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

### Inni boken

Resultat 1-5 av 35

Side 10

...

to the angle DEF , and the angle ACB to ... If the equal sides AB and DE be

produced to G and H , the angles GBC and HEF below the base will also be

equal .

...

**remaining angles**of the other , and be equal to them ; namely , the angle ABCto the angle DEF , and the angle ACB to ... If the equal sides AB and DE be

produced to G and H , the angles GBC and HEF below the base will also be

equal .

Side 15

Given the straight line AB making with CD , upon one side of it , the angles CBA

and ABD ; to prove that these are either ... Hence if one of a pair of

supplementary angles be equal to one of another pair , the

must be equal .

Given the straight line AB making with CD , upon one side of it , the angles CBA

and ABD ; to prove that these are either ... Hence if one of a pair of

supplementary angles be equal to one of another pair , the

**remaining angles**must be equal .

Side 16

Take away the common angle AED , and the

3 ) to the

the angles CEB and AED are equal . COR . 1 . – From this it is manifest that , if ...

Take away the common angle AED , and the

**remaining angle**CEA is equal ( Ax .3 ) to the

**remaining angle**DEB . In the same manner it can be demonstrated thatthe angles CEB and AED are equal . COR . 1 . – From this it is manifest that , if ...

Side 17

... each to each ; and the angle AEB is equal to the angle CEF ( I . 15 ) , because

they are opposite vertical angles ; therefore ( I . 4 ) the base AB is equal to the

base CF , and the triangle AEB to the triangle CEF , and the

the ...

... each to each ; and the angle AEB is equal to the angle CEF ( I . 15 ) , because

they are opposite vertical angles ; therefore ( I . 4 ) the base AB is equal to the

base CF , and the triangle AEB to the triangle CEF , and the

**remaining angles**tothe ...

Side 35

the given triangle ABC , and having one of ... equal to the whole ADC ; therefore

the

.

**angle**D ; wherefore there has been described a parallelogram FECG equal tothe given triangle ABC , and having one of ... equal to the whole ADC ; therefore

the

**remaining**complement BK is equal to the**remaining**complement KD ( Ax . 3 ).

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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 angle ABC angle BAC base BC is equal bisected called centre chord circle ABC circumference common Const cosine demonstrated described diameter difference divided double draw equal equal angles equiangular equilateral equimultiples extremities figure fore four fourth given given point given straight line greater half hence inscribed join less magnitudes manner mean 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 sine square square on AC straight line Take taken tangent THEOREM third triangle ABC vertical 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.