A companion to Euclid: being a help to the understanding and remembering of the first four books. With a set of improved figures, and an original demonstration of the proposition called in Euclid the twelfth axiom, by a graduate |
Inni boken
Resultat 1-5 av 5
Side 84
To inscribe an equilateral and equiangular pentagon in a given circle . ai ni Steps
of the Demonstration . 1 . Prove that 5 ZS DAC , ACE , ECD , CDB , and BDA =
each other , that AB , BC , CD , DE , EA = each other , that the pentagon ABCDE ...
To inscribe an equilateral and equiangular pentagon in a given circle . ai ni Steps
of the Demonstration . 1 . Prove that 5 ZS DAC , ACE , ECD , CDB , and BDA =
each other , that AB , BC , CD , DE , EA = each other , that the pentagon ABCDE ...
Side 85
PROPOSITION XII . DM lateral Problem . To describe an equilateral and
equiangular pentagon about a given circle . aj ti 200 r Steps of the Demonstration
. 1 . Prove that each of Xs at c is a right Z , that , similarly , each of sat B and d are
rt .
PROPOSITION XII . DM lateral Problem . To describe an equilateral and
equiangular pentagon about a given circle . aj ti 200 r Steps of the Demonstration
. 1 . Prove that each of Xs at c is a right Z , that , similarly , each of sat B and d are
rt .
Side 86
To inscribe a circle Le in a given equilateral and equi☺ angular pentagon . si ai i
Steps of the Demonstration . 1 . Prove that , in As BCF , DCF , { base BF = base
FD , DCF , and < CBF = Z CDF , that CBA = 2 % CBF , that _ ABC is bisected by ...
To inscribe a circle Le in a given equilateral and equi☺ angular pentagon . si ai i
Steps of the Demonstration . 1 . Prove that , in As BCF , DCF , { base BF = base
FD , DCF , and < CBF = Z CDF , that CBA = 2 % CBF , that _ ABC is bisected by ...
Side 87
FC = FD , that the five right lines , FA , FB , FC , FD , FE = each other ; and : . a
described from centre F and distance any one of them , will pass through the
points A , B , C , D , E and bedescribed about the given pentagon .
PROPOSITION XV .
FC = FD , that the five right lines , FA , FB , FC , FD , FE = each other ; and : . a
described from centre F and distance any one of them , will pass through the
points A , B , C , D , E and bedescribed about the given pentagon .
PROPOSITION XV .
Side 88
... an equilateral and equiangular pentagon will be inscribed in it . ai aj
PUBLISHED BY JOHN W . PARKER , WEST STRAND ,. * It should be observed
that the triangle and pentagon must be so placed in the circle , that one angle of
each may ...
... an equilateral and equiangular pentagon will be inscribed in it . ai aj
PUBLISHED BY JOHN W . PARKER , WEST STRAND ,. * It should be observed
that the triangle and pentagon must be so placed in the circle , that one angle of
each may ...
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Vanlige uttrykk og setninger
alternate angle contained angle equal applied Argument ad absurdum base bisect BOOK centre circumference coincides construction Demonstration described diameter directed divided draw drawn Edition Engravings equal equiangular equilateral Euclid extremities fall figure given circle given point given rectilineal given straight line greater HISTORY impossible inscribe interior joins learner least less meet Nature necessary opposite parallel parallelogram pass pentagon point of contact Problem produced proof PROPOSITION PROPOSITION VIII PROPOSITION XV Proved by showing READINGS rectangle contained right angles right line right Zs segment shows the supposition sides similarly square Steps straight line Suppose supposition is false Theorem touch triangle VOLUME whole whole line YOUNG
Populære avsnitt
Side 24 - If two triangles have two angles of the [one equal to two angles of the other, each to each, and one side equal to one side, namely, either t}le sides adjacent to the equal...
Side 45 - To divide a given straight line into two parts, so that the rectangle contained by the whole and one of the parts, shall be equal to the square of the other part.
Side 18 - If, at a point in a straight line, two other straight lines, upon the opposite sides of it, make the adjacent angles together equal to two right angles, these two straight lines shall be in one and the same straight line.
Side 61 - From this it is manifest that the straight line which is drawn at right angles to the diameter of a circle from the extremity of it, touches the circle...
Side 37 - In any right-angled triangle, the square which is described upon the side subtending the right angle, is equal to the squares described upon the sides which contain the right angle.
Side 76 - IF from any point without a circle two straight lines be drawn, one of which cuts the circle, and the other touches it ; the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, shall be equal to the square of the line which touches it.
Side 77 - If from a point without a circle there be drawn two straight lines, one of which cuts the circle, and the other meets it, and if the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, be equal to the square on GEOMETRY.
Side 72 - If a straight line touch a circle, and from the point of contact a straight line be drawn at right angles to the touching line, the centre of the circle shall be in that line.
Side 27 - If a straight line fall on two parallel straight lines, it makes the alternate angles equal to one another, and the exterior angle equal to the interior and opposite angle on the same side; and also the two interior angles on the same side together equal to two right angles.
Bibliografisk informasjon
| Tittel | A companion to Euclid: being a help to the understanding and remembering of the first four books. With a set of improved figures, and an original demonstration of the proposition called in Euclid the twelfth axiom, by a graduate |
| Forfatter | Euclides |
| Utgiver | John W. Parker, 1837 |
| Original fra | Oxford University |
| Digitalisert | 16. okt 2006 |
| Lengde | 88 sider |
| Eksporter sitat | BiBTeX EndNote RefMan |