## Elements of geometry, based on Euclid, books i-iii |

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Resultat 1-5 av 34

Side 18

AE = AD , DL Join DE . A DEF e1 Upon DE , on the side remote from A ,

dequilateral . scribe an equilateral triangle DEF ( I . 1 ) . D Join AF . Then the

straight line AF shall bisect the angle BAC . PROOF . — Because AD is equal to

AE (

AE = AD , DL Join DE . A DEF e1 Upon DE , on the side remote from A ,

dequilateral . scribe an equilateral triangle DEF ( I . 1 ) . D Join AF . Then the

straight line AF shall bisect the angle BAC . PROOF . — Because AD is equal to

AE (

**Const**. ) ... Side 19

Because AC is equal to CB (

, BCD : The two sides AC , CD are equal to the two sides BC , CD , each to each ;

And the angle ACD is equal to the angle BCD (

Because AC is equal to CB (

**Const**. ) , and CD common to the two triangles ACD, BCD : The two sides AC , CD are equal to the two sides BC , CD , each to each ;

And the angle ACD is equal to the angle BCD (

**Const**. ) ; Therefore the base ... Side 20

Join CF , CH , CG . Then CH shall be perpendicular to AB . PROOF . - Because

FH is equal to HG

CD as radius . in II . The two sides FH , HC are equal to the 20 GEOMETRY .

Join CF , CH , CG . Then CH shall be perpendicular to AB . PROOF . - Because

FH is equal to HG

**Const**. ) , and HC common to the two triangles FHC , GHC .CD as radius . in II . The two sides FH , HC are equal to the 20 GEOMETRY .

Side 24

Because AE is equal to EC , and BE equal to EF (

CE , EF , each to each ; And the angle AEB is equal to the angle CEF , because

they are opposite vertical angles ( I . 15 ) . Therefore the base AB is equal to the ...

Because AE is equal to EC , and BE equal to EF (

**Const**. ) , AE , EB are equal toCE , EF , each to each ; And the angle AEB is equal to the angle CEF , because

they are opposite vertical angles ( I . 15 ) . Therefore the base AB is equal to the ...

Side 28

But FD is equal to A (

Again , because the point G is the centre of the circle IILK , GH is equal to GK (

Def . 15 ) . But GH is equal to C (

Ax . 1 ) ...

But FD is equal to A (

**Const**. ) ; FK = A . Therefore FK is equal to A ( Ax . 1 ) .Again , because the point G is the centre of the circle IILK , GH is equal to GK (

Def . 15 ) . But GH is equal to C (

**Const**. ) ; GK = C . Therefore GK is equal to C (Ax . 1 ) ...

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Elements of Geometry, Based on Euclid, Bøker 1-3 Edward Atkins Ingen forhåndsvisning tilgjengelig - 2016 |

### Vanlige uttrykk og setninger

ABCD angle ABC angle BAC angle BCD angle equal base BC BC is equal bisect centre chord circle ABC circumference coincide common Const CONSTRUCTION cqual describe diagonal diameter difference divided double draw drawn equal to CD equal to twice exterior angle extremities fall figure four given point given rectilineal given straight line gnomon greater impossible join length less Let ABC Let the straight manner meet opposite angles parallel parallelogram pass perpendicular possible produced PROOF PROOF.—Because Proposition proved rectangle contained right angles segment semicircle shown side BC sides square on AC Take taken third touches the circle triangle ABC twice the rectangle unequal whole

### Populære avsnitt

Side 35 - IF a side of any triangle be produced, the exterior angle is equal to the two interior and opposite angles; and the three interior angles of every triangle are equal to two right angles.

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Side 67 - If a straight line be bisected, and produced to any point; the rectangle contained by the whole line thus produced, and the part of it produced, together with the square of half the line bisected, is equal to the square of the straight line, which is made up of the half and the part produced.