## Euclid's Elements of plane geometry [book 1-6] explicitly enunciated, by J. Pryde. [With] Key |

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Resultat 6-10 av 37

Side 78

The

nearer to the centre is always greater than one more remote ; and the greater is

nearer to the centre than the less . Given a circle ABCD , of which the

AD ...

The

**diameter**is the greatest chord in a circle ; and , of all others , that which isnearer to the centre is always greater than one more remote ; and the greater is

nearer to the centre than the less . Given a circle ABCD , of which the

**diameter**isAD ...

Side 79

The straight line drawn at right angles to the

extremity of it , falls without the circle ; ( 2 ) and a straight line making an acute

angle with the

centre of ...

The straight line drawn at right angles to the

**diameter**of a circle , from theextremity of it , falls without the circle ; ( 2 ) and a straight line making an acute

angle with the

**diameter**at its extremity cuts the circle . Given a circle ABC , thecentre of ...

Side 88

Given a circle ABCD , of which the

dividing the circle into the segments ABC and ADC , and join BA , AD , and DC ;

to prove that the angle in the semicircle BAC is a right angle ; and the angle in the

...

Given a circle ABCD , of which the

**diameter**is BC , and centre E ; and draw CAdividing the circle into the segments ABC and ADC , and join BA , AD , and DC ;

to prove that the angle in the semicircle BAC is a right angle ; and the angle in the

...

Side 90

And because from the point A , the extremity of the

right angles to AE , therefore AD touches the circle ( III . 16 , Cor . ) ; and because

AB drawn from the point of contact A cuts the circle , the angle DAB is equal to ...

And because from the point A , the extremity of the

**diameter**AE , AD is drawn atright angles to AE , therefore AD touches the circle ( III . 16 , Cor . ) ; and because

AB drawn from the point of contact A cuts the circle , the angle DAB is equal to ...

Side 92

... is therefore equal to the remaining rectangle H BEED , Case Fourth . - - Let

neither of the straight lines AC , BD pass through the centre ; take the centre F ,

and through E , the Af intersection of the straight lines AC and DB , draw the

... is therefore equal to the remaining rectangle H BEED , Case Fourth . - - Let

neither of the straight lines AC , BD pass through the centre ; take the centre F ,

and through E , the Af intersection of the straight lines AC and DB , draw the

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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 - 2018 |

Euclid's Elements of Plane Geometry [Book 1-6] Explicitly Enunciated, by J ... Euclides,James Pryde Ingen forhåndsvisning tilgjengelig - 2015 |

### Vanlige uttrykk og setninger

ABCD alternate angle ABC angle BAC base BC is equal bisected called centre chord circle ABC circumference common Const demonstrated described diameter difference divided double draw equal equal angles equiangular equilateral equimultiples extremities fall figure fore four fourth given given line given point given straight line greater half hence inscribed join less magnitudes manner meet multiple namely parallel parallelogram pass perpendicular polygon PROBLEM produced proportional PROPOSITION prove radius ratio reason rectangle contained rectilineal figure remaining angle respectively right angles segment shewn sides similar square square on AC straight line Take taken tangent THEOREM third triangle ABC twice the rectangle wherefore whole

### 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.