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 graduateJohn W. Parker, 1837 - 88 Seiten |
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Seite 84
... pentagon in a given circle . Steps of the Demonstration . 1. Prove that 5 ≤S DAC , ACE , ECD , CDB , and BDA = each other , 2 . 3 . 4 . 5 . 6 . that AB , BC , CD , DE , EA = each other , that the pentagon ABCDE is equilateral , that ...
... pentagon in a given circle . Steps of the Demonstration . 1. Prove that 5 ≤S DAC , ACE , ECD , CDB , and BDA = each other , 2 . 3 . 4 . 5 . 6 . that AB , BC , CD , DE , EA = each other , that the pentagon ABCDE is equilateral , that ...
Seite 85
... pentagon about a given circle . 2 . 3 . Steps of the Demonstration . 1. Prove that each of s at c is a right , that ... pentagon is equilateral , j that HKL = / KLM , that the pentagon is equiangular , and is described about the given ...
... pentagon about a given circle . 2 . 3 . Steps of the Demonstration . 1. Prove that each of s at c is a right , that ... pentagon is equilateral , j that HKL = / KLM , that the pentagon is equiangular , and is described about the given ...
Seite 86
... pentagon . Steps of the Demonstration . K ( base BF base FD , CBF = CDF , 1. Prove that , in As BCF , DCF , and that △ CBA = 2 ≤ cbf , ABC is bisected by BF , that , similarly , S BAE , AED , are bisected by AF , FE , respectively , 2 ...
... pentagon . Steps of the Demonstration . K ( base BF base FD , CBF = CDF , 1. Prove that , in As BCF , DCF , and that △ CBA = 2 ≤ cbf , ABC is bisected by BF , that , similarly , S BAE , AED , are bisected by AF , FE , respectively , 2 ...
Seite 87
... pentagon . B PROPOSITION XV . Problem . To inscribe an equila- teral and equiangular hexagon in a given circle . Steps of the Demonstration . 1. Prove that A EGD is equilateral , that EGD of 2 rights , and similarly DGC = of 2 rights ...
... pentagon . B PROPOSITION XV . Problem . To inscribe an equila- teral and equiangular hexagon in a given circle . Steps of the Demonstration . 1. Prove that A EGD is equilateral , that EGD of 2 rights , and similarly DGC = of 2 rights ...
Seite 88
... pentagon must be so placed in the circle , that one angle of each may meet in the same point A. THE END . LONDON -- JOHN W. PARKER , ST . MARTIN'S LANE . PUBLISHED BY JOHN W. PARKER , WEST STRAND , LONDON 88 FOURTH BOOK .
... pentagon must be so placed in the circle , that one angle of each may meet in the same point A. THE END . LONDON -- JOHN W. PARKER , ST . MARTIN'S LANE . PUBLISHED BY JOHN W. PARKER , WEST STRAND , LONDON 88 FOURTH BOOK .
Häufige Begriffe und Wortgruppen
AB² AC² AD² AEX EC angle contained angle equal Argument ad absurdum base DF BC² BD² bisect CB² cuts the circle DC² Demonstration itself consists diameter EB² EF² EG² Engravings equal straight lines equi equiangular equilateral Euclid F Steps fall figure GF² given circle given point given rectilineal angle given straight line given triangle i. e. less inscribe interior angles learner less greater line be divided line drawn parallel parallelogram PARKER pass pentagon point of contact Problem proof PROPOSITION IX PROPOSITION VIII Proved by showing rectangle contained right angles right line shows the supposition similarly Suppose supposition is false Theorem WEST STRAND whole line
Beliebte Passagen
Seite 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...
Seite 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.
Seite 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.
Seite 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...
Seite 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.
Seite 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.
Seite 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.
Seite 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.
Seite 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.