## The Elements of Euclid |

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Resultat 1-5 av 12

Side 89

It is also

the angle HKL is double of the angle FKC, and KLM double of FLC, as was

before demonstrated, the angle HKL is equal to KLM : and in like manner it may

be ...

It is also

**equiangular**; for, since the angle FKC is equal to the angle FLC, and thatthe angle HKL is double of the angle FKC, and KLM double of FLC, as was

before demonstrated, the angle HKL is equal to KLM : and in like manner it may

be ...

Side 91

Euclid Robert Simson. points B, F; and join AB, BC, CD, DE, EF, FA : the hexagon

ABCDEF, is equilateral and

ABCDEF, GE is equal to GD : and because D is the centre of the circle EGCH, DE

...

Euclid Robert Simson. points B, F; and join AB, BC, CD, DE, EF, FA : the hexagon

ABCDEF, is equilateral and

**equiangular**. Because G is the centre of the circleABCDEF, GE is equal to GD : and because D is the centre of the circle EGCH, DE

...

Side 92

To inscribe an equilateral and

ABCD be the given circle; it is required to inscribe an equilateral and

quindecagon in the circle ABCD. Let AC be the side of an equilateral triangle ...

To inscribe an equilateral and

**equiangular**quindecagon, in a given circle.* LetABCD be the given circle; it is required to inscribe an equilateral and

**equiangular**quindecagon in the circle ABCD. Let AC be the side of an equilateral triangle ...

Side 131

Next, let each of the angles at C, F, be not less than a right angle: the triangle

ABC is also in this case

being A made, it may be proved in like D manner that BC is equal to BG, and the

...

Next, let each of the angles at C, F, be not less than a right angle: the triangle

ABC is also in this case

**equiangular**to the triangle DEF. The same constructionbeing A made, it may be proved in like D manner that BC is equal to BG, and the

...

Side 138

the angle BGH equal N to the angle DFE, and the angle GBH equal to FDE: L N|

K therefore the remaining an- NJ' gle FED is equal to the re- A B C D maining

angle GHB; and the triangle FDE

...

the angle BGH equal N to the angle DFE, and the angle GBH equal to FDE: L N|

K therefore the remaining an- NJ' gle FED is equal to the re- A B C D maining

angle GHB; and the triangle FDE

**equiangular**to the triangle GBH; then, because...

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### Andre utgaver - Vis alle

The Elements of Euclid: Viz, the First Six Books, Together with the Eleventh ... Euclid,Robert Simson Uten tilgangsbegrensning - 1834 |

The Elements of Euclid: Viz. The First Six Books, Together With the Eleventh ... Robert Simson Ingen forhåndsvisning tilgjengelig - 2017 |

The Elements of Euclid: Viz. The First Six Books, Together With the Eleventh ... Robert Simson Ingen forhåndsvisning tilgjengelig - 2017 |

### 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.