## The Elements of Euclid |

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Resultat 1-5 av 35

Side 3

Peter Ramus ascribes the Propositions, as well as their Demonstrations, to

Theon; others think the Propositions to be

are Theon's; and others maintain, that all the Propositions and their

Demonstrations are ...

Peter Ramus ascribes the Propositions, as well as their Demonstrations, to

Theon; others think the Propositions to be

**Euclid's**, but that the Demonstrationsare Theon's; and others maintain, that all the Propositions and their

Demonstrations are ...

Side 25

equal, each to each; and also the third angle of the one to the third angle of the

other. Let ABC, DEF be two triangles which have the angles ABC, BCA equal to

the ...

**Euclid**Robert Simson. o to equal angles in each; then shall the other sides beequal, each to each; and also the third angle of the one to the third angle of the

other. Let ABC, DEF be two triangles which have the angles ABC, BCA equal to

the ...

Side 36

HB. Then, because the straight line HF falls upon the parallels AH, EF, the ahgles

AHF, HFE are together equal (29. 1.) to two right angles: wherefore the angles ...

**Euclid**Robert Simson. through A draw (31. 1.) AH parallel to BG or EF, and joinHB. Then, because the straight line HF falls upon the parallels AH, EF, the ahgles

AHF, HFE are together equal (29. 1.) to two right angles: wherefore the angles ...

Side 86

equal to AB, and AC common to the triangles DAC, BAC, the two sides DA, AC

are equal to the two BA, AC ; and the base DC is equal to the base BC; wherefore

...

**Euclid**Robert Simson. Join AC, BD cutting one another in E; and because DA isequal to AB, and AC common to the triangles DAC, BAC, the two sides DA, AC

are equal to the two BA, AC ; and the base DC is equal to the base BC; wherefore

...

Side 89

of KC: in the same manner, it may be shown that HK is double of BK : and

because BK is equal to KC, as was demonstrated, and that KL is double of KC,

and HK ...

**Euclid**Robert Simson. angle FLC: and because KC is equal to CL, KL is doubleof KC: in the same manner, it may be shown that HK is double of BK : and

because BK is equal to KC, as was demonstrated, and that KL is double of KC,

and HK ...

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The Elements of Euclid: Viz. the First Six Books, Together with the Eleventh ... Euclid Ingen forhåndsvisning tilgjengelig - 2016 |

The Elements of Euclid: Viz, the First Six Books, Together with the Eleventh ... Euclid Ingen forhåndsvisning tilgjengelig - 2016 |

### Vanlige uttrykk og setninger

altitude angle ABC angle BAC base BC BC is equal BC is given bisected centre circle ABCD circumference cone cylinder demonstrated described diameter draw drawn equal angles equiangular equimultiples Euclid excess fore given angle given in magnitude given in position given in species given magnitude given point given ratio given straight line gles gnomon join less Let ABC meet multiple parallel parallelogram AC perpendicular point F polygon prism proportionals proposition pyramid Q. E. D. PROP radius ratio of AE rectangle CB rectangle contained rectilineal figure remaining angle right angles segment sides BA similar sine solid angle solid parallelopiped square of AC straight line AB straight line BC tangent THEOR third three plane angles triangle ABC triplicate ratio vertex wherefore

### Populære avsnitt

Side 45 - Ir a straight line be divided into any two parts, four times the rectangle contained by the whole line, and one of the parts, together with the square of the other part, is equal to the square of the straight line which is made up of the whole and that part.

Side 41 - If a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts, is equal to the rectangle contained by the two parts, together with the square of the aforesaid part.

Side 54 - Ir any two points be taken in the circumference of a circle, the straight line which joins them shall fall within the circle. Let ABC be a circle, and A, B any two points in the circumference ; the straight line drawn from A to B shall fall within the circle.

Side 18 - ABD, the less to the greater, which is impossible ; therefore BE is not in the same straight line with BC.

Side 10 - From a given point to draw a straight line equal to a given straight line. Let A be the given point, and BC the given straight line: it is required to draw from the point A a straight line equal to BC.

Side 8 - Let it be granted that a straight line may be drawn from any one point to any other point.

Side 256 - Again ; the mathematical postulate, that " things which are equal to the same are equal to one another," is similar to the form of the syllogism in logic, which unites things agreeing in the middle term.

Side 129 - If two triangles have one angle of the one equal to one angle of the other and the sides about these equal angles proportional, the triangles are similar.

Side 23 - At a given point in a given straight line, to make a rectilineal angle equal to a given rectilineal angle. Let AB be the given straight line, and A...

Side 20 - ANY two angles of a triangle are together less than two right angles.