Euclid's Elements of plane geometry [book 1-6] explicitly enunciated, by J. Pryde. [With] Key1860 |
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Side 37
... ABCD is equal to the whole parallelogram FKML ; therefore the parallelogram FKML has been described equal to the given rec- tilineal figure ABCD , having the angle FKM equal to the given angle E. COR . - From this it is manifest how to ...
... ABCD is equal to the whole parallelogram FKML ; therefore the parallelogram FKML has been described equal to the given rec- tilineal figure ABCD , having the angle FKM equal to the given angle E. COR . - From this it is manifest how to ...
Side 62
... ABCD , of which the diameters are AC and BD ; to prove that the sum of the squares on AC and BD is equal to the sum of the squares on AB , BC , CD , and DA . ( Dem . ) Let AC and BD intersect one another in E ; because the vertical ...
... ABCD , of which the diameters are AC and BD ; to prove that the sum of the squares on AC and BD is equal to the sum of the squares on AB , BC , CD , and DA . ( Dem . ) Let AC and BD intersect one another in E ; because the vertical ...
Side 70
... ABCD , centre ; to prove that and AC , BD , two chords in it , which cut one another in the point E , and do not both pass through the AC and BD do not bisect one another . and ( Dem . ) For , if it be possible , let AE be equal to EC ...
... ABCD , centre ; to prove that and AC , BD , two chords in it , which cut one another in the point E , and do not both pass through the AC and BD do not bisect one another . and ( Dem . ) For , if it be possible , let AE be equal to EC ...
Side 72
... ABCD , and AD its diameter , and E the centre , in which any point F is taken which is not the centre . To prove that of all the straight lines FB , FC , FG , & c . , that can be drawn from F to the circumference , FA is the greatest ...
... ABCD , and AD its diameter , and E the centre , in which any point F is taken which is not the centre . To prove that of all the straight lines FB , FC , FG , & c . , that can be drawn from F to the circumference , FA is the greatest ...
Side 78
... ABCD , of which the diameter is AD , and the centre E ; and BC a chord nearer to the centre than FG ; to prove that AD is greater than any chord BC which is not a diameter , and BC , which is nearer the centre , is greater than FG ...
... ABCD , of which the diameter is AD , and the centre E ; and BC a chord nearer to the centre than FG ; to prove that AD is greater than any chord BC which is not a diameter , and BC , which is nearer the centre , is greater than FG ...
Andre utgaver - Vis alle
Euclid's Elements of plane geometry [book 1-6] explicitly enunciated, by J ... Euclides Uten tilgangsbegrensning - 1860 |
Euclid's Elements of Plane Geometry [Book 1-6] Explicitly Enunciated, by J ... Euclides,James Pryde Ingen forhåndsvisning tilgjengelig - 2023 |
Euclid's Elements of Plane Geometry [book 1-6] Explicitly Enunciated, by J ... Euclides,James Pryde Ingen forhåndsvisning tilgjengelig - 2018 |
Vanlige uttrykk og setninger
ABCD adjacent angles angle ABC angle ACB angle BAC apothem base BC BC is equal bisected centre Chambers's chord circle ABC circumference Const cosec cosine described diameter divided double draw equal angles equal to twice equiangular equilateral equilateral polygon equimultiples exterior angle fore given line given point given straight line gnomon greater hence hypotenuse inscribed isosceles triangle less line drawn multiple number of sides opposite angle parallel parallelogram perimeter perpendicular polygon produced proportional PROPOSITION prove radius ratio rectangle contained rectilineal figure regular polygon remaining angle right angles right-angled triangle segment semiperimeter shewn similar sine square on AC straight line AC tangent THEOREM third touches the circle triangle ABC triangle DEF twice the rectangle vertical angle wherefore
Populære avsnitt
Side 23 - 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, viz. either the sides adjacent to the equal...
Side 52 - If a straight line be bisected, and produced to any point; the rectangle contained by the whole line thus produced, and the part of it produced, together with the square of half the line bisected, is equal to the square of the straight line which is made up of the half and the part produced.
Side 51 - If a straight line be divided into two equal parts, and also into two unequal parts, the rectangle contained by the unequal parts, together with the square on the line between the points of section, is equal to the square on half the line.
Side 53 - If a straight line be divided into any two parts, the squares of the whole line, and of one of the parts, are equal to twice the rectangle contained by the whole and that part, together with the square of the other part. Let the straight line AB be divided into any two parts in the point C ; the squares of AB, BC are equal to twice the rectangle AB, BC...
Side 3 - When a straight line standing on another straight line makes the adjacent angles equal to one another, each of the angles is called a right angle ; and the straight line which stands on the other is called a perpendicular to it.
Side 29 - Therefore all the angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has sides.
Side 117 - And the same thing is to be understood when it is more briefly expressed by saying, a has to d the ratio compounded of the ratios of e to f, g to h, and k to l. In like manner, the same things being supposed, if m has to n the same ratio which a has to d ', then, for shortness...
Side 13 - Upon the same base, and on the same side of it, there cannot be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise those which are terminated in the other extremity.
Side 159 - From the point A draw a straight line AC, making any angle with AB ; and in AC take any point D, and take AC the same multiple of AD, that AB is of the part which is to be cut off from it : join BC, and draw DE parallel to it : then AE is the part required to be cut off. Because ED is parallel to one of the sides of the triangle ABC, viz. to BC ; as CD is to DA, so is (2.
Side 60 - CB, BA, by twice the rectangle CB, BD. Secondly, Let AD fall without the triangle ABC. Then, because the angle at D is a right angle, the angle ACB is greater than a right angle ; (i.