## 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 90

AG is equal a to GC ; wherefore the

of EG , is equal o to the square of AG . to each of these equals add the square of

GF , therefore the

AG is equal a to GC ; wherefore the

**rectangle**AE , EC , together with the squareof EG , is equal o to the square of AG . to each of these equals add the square of

GF , therefore the

**rectangle**AE , b . So 2 . EC , together with the squares of EG ... Side 328

Book VI . tangle AH is equal to the given square upon the straight line C.

wherefore the

to the given straight line AB , deficient by the square GK . Which was to be done .

2.

Book VI . tangle AH is equal to the given square upon the straight line C.

wherefore the

**rectangle**AH equal to the given square upon C , has been appliedto the given straight line AB , deficient by the square GK . Which was to be done .

2.

Side 329

3 . the

AL the half of AB , wherefore the

of AL , that is of KG . add to each the square of KE , therefore the square bof AK is

...

3 . the

**rectangle**C , D , from the determination , is not greater than the square ofAL the half of AB , wherefore the

**rectangle**EA , AH is not greater than the squareof AL , that is of KG . add to each the square of KE , therefore the square bof AK is

...

Side 440

D than GH , GL , that is than the square of AC ; wherefore twice the

AF is less than the square of AC , and the

the square of AC , that is than the

...

D than GH , GL , that is than the square of AC ; wherefore twice the

**rectangle**CA ,AF is less than the square of AC , and the

**rectangle**CA , AF itself less than halfthe square of AC , that is than the

**rectangle**contained by the diameter AC and its...

Side 441

The first six books, together with the eleventh and twelfth Robert Simson. BC ,

and because the angle ABL is equal to EFG , the triangle ABL is equiangular to

EFG . and the parallelogram AC , that is the

AB ...

The first six books, together with the eleventh and twelfth Robert Simson. BC ,

and because the angle ABL is equal to EFG , the triangle ABL is equiangular to

EFG . and the parallelogram AC , that is the

**rectangle**AL , BC is to the**rectangle**AB ...

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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.