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

### Inni boken

Side

5 O SCHOOL ATLAS of Modern & Ancient Geography ; containing 34 quarto

Writing and

Copy - Books , each 0 6 SMALL SCHOOL AND FAMILY MAPS ; - 13 Fine Post

Copy ...

5 O SCHOOL ATLAS of Modern & Ancient Geography ; containing 34 quarto

Writing and

**Drawing**. Maps , coloured with Index . . . . . . . 10 6 Writing : 15 PostCopy - Books , each 0 6 SMALL SCHOOL AND FAMILY MAPS ; - 13 Fine Post

Copy ...

Side 7

... the circle ACE ; and from the point C , in which the circles cut one another ,

be the equilateral FIRST BOOK . EXPLANATION OF SIGNS, ABBREVIATIONS,

... the circle ACE ; and from the point C , in which the circles cut one another ,

**draw**the straight lines ( Post . 1 ) CA and CB , to the . points A and B ; ABC shallbe the equilateral FIRST BOOK . EXPLANATION OF SIGNS, ABBREVIATIONS,

Side 8

From a given point , to

the point A , and the straight line BC ; it is required to

straight line equal to BC . ( Const . ) From the point A to B

straight ...

From a given point , to

**draw**a straight line equal to a given straight line . Given ,the point A , and the straight line BC ; it is required to

**draw**from the point A astraight line equal to BC . ( Const . ) From the point A to B

**draw**( Post . 1 ) thestraight ...

Side 9

From the point A

A , and with the radius AD , describe ( Post . 3 ) the circle DEF . ( Denn . )

Because A is the centre of the circle DEF , AE is equal to AD ; but the straight line

C is ...

From the point A

**draw**( I . 2 ) thé straight line AD equal to C ; and from the centreA , and with the radius AD , describe ( Post . 3 ) the circle DEF . ( Denn . )

Because A is the centre of the circle DEF , AE is equal to AD ; but the straight line

C is ...

Side 13

... but the angle ACD is also equal to the angle BCD ; therefore the base AD is

equal to the base DB ( I . 4 ) , and the * straight line AB is divided into two equal

parts in the point D . PROPOSITION XI . PROBLEM . To

right ...

... but the angle ACD is also equal to the angle BCD ; therefore the base AD is

equal to the base DB ( I . 4 ) , and the * straight line AB is divided into two equal

parts in the point D . PROPOSITION XI . PROBLEM . To

**draw**a straight line atright ...

### Hva folk mener - Skriv en omtale

Vi har ikke funnet noen omtaler på noen av de vanlige stedene.

### 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 - 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.