Elements of Geometry and Trigonometry |
From inside the book
Results 1-5 of 32
Page 32
Let ABCD be a parallelogram : then will AB = DC , AD = BC , A = C , and ADC = ABC . PROPOSITION XXVIII . THEOREM . B For , draw the diagonal BD . The triangles ABD , DBC , have a common side BD ; and since AD , BC , are parallel ...
Let ABCD be a parallelogram : then will AB = DC , AD = BC , A = C , and ADC = ABC . PROPOSITION XXVIII . THEOREM . B For , draw the diagonal BD . The triangles ABD , DBC , have a common side BD ; and since AD , BC , are parallel ...
Page 33
For a like reason AB is parallel to CD therefore the quadrilateral ABCD is a parallelogram . : PROPOSITION XXX . THEOREM . If two opposite sides of a quadrilateral are equal and parallel , the remaining sides will also be equal and ...
For a like reason AB is parallel to CD therefore the quadrilateral ABCD is a parallelogram . : PROPOSITION XXX . THEOREM . If two opposite sides of a quadrilateral are equal and parallel , the remaining sides will also be equal and ...
Page 55
The opposite angles A and C , of an inscribed quadrilateral ABCD , are together equal to two right angles : for the angle BAD is measured by half the arc BCD , the angle BCD is measured by half the arc BAD ; hence the two angles BAD ...
The opposite angles A and C , of an inscribed quadrilateral ABCD , are together equal to two right angles : for the angle BAD is measured by half the arc BCD , the angle BCD is measured by half the arc BAD ; hence the two angles BAD ...
Page 69
Let AB be the common base of D CF EDF CE the two parallelograms ABCD , ABEF : and since they are supposed to have the same altitude , their upper bases DC , FE , will be both situated in one straight line parallel to AB .
Let AB be the common base of D CF EDF CE the two parallelograms ABCD , ABEF : and since they are supposed to have the same altitude , their upper bases DC , FE , will be both situated in one straight line parallel to AB .
Page 70
Hence these two parallelograms ABCD , ABEF , which havẽ the same base and altitude , are equivalent Cor . Every parallelogram is equivalent to the rectangle which has the same base and the same altitude . PROPOSITION II . THEOREM .
Hence these two parallelograms ABCD , ABEF , which havẽ the same base and altitude , are equivalent Cor . Every parallelogram is equivalent to the rectangle which has the same base and the same altitude . PROPOSITION II . THEOREM .
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 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 |