements of Geometry and Trigonometry |
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Page 15
From the point C draw the line CF , making with AC the right angle ACF . Now , since ACD is a straight line , the angle FCD will be a right angle ( Prop . I. Cor . 1. ) ; and since ACE is a straight line , the angle FCE will likewise be ...
From the point C draw the line CF , making with AC the right angle ACF . Now , since ACD is a straight line , the angle FCD will be a right angle ( Prop . I. Cor . 1. ) ; and since ACE is a straight line , the angle FCE will likewise be ...
Page 18
If from any point within a triangle , two straight lines be drawn to the extremities of either side , their sum will be less than the sum of the two other sides of the triangle . Let any point , as O , be taken within the trian- gle BAC ...
If from any point within a triangle , two straight lines be drawn to the extremities of either side , their sum will be less than the sum of the two other sides of the triangle . Let any point , as O , be taken within the trian- gle BAC ...
Page 21
Then , take BD equal to AC , and draw CD . Now , in the two triangles BDC , BAC , we have BD - AC , by construction ; the angle B equal to the angle ACB , by hypothesis ; B and the side BC common : therefore , the two triangles ...
Then , take BD equal to AC , and draw CD . Now , in the two triangles BDC , BAC , we have BD - AC , by construction ; the angle B equal to the angle ACB , by hypothesis ; B and the side BC common : therefore , the two triangles ...
Page 22
From a given point , without a straight line , only one perpendicu lar can be drawn to that line . Let A be the point , and DE the given line .. B E Let us suppose that we can draw two perpendiculars , AB , AC .
From a given point , without a straight line , only one perpendicu lar can be drawn to that line . Let A be the point , and DE the given line .. B E Let us suppose that we can draw two perpendiculars , AB , AC .
Page 23
Produce the perpendicular AB till BF is equal to AB , and draw FC , FD . First . The triangle BCF , is equal to the triangle BCA , for they have the right angle CBF CBA , the side CB common , and the - D side BF = BA ; hence the third ...
Produce the perpendicular AB till BF is equal to AB , and draw FC , FD . First . The triangle BCF , is equal to the triangle BCA , for they have the right angle CBF CBA , the side CB common , and the - D side BF = BA ; hence the third ...
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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 |