## The Elements of Euclid: Also the Book of Euclid's Data ...cor. viz. The first six books, together with the eleventh and twelfth |

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Resultat 1-5 av 5

Side 262

Book XII . greater than the

erect a prism of the fame altitude with the cylinder ; this prism is greater than

of the cylinder ; because if a square be described about the circle , and a prism ...

Book XII . greater than the

**half**of the circle ABCD + . upon the square ABCD boperect a prism of the fame altitude with the cylinder ; this prism is greater than

**half**of the cylinder ; because if a square be described about the circle , and a prism ...

Side 5

N a plain triangle , the sum of any two fides is to their difference , as the tangent of

ABC be a plain triangle , the sum of any two sides AB , AC will be to their ...

N a plain triangle , the sum of any two fides is to their difference , as the tangent of

**half**che sum of the angles at the base , to the tangent of**half**their difference . LetABC be a plain triangle , the sum of any two sides AB , AC will be to their ...

Side 9

Let AD be

, then AC will be equal to BD , and DC together ; that is , to BC , and twice DC ;

confequently twice DC is the difference , and DC

Let AD be

**half**the given sum ; and to AD , DB , which are equa- , let DC be added, then AC will be equal to BD , and DC together ; that is , to BC , and twice DC ;

confequently twice DC is the difference , and DC

**half**that difference ; but AC the ... Side 34

WHE rectangle contained by

fines of two arches , is equal to the rectangle contained by the fines of

fum , and

and let ...

WHE rectangle contained by

**half**of the radius , and the excess of the versedfines of two arches , is equal to the rectangle contained by the fines of

**half**thefum , and

**half**the difference of the fame arches . Let AB , AC be any two arches ,and let ...

Side 35

BE or AE therefore is the fine of the arch DB or AD , the

perpendicular to AC , and AF will be the versed fine of the arch BA ; but , because

of the similar triangles CAE , BAF , CA is to AE as AB , that is , twice AE to AF ;

and ...

BE or AE therefore is the fine of the arch DB or AD , the

**half**of AB : let BF beperpendicular to AC , and AF will be the versed fine of the arch BA ; but , because

of the similar triangles CAE , BAF , CA is to AE as AB , that is , twice AE to AF ;

and ...

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

### Populære avsnitt

Side 157 - D ; wherefore the remaining angle at C is equal to the remaining angle at F ; Therefore the triangle ABC is equiangular to the triangle DEF. 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.

Side 24 - ... be 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, ECA equal to the angles DEF, EFD, viz.

Side 45 - 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 73 - THE straight line drawn at right angles to the diameter of a circle, from the extremity of...

Side 163 - 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 81 - ELF. Join BC, EF ; and because the circles ABC, DEF are equal, the straight lines drawn from their centres are equal: therefore the two sides BG, GC, are equal to the two EH, HF ; and the angle at G is equal to the angle at H ; therefore the base BC is equal (4.

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

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 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.