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 páginas |
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Página 11
Euclides. FIRST BOOK . PROPOSITION I. Problem . To describe an equilateral triangle upon a given finite straight line . K B D H PROPOSITION II . B E Problem . From a given point to draw a straight line equal to a given straight line ...
Euclides. FIRST BOOK . PROPOSITION I. Problem . To describe an equilateral triangle upon a given finite straight line . K B D H PROPOSITION II . B E Problem . From a given point to draw a straight line equal to a given straight line ...
Página 82
... equilateral , that AGB is a right 4 , that fig . GK is rectangular , that : . GK is a square , and described about ABCD . PROPOSITION VIII . Problem . To inscribe a circle in a given square . F A E B H K Steps of the Demonstration . 1 ...
... equilateral , that AGB is a right 4 , that fig . GK is rectangular , that : . GK is a square , and described about ABCD . PROPOSITION VIII . Problem . To inscribe a circle in a given square . F A E B H K Steps of the Demonstration . 1 ...
Página 84
... equilateral and equi- angular 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 ...
... equilateral and equi- angular 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 ...
Página 85
... 2 KC ; and , similarly , HK = 2 BK , that HK KL , that the pentagon is equilateral , j that HKL = / KLM , that the pentagon is equiangular , and is described about the given . B M PROPOSITION XIII . E Problem . To inscribe FOURTH BOOK . 85.
... 2 KC ; and , similarly , HK = 2 BK , that HK KL , that the pentagon is equilateral , j that HKL = / KLM , that the pentagon is equiangular , and is described about the given . B M PROPOSITION XIII . E Problem . To inscribe FOURTH BOOK . 85.
Página 86
... equilateral and equi- angular 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 ...
... equilateral and equi- angular 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 ...
Términos y frases comunes
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
Pasajes populares
Página 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...
Página 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.
Página 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.
Página 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...
Página 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.
Página 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.
Página 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.
Página 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.
Página 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.