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

### Inni boken

Resultat 1-5 av 5

Side 28

TRAIGHT lines which are

another . Let AB , CD be each of them

Let the straight line GHK cut AB , EF , CD ; and because GHK cuts the

TRAIGHT lines which are

**parallel**to the same straight line , are**parallel**to oneanother . Let AB , CD be each of them

**parallel**to EF ; AB is also**parallel**to CD .Let the straight line GHK cut AB , EF , CD ; and because GHK cuts the

**parallel**... Side 35

equal 6 to the triangle EBC , because it is upon the same base BC , Book I. and

between the fame

same manner it can be demonstrated B that no other line but AD is

...

equal 6 to the triangle EBC , because it is upon the same base BC , Book I. and

between the fame

**parallels**BC , AE . ... therefore AE is not**parallel**to BC . in thesame manner it can be demonstrated B that no other line but AD is

**parallel**to BC...

Side 197

THEOR . wo straight lines which are each of them

line , and not in the same plane with it , are

CD be each of them

...

THEOR . wo straight lines which are each of them

**parallel**to the same straightline , and not in the same plane with it , are

**parallel**to one another . TWO Let AB ,CD be each of them

**parallel**to EF , and not in the fame plane with it ; AB shall be...

Side 198

Book XI . fore AD is a both equal and

is equal a . 33. I. and

equal and

straight ...

Book XI . fore AD is a both equal and

**parallel**to B BE . For the famé reason , CFis equal a . 33. I. and

**parallel**to BE . Therefore AD and CF are each of themequal and

**parallel**A C to BE . But straight lines that are**parallel**to the samestraight ...

Side 202

See N. [ F two

planes AB , CD be cut by the plane EFHG , and let their common sections with it

be EF , GH . EF is

...

See N. [ F two

**parallel**planes be cut by another plane , their Let the**parallel**planes AB , CD be cut by the plane EFHG , and let their common sections with it

be EF , GH . EF is

**parallel**to GH . For , if it is not , EF , GH shall meet , if produced...

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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.