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

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Resultat 1-5 av 5

Side 58

Thus the parallelogram HG , together with the complements AF , FC , is a

opposite angles of the parallelograms which make the

rectangle ...

Thus the parallelogram HG , together with the complements AF , FC , is a

**gnomon**, which is briefly expressed by the letters AGK , or EHC , which are at theopposite angles of the parallelograms which make the

**gnomon**. É G Therectangle ...

Side 64

To each of these add CM ; therefore the whole AM i3 cqual to the

Ax . 2 ) . But AM is the rectangle contained by AD and DB , since DM is equal to

DB ( II . 4 , cor . ) ; Therefore the

To each of these add CM ; therefore the whole AM i3 cqual to the

**gnomon**CMG (Ax . 2 ) . But AM is the rectangle contained by AD and DB , since DM is equal to

DB ( II . 4 , cor . ) ; Therefore the

**gnomon**CMG is equal to the rectangle AD , DB ... Side 65

But AK and CE are the

rectangle AB , BC is also double of AK , for BK is equal to BC ( II . 4 , cor . ) ...

But AK and CE are the

**gnomon**AKF , together with the square CK ; Therefore the**gnomon**AKF , together with the square CK , . AKF is double of AK . But twice therectangle AB , BC is also double of AK , for BK is equal to BC ( II . 4 , cor . ) ...

Side 66

And it was demonstrated that the four CK , BN , GR , and RN are quadruple of CK

; Therefore the eight rectangles which make up the

of AK , are together = 4 AG , AOH = 4 ( CK + AG ) And because 66 GEOMETP ...

And it was demonstrated that the four CK , BN , GR , and RN are quadruple of CK

; Therefore the eight rectangles which make up the

**gnomon**AOH are quadrupleof AK , are together = 4 AG , AOH = 4 ( CK + AG ) And because 66 GEOMETP ...

Side 67

AOH = 4 ( CK + AG ) And because AK is the rectangle contained by AB and

quadruple of EA AKP AK . But the

quadruple ...

AOH = 4 ( CK + AG ) And because AK is the rectangle contained by AB and

**Gnomon**BC , for BK is equal to BC ; Therefore four times the rectangle AB , BC isquadruple of EA AKP AK . But the

**gnomon**AOH was demonstrated to bequadruple ...

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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.

Side 13 - THE angles at the base of an isosceles triangle are equal to one another : and, if the equal sides be produced, the angles upon the other side of the base shall be equal. Let ABC be an isosceles triangle, of which the side AB is equal to AC, and let the straight lines AB, AC be produced to D and E.

Side 7 - A diameter of a circle is a straight line drawn through the centre, and terminated both ways by the circumference.

Side 17 - If two triangles have two sides of the one equal to two sides of the...

Side 51 - If the square described upon one of the sides of a triangle, be equal to the squares described upon the other two sides of it ; the angle contained by these two sides is a right angle.

Side 9 - If a straight line meet two straight lines, so as to make the two interior angles on the same side of it taken together less than two right angles...

Side 69 - To divide a given straight line into two parts, so that the rectangle contained by the whole, and one of the parts, may be equal to the square of the other part.

Side 9 - Things which are double of the same, are equal to one another. 7. Things which are halves of the same, are equal to one another.

Side 32 - Wherefore, if a straight line, &c. QED PROPOSITION XXVIII. THEOREM. If a straight line falling upon two other straight lines, make the exterior angle equal to the interior and opposite upon the same side of the line ; or make the interior angles upon the same side together equal to two right angles ; the two straight lines shall be parallel to one another.

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.