Euclid's Elements of plane geometry [book 1-6] explicitly enunciated, by J. Pryde. [With] Key1860 |
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Seite 19
... reason , the exterior angle CEB of the triangle ABE is greater than BAC ; and it has been demonstrated that the angle BDC is greater than the angle CEB ; still more ( Ax . 16 ) , then , is the angle BDC greater than the angle BAC ...
... reason , the exterior angle CEB of the triangle ABE is greater than BAC ; and it has been demonstrated that the angle BDC is greater than the angle CEB ; still more ( Ax . 16 ) , then , is the angle BDC greater than the angle BAC ...
Seite 30
... reason EF is equal to BC ; wherefore AD is equal to EF ( Ax . 1 ) ; and DE is common ; there- fore ( Ax . 2 or 3 ) the whole or the remainder AE is equal to the whole or the remainder DF ; AB also is equal to DC ; the two EA and AB are ...
... reason EF is equal to BC ; wherefore AD is equal to EF ( Ax . 1 ) ; and DE is common ; there- fore ( Ax . 2 or 3 ) the whole or the remainder AE is equal to the whole or the remainder DF ; AB also is equal to DC ; the two EA and AB are ...
Seite 31
... reason , the paral- lelogram EFGH is equal to the same parallelogram EBCH ; therefore ( Ax . 1 ) the parallelogram ABCD is equal to EFGH . PROPOSITION XXXVII . THEOREM . Triangles upon the same base , and between the same parallels ...
... reason , the paral- lelogram EFGH is equal to the same parallelogram EBCH ; therefore ( Ax . 1 ) the parallelogram ABCD is equal to EFGH . PROPOSITION XXXVII . THEOREM . Triangles upon the same base , and between the same parallels ...
Seite 35
... reason , the triangle KGC is equal to the triangle KFC . Then , because the triangle AEK is equal to the triangle AHK , and the triangle KGC to KFC ; the triangle AEK , together with the triangle KGC , is equal to the triangle AHK ...
... reason , the triangle KGC is equal to the triangle KFC . Then , because the triangle AEK is equal to the triangle AHK , and the triangle KGC to KFC ; the triangle AEK , together with the triangle KGC , is equal to the triangle AHK ...
Seite 38
... reason , AB and AH are in the same straight line ; ملو 30 B C DAN ربراون 8.000 . 891 92 : D L and because the angle DBC is equal to the angle FBA , each of them being a right angle , add to each the angle ABC , and the whole angle ABD ...
... reason , AB and AH are in the same straight line ; ملو 30 B C DAN ربراون 8.000 . 891 92 : D L and because the angle DBC is equal to the angle FBA , each of them being a right angle , add to each the angle ABC , and the whole angle ABD ...
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Euclid's Elements of Plane Geometry [Book 1-6] Explicitly Enunciated, by J ... Euclides,James Pryde Keine Leseprobe verfügbar - 2023 |
Euclid's Elements of Plane Geometry [book 1-6] Explicitly Enunciated, by J ... Euclides,James Pryde Keine Leseprobe verfügbar - 2018 |
Häufige Begriffe und Wortgruppen
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
Beliebte Passagen
Seite 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...
Seite 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.
Seite 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.
Seite 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...
Seite 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.
Seite 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.
Seite 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...
Seite 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.
Seite 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.
Seite 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.