## 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 6

Side 40

From the point A draw a AC at

and thro ' the point D draw DE parallel to it , and thro ' B draw BE parallel to AD .

therefore ADEB is a parallelogram ; whence AB is equal d to DE , and AD to BE ...

From the point A draw a AC at

**right angles**to AB ; and make b AD equal to AB ,and thro ' the point D draw DE parallel to it , and thro ' B draw BE parallel to AD .

therefore ADEB is a parallelogram ; whence AB is equal d to DE , and AD to BE ...

Side 85

IN D N a circle , the angle in a semicircle is a

segment greater than a semicircle is less than a

segment less than a femicircle is greater than a

, of ...

IN D N a circle , the angle in a semicircle is a

**right angle**; but the angle in asegment greater than a semicircle is less than a

**right angle**; and the angle in asegment less than a femicircle is greater than a

**right angle**. Let ABCD be a circle, of ...

Side 194

BE , it is also at

Def.st. fore makes

BF which is in that plane meets F it . therefore the angle ABF is a

...

BE , it is also at

**right angles**to the plane passing thro ' them ; and there . A c . 3.Def.st. fore makes

**right angles**with every straight line meeting it in tliat plane ; butBF which is in that plane meets F it . therefore the angle ABF is a

**right angle**; but...

Side 196

Book XI . с Let AB , CD be two parallel straight lines , and let one of them AB be at

, CD , meet the plane in the points B , D , and join BD . therefore AB , CD , BD ...

Book XI . с Let AB , CD be two parallel straight lines , and let one of them AB be at

**right angles**to a plane ; the other CD is at**right angles**to the same plane . Let AB, CD , meet the plane in the points B , D , and join BD . therefore AB , CD , BD ...

Side 269

5 . are about equal angles AKQ , EMS , because these angles are , Book XII .

each of them , the fame part of four

triangle AKQ is similar to the triangle EMS . c . 6.6 . and because it has been

shewn ...

5 . are about equal angles AKQ , EMS , because these angles are , Book XII .

each of them , the fame part of four

**right angles**at the centers K , M ; therefore thetriangle AKQ is similar to the triangle EMS . c . 6.6 . and because it has been

shewn ...

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### Vanlige uttrykk og setninger

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.