## 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 196

And because AB is

straight 8. 3. Def.11 . line which meets it , and is in that plane ! therefore each of

the angles ABD , ABE , is a right angle . and because the straight line BD meets

the ...

And because AB is

**perpendicular**to the plane , it is**perpendicular**to everystraight 8. 3. Def.11 . line which meets it , and is in that plane ! therefore each of

the angles ABD , ABE , is a right angle . and because the straight line BD meets

the ...

Side 198

I. AC to the base DF ; the angle ABC is equal d to the angle DEF . Therefore if two

straight lines , & c . Q. E. D. PROP . XI . PROB . To a 10 draw a straight line

above ...

I. AC to the base DF ; the angle ABC is equal d to the angle DEF . Therefore if two

straight lines , & c . Q. E. D. PROP . XI . PROB . To a 10 draw a straight line

**perpendicular**to a plane , from a given point above it . Let A be the given pointabove ...

Side 199

therefore GH is

right angles to each ...

therefore GH is

**perpendicular**to AF , and consequently AF is Book XI .**perpendicular**to GH . and AF is**perpendicular**to DE ; therefore AF is**perpendicular**to each of the straight lines GH , DE . but if a straight line stands atright angles to each ...

Side 201

From the point B draw BG

through DE , EF , and let it meet that plane in G ; and through G draw . GH

parallel o to ED , and GK parallel to EF . b . 31. t . and because BG is

From the point B draw BG

**perpendicular**to the plane which a . 11. it . passesthrough DE , EF , and let it meet that plane in G ; and through G draw . GH

parallel o to ED , and GK parallel to EF . b . 31. t . and because BG is

**perpendicular**to the ... Side 204

Def.11 . right angles to CE the common section of the planes , has been proved

to be

angles to the plane CK . In like manner , it may be proved that all the planes

which ...

Def.11 . right angles to CE the common section of the planes , has been proved

to be

**perpendicular**to the other plane CK ; therefore the plane DE is at rightangles to the plane CK . In like manner , it may be proved that all the planes

which ...

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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.