## 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 |

### Inni boken

Resultat 1-5 av 20

Side 33

PARALLELOGRAMS upon equal bases and between the

equal to one another . G 4.3 Let ABCD , EFGH be parallelograms upon equal

bafcs BC , FG , and between the same parallels AH , BG ; the paral- A D E H

lelogram ...

PARALLELOGRAMS upon equal bases and between the

**fame**parallels , areequal to one another . G 4.3 Let ABCD , EFGH be parallelograms upon equal

bafcs BC , FG , and between the same parallels AH , BG ; the paral- A D E H

lelogram ...

Side 35

EQUAL triangles upon equal bases , and towards the

the

EF , and towards the

EQUAL triangles upon equal bases , and towards the

**fame**parts , are betweenthe

**fame**parallels . Let the equal triangles ABC , DEF be upon equal bases BC ,EF , and towards the

**fame**parts ; A D they are between the**fame**paralléls . Side 141

The first six books, together with the eleventh and t Robert Simson. Next , Let

there be four magnitudes A , B , C , D , Book V. and other four E , F , G , H , which ,

taken two and A.B. C.D. two in a cross order , have the

E. F. G. H. ...

The first six books, together with the eleventh and t Robert Simson. Next , Let

there be four magnitudes A , B , C , D , Book V. and other four E , F , G , H , which ,

taken two and A.B. C.D. two in a cross order , have the

**fame**ratio , viz . AE. F. G. H. ...

Side 167

... GH is to HL , as FE to EK . for the

DK , and BL is to LH , as DK to KE , because the triangles BLH , DKE are

equiangular . therefore because the five fided figures AGHLB , CFEKD are

equiangular ...

... GH is to HL , as FE to EK . for the

**fame**reafon , AB is d . 22. 5 , to BL , as CD toDK , and BL is to LH , as DK to KE , because the triangles BLH , DKE are

equiangular . therefore because the five fided figures AGHLB , CFEKD are

equiangular ...

Side 184

... the triangle GCK . and because the circumference BC is equal to the

circumference CK , the remaining part of the whole circumference of the circle

ABC , is equal to the remaining part of the whole circumference of the

. wherefore ...

... the triangle GCK . and because the circumference BC is equal to the

circumference CK , the remaining part of the whole circumference of the circle

ABC , is equal to the remaining part of the whole circumference of the

**fame**circle. wherefore ...

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The Elements of Euclid, Viz: The Errors, by which Theon, Or Others, Have ... Robert Simson Uten tilgangsbegrensning - 1775 |

The Elements of Euclid: The Errors, by which Theon, Or Others, Have Long Ago ... Robert Simson Uten tilgangsbegrensning - 1762 |

The Elements of Euclid: The Errors by which Theon, Or Others, Have Long ... Robert Simson Uten tilgangsbegrensning - 1827 |

### Vanlige uttrykk og setninger

added alſo altitude angle ABC angle BAC baſe becauſe Book caſe circle circle ABCD circumference common cone cylinder Definition demonſtrated deſcribed diameter divided double draw drawn equal equal angles equiangular equimultiples exceſs fame fides figure firſt folid angle fore four fourth given angle given in poſition given magnitude given ratio given ſtraight line greater half join leſs likewiſe magnitude manner meet multiple muſt oppoſite parallel parallelogram perpendicular plane produced PROP proportionals Propoſition pyramid radius reaſon rectangle rectangle contained rectilineal remaining right angles ſame ſecond ſegment ſhall ſides ſimilar ſolid ſphere ſquare ſquare of AC taken THEOR theſe third thro triangle ABC wherefore whole

### Populære avsnitt

Side 156 - 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 3 - A circle is a plane figure contained by one line, which is called the circumference, and is such that all straight lines drawn from a certain point within the figure to the circumference, are equal to one another.

Side 323 - Equiangular parallelograms have to one another the ratio which is compounded of the ratios of their sides.

Side 92 - If from a point without a circle there be drawn two straight lines, one of which cuts the circle, and the other meets it ; if the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, be equal to the square- of the line which meets it, the line which meets shall touch the circle.

Side 80 - EA : and because AD is equal to DC, and DE common to the triangles ADE, CDE, the two sides AD, DE are equal to the two CD, DE, each to each ; and the angle ADE is equal to the angle CDE, for each of them is a right angle ; therefore the base AE is equal (4.

Side 52 - If a straight line be bisected, and produced to any point, the square of the whole line thus produced, and the square of the part of it produced, are together double of the square of half the line bisected, and of the square of the line made up of the half and the part produced.

Side 36 - To a given straight line to apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.

Side 2 - 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 54 - AB be the given straight line ; it is required to divide it into two parts, so that the rectangle contained by the whole, and one of the parts, shall be equal to the square of the other part.

Side 74 - The straight line drawn at right angles to the diameter of a circle, from the extremity of it, falls without the circle...