## 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 34

Book T. rallelogram EBCA , because the diameter AB bisects cit ; and the triangle

DBC is the half of the parallelogram ... Let the

bases BC , EF , and between the same parallels BF , AD . the

Book T. rallelogram EBCA , because the diameter AB bisects cit ; and the triangle

DBC is the half of the parallelogram ... Let the

**triangles ABC**, DEF be upon equalbases BC , EF , and between the same parallels BF , AD . the

**triangle ABC**is ... Side 148

F Let the

altitude , viz . the perpendicular drawn from ... BC is to the base CD , so is the

CF.

F Let the

**triangles ABC**, ACD , and the parallelograms EC , CF have the samealtitude , viz . the perpendicular drawn from ... BC is to the base CD , so is the

**triangle ABC**to the triangle ACD , and the parallelogram EC to the parallelogramCF.

Side 418

IF : a triangle has a given obtuse angle ; the excess of the square of the side

which fubtends the obtufe angle above the squares of the sides which contain it ,

thall have a given ratio to the triangle . Let the

IF : a triangle has a given obtuse angle ; the excess of the square of the side

which fubtends the obtufe angle above the squares of the sides which contain it ,

thall have a given ratio to the triangle . Let the

**triangle ABC**have a given obtuse ... Side 424

F a triangle have one angle given , and if the ratio of the rectangle of the fides

which contain the given angle to the square of the third fide be given ; the triangle

is given in species . Let the

F a triangle have one angle given , and if the ratio of the rectangle of the fides

which contain the given angle to the square of the third fide be given ; the triangle

is given in species . Let the

**triangle ABC**have the given angle BAC , and let the ... Side 425

Dat , OPQ . because , as has been shewn , the ratio of AD to BC is the fame with

the ratio of ( HF to FL , that is , by the construction , with the ratio of ) OR to PQ ;

and the angle BAC is equal to the angle POO . therefore the

similar ...

Dat , OPQ . because , as has been shewn , the ratio of AD to BC is the fame with

the ratio of ( HF to FL , that is , by the construction , with the ratio of ) OR to PQ ;

and the angle BAC is equal to the angle POO . therefore the

**triangle ABC**issimilar ...

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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.