Elements of Geometry and Trigonometry |
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Page 26
Opposite exterior and interior angles lie on the same side of the secant line ... other within the parallels , but not adjacent : thus , EGB , GOD , are opposite exterior and interior angles ; and so also , are the angles AGE , GOC .
Opposite exterior and interior angles lie on the same side of the secant line ... other within the parallels , but not adjacent : thus , EGB , GOD , are opposite exterior and interior angles ; and so also , are the angles AGE , GOC .
Page 27
I. ) . Taking from each , the angle OGB , and there remains OGA = GOD . In the same manner we may prove that GOC = OGB . B D Cor . 3. If a straight line meet two parallel lines , the opposite exterior ...
I. ) . Taking from each , the angle OGB , and there remains OGA = GOD . In the same manner we may prove that GOC = OGB . B D Cor . 3. If a straight line meet two parallel lines , the opposite exterior ...
Page 29
Then , since AE , CB , are parallel , and CAD cuts them , the exterior angle DAE will be equal to its inte - C rior opposite one ACB ( Prop . XX . Cor . 3. ) ; in like manner , since AE , CB , are parallel , and AB cuts them ...
Then , since AE , CB , are parallel , and CAD cuts them , the exterior angle DAE will be equal to its inte - C rior opposite one ACB ( Prop . XX . Cor . 3. ) ; in like manner , since AE , CB , are parallel , and AB cuts them ...
Page 30
In every triangle ABC , the exterior angle BAD is equal to the sum of the two interior opposite angles B and C. For , AE being parallel to BC , the part BAE is equal to the angle B , and the other part DAE is equal to the angle C.
In every triangle ABC , the exterior angle BAD is equal to the sum of the two interior opposite angles B and C. For , AE being parallel to BC , the part BAE is equal to the angle B , and the other part DAE is equal to the angle C.
Page 32
In every parallelogram , the opposite sides and angles are equal . 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 .
In every parallelogram , the opposite sides and angles are equal . 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 .
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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 |