## The Elements of Euclid: The Errors, by which Theon, Or Others, Have Long Ago Vitiated These Books are Corrected, and Some of Euclid's Demonstrations are Restored. Also, the Book of Euclid's Data, in Like Manner Corrected. viz. the first six books, together with the eleventh and twelfth |

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Resultat 1-5 av 8

Side 48

EG is double of the

of EG is double of the

EA is double of the

...

EG is double of the

**square**of EF : and EF is equal to CD ; wherefore the**square**of EG is double of the

**square**of CD : but it was de monstrated , that the**square**ofEA is double of the

**square**of AC ; therefore the squares of AE , EG are double of...

Side 78

to the

together with the

ED is equal ( 47. 1. ) to the squares of EB , BD because EBD is a right angle : E ...

to the

**square**С of ED , and CE is equal to EB : therefore the rectangle AD , DC ,together with the

**square**of В. EB , is equal to the**square**of ED : but the**square**ofED is equal ( 47. 1. ) to the squares of EB , BD because EBD is a right angle : E ...

Side 86

it in the point C , so that the rectangle AB , BC be equal to the

from the centre A , at the distance AB , describe the circle BDE , in which place ( 1

. 4. ) the straight line BD is equal to AC , which is not greater than the diameter ...

it in the point C , so that the rectangle AB , BC be equal to the

**square**of CA ; andfrom the centre A , at the distance AB , describe the circle BDE , in which place ( 1

. 4. ) the straight line BD is equal to AC , which is not greater than the diameter ...

Side 217

is the polygon ATBYCVDQ to the polygon EOFPGRHS ; and as the

to the

the circle ABCD ( 11. 5. ) is to the circle EFGH , as the polygon ATBYCVDQ to ...

is the polygon ATBYCVDQ to the polygon EOFPGRHS ; and as the

**square**of ACto the

**square**of EG , so is ( 2. 12. ) the circle ABCD to the circle EFGH ; thereforethe circle ABCD ( 11. 5. ) is to the circle EFGH , as the polygon ATBYCVDQ to ...

Side 341

Let the triangle ABC have a given obtuse angle ABC ; and produce the straight

line CB , and from the point A draw AD perpendicular to BC : the excess of the

Let the triangle ABC have a given obtuse angle ABC ; and produce the straight

line CB , and from the point A draw AD perpendicular to BC : the excess of the

**square**of AC above the squares of AB , BC , that is ( 12. 2. ) , the double of the ...### Hva folk mener - Skriv en omtale

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The Elements of Euclid: Viz. the First Six Books, Together with the Eleventh ... Euclid,Robert Simson Uten tilgangsbegrensning - 1825 |

The Elements of Euclid: Viz. the First Six Books, Together with the Eleventh ... Euclid Uten tilgangsbegrensning - 1892 |

### Vanlige uttrykk og setninger

added altitude angle ABC angle BAC base centre circle circle ABCD circumference common cone contained cylinder definition demonstrated described diameter divided double draw drawn equal equal angles equiangular equimultiples Euclid excess fore four fourth given angle given in position given in species given magnitude given ratio given straight line gles greater Greek half join less likewise manner meet multiple Note opposite parallel parallelogram pass perpendicular plane prism produced PROP proportionals proposition pyramid Q. E. D. PROP radius reason rectangle rectangle contained rectilineal figure remaining right angles segment shown sides similar sine solid solid angle sphere square square of AC taken THEOR third triangle ABC wherefore whole

### 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.