Elements of Geometry and Trigonometry |
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Page 13
11 From one point to another only one straight line can be drawn . 12. Through the same point , only one straight line can be drawn which shall be parallel to a given line . 13. Magnitudes , which being applied to each other , coincide ...
11 From one point to another only one straight line can be drawn . 12. Through the same point , only one straight line can be drawn which shall be parallel to a given line . 13. Magnitudes , which being applied to each other , coincide ...
Page 18
Let any point , as O , be taken within the triangle BAC , and let the lines OB , OC , be drawn to the extremities of either side , as BC ; then will OB + OC < BA + AC . PROPOSITION VIII . THEOREM . If from any point within a triangle ...
Let any point , as O , be taken within the triangle BAC , and let the lines OB , OC , be drawn to the extremities of either side , as BC ; then will OB + OC < BA + AC . PROPOSITION VIII . THEOREM . If from any point within a triangle ...
Page 20
The equality of the triangles BAD , DAC , proves also that the angle BAD , is equal to DAC , and BDA to ADC , hence the latter two are right angles ; therefore , the line drawn from the vertex of an isosceles triangle to the middle ...
The equality of the triangles BAD , DAC , proves also that the angle BAD , is equal to DAC , and BDA to ADC , hence the latter two are right angles ; therefore , the line drawn from the vertex of an isosceles triangle to the middle ...
Page 22
From a given point , without a straight line , only one perpendicular can be drawn to that line . Let A be the point , and DE the given line . A Let us suppose that we can draw two perpendiculars , AB , AC .
From a given point , without a straight line , only one perpendicular can be drawn to that line . Let A be the point , and DE the given line . A Let us suppose that we can draw two perpendiculars , AB , AC .
Page 23
A Produce the perpendicular AB till BF is equal to AB , and draw FC , FD . D F First . The triangle BCF , is equal to the triangle BCA , for they have the right angle CBF CBA , the side CB common , and the side BF - BA ; hence the third ...
A Produce the perpendicular AB till BF is equal to AB , and draw FC , FD . D F First . The triangle BCF , is equal to the triangle BCA , for they have the right angle CBF CBA , the side CB common , and the 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 contained Cosine Cotang cylinder described determine diameter difference distance divided draw drawn equal equations equivalent expressed extremities faces feet figure follows formed four frustum give given gles greater half hence homologous hypothenuse included inscribed intersection less let fall logarithm manner means measured meet middle multiplied number of sides opposite parallel parallelogram pass perpendicular plane polygon prism PROBLEM Prop proportional PROPOSITION pyramid quadrant quantities radii radius ratio reason rectangle regular remaining right angles Scholium segment sides similar Sine solid solid angle sphere spherical triangle square straight line suppose taken Tang tangent THEOREM third triangle triangle ABC unit vertex whole
Bibliographic information
| Title | Elements of Geometry and Trigonometry DAVIES' COURSE OF MATHEMATICS |
| Author | Adrien Marie Legendre |
| Editor | Charles Davies |
| Translated by | David Brewster |
| Publisher | A.S. Barnes and Company, 1839 |
| Original from | Harvard University |
| Digitized | 14 Jan 2008 |
| Length | 359 pages |
| Export Citation | BiBTeX EndNote RefMan |