## 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 14

Book I. Because AC is equal to CB , and CD

BCD ; the two fides AC , CD are equal to BC , CD , each to each ; and the angle

ACD is equal to the angle BCD ; therefore the base AD is equal to the base o DB

...

Book I. Because AC is equal to CB , and CD

**common**to the two triangles ACD ,BCD ; the two fides AC , CD are equal to BC , CD , each to each ; and the angle

ACD is equal to the angle BCD ; therefore the base AD is equal to the base o DB

...

Side 204

Book XI . to their

and any straight line FG in the plane DE , which is at 4.4.Def.11 . right angles to

CE the

the ...

Book XI . to their

**common**section , are also at right angles to the other plane d ;and any straight line FG in the plane DE , which is at 4.4.Def.11 . right angles to

CE the

**common**section of the planes , has been proved to be perpendicular tothe ...

Side 334

F fame manner the triangle EBC is similar to the triangle FBC , and the triangle

EAC to FAC . therefore there are two solid figures each of which is contained by

six triangles , one of them by three triangles the

F fame manner the triangle EBC is similar to the triangle FBC , and the triangle

EAC to FAC . therefore there are two solid figures each of which is contained by

six triangles , one of them by three triangles the

**common**vertex of which is the ... Side 340

Pouk XI . these points

, that they cannot have a

meet in two points , from which their not having a

Pouk XI . these points

**common**to both . Now , as it has not been shewn in Euclid, that they cannot have a

**common**segment , this does not prove that they cannotmeet in two points , from which their not having a

**common**segment is deduced ... Side 15

in the same manner the plane FHG is perpendicular to the plane ADB ; and

therefore GF the .

the plane ADB ; ( 19. II . ) and because the angle FHG is the inclination of the

planes ...

in the same manner the plane FHG is perpendicular to the plane ADB ; and

therefore GF the .

**common**section of the planes FEG , FHG is perpendicular tothe plane ADB ; ( 19. II . ) and because the angle FHG is the inclination of the

planes ...

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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.