## 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 7

Side 8

Ax Book I. Because the point B is the center of the circle CGH , BC is

. and because D is the center of the circle GKL , DL e . 15. Def . is

and DA , DB parts of them are

the ...

Ax Book I. Because the point B is the center of the circle CGH , BC is

**equal**to BG. and because D is the center of the circle GKL , DL e . 15. Def . is

**equal**to DG ,and DA , DB parts of them are

**equal**; therefore f . the remainder AL is**equal**tothe ...

Side 10

G Book I. to AC , and let the straight lines AB , AC be produced to D and m E. the

angle ABC shall be

BCE . In BD take any point F , and from AE , the greater , cut off AG a . 3. 1 .

to ...

G Book I. to AC , and let the straight lines AB , AC be produced to D and m E. the

angle ABC shall be

**equal**to the angle ACB , and the angle CBD to the angleBCE . In BD take any point F , and from AE , the greater , cut off AG a . 3. 1 .

**equal**to ...

Side 25

DE , EF , each to each ; and the angle GBC is

the base GC is

...

**equal**t ) DE , and BC to EF , the two sides GB , BC are**equal**to Book I. the twoDE , EF , each to each ; and the angle GBC is

**equal**to the angle DEF ; thereforethe base GC is

**equal**to the base DF , and a . 4. I. the triangle GBC to the triangle...

Side 219

II . to DG , and from the point K erect « KH at right angles to the Book XI . plane

BAL ; and make KH

contained by the three plane angles BAL , BAH , HAL is

at ...

II . to DG , and from the point K erect « KH at right angles to the Book XI . plane

BAL ; and make KH

**equal**to GF , and join AH . then the solid angle at A which iscontained by the three plane angles BAL , BAH , HAL is

**equal**to the solid angleat ...

Side 236

Def.11 . angles d with every straight line meeting it in that plane . but BH meets it

in that plane ; therefore ABH is a right angle . for the same reason , DEM is a right

angle , and is therefore

Def.11 . angles d with every straight line meeting it in that plane . but BH meets it

in that plane ; therefore ABH is a right angle . for the same reason , DEM is a right

angle , and is therefore

**equal**to the angle ABH . and the angle HAB is**equal**to ...### Hva folk mener - Skriv en omtale

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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.