## The Elements of Euclid |

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Side 84

It is also rect- N // angular; for the straight line BD, being N 2^ / the diameter of the

31. 3.) angle; for the same reason C each of the angles ABC, BCD, CDA is a right

...

It is also rect- N // angular; for the straight line BD, being N 2^ / the diameter of the

**circle ABCD**, BAD is a N / / semicircle; wherefore the angle BAD is a N}^ - right (31. 3.) angle; for the same reason C each of the angles ABC, BCD, CDA is a right

...

Side 86

... and be described about the square

straight line BD is equal to AC, which is not greater than the diameter of the

BDE; join DA, DC, and about the triangle ADC describe (5. 4.) the

...

... and be described about the square

**ABCD**. Which was to be done. ... thestraight line BD is equal to AC, which is not greater than the diameter of the

**circle**BDE; join DA, DC, and about the triangle ADC describe (5. 4.) the

**circle**ACD; the...

Side 154

Describe (5.4) the

join EC; because the right angle BDA is ... Let

inscribed in a

to the ...

Describe (5.4) the

**circle**ACB about the triangle, and draw its diameter AE, andjoin EC; because the right angle BDA is ... Let

**ABCD**be any quadrilateralinscribed in a

**circle**, and join AC, BD; the rectangle contained by AC, BD is equalto the ...

Side 202

CIRCLEs are to one another as the squares of their diameters.” Let ABCD, EFGH

be two circles, and BD, FH their diameters: as the square of BD to the square of

FH, so is the

CIRCLEs are to one another as the squares of their diameters.” Let ABCD, EFGH

be two circles, and BD, FH their diameters: as the square of BD to the square of

FH, so is the

**circle ABCD**, to the circle EFGH. - For, if it be not so, the square of ... Side 203

excess of the circle EFGH above the space S : because, by the preceding lemma,

if from the greater of two unequal ... the polygon AXBOCPDR to the polygon

EKFLGMHN: but the

excess of the circle EFGH above the space S : because, by the preceding lemma,

if from the greater of two unequal ... the polygon AXBOCPDR to the polygon

EKFLGMHN: but the

**circle ABCD**is greater than the polygon contained in it: ...### Hva folk mener - Skriv en omtale

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### Andre utgaver - Vis alle

The Elements of Euclid: Viz, the First Six Books, Together with the Eleventh ... Euclid,Robert Simson Uten tilgangsbegrensning - 1834 |

The Elements of Euclid: Viz. The First Six Books, Together With the Eleventh ... Robert Simson Ingen forhåndsvisning tilgjengelig - 2017 |

The Elements of Euclid: Viz. The First Six Books, Together With the Eleventh ... Robert Simson Ingen forhåndsvisning tilgjengelig - 2017 |

### Vanlige uttrykk og setninger

altitude angle ABC angle BAC base BC BC is equal BC is given bisected centre circle ABCD circumference cone cylinder demonstrated described diameter draw drawn equal angles equiangular equimultiples Euclid excess fore given angle given in magnitude given in position given in species given magnitude given point given ratio given straight line gles gnomon join less Let ABC meet multiple parallel parallelogram AC perpendicular point F polygon prism proportionals proposition pyramid Q. E. D. PROP radius ratio of AE rectangle CB rectangle contained rectilineal figure remaining angle right angles segment sides BA similar sine solid angle solid parallelopiped square of AC straight line AB straight line BC tangent THEOR third three plane angles triangle ABC triplicate ratio vertex wherefore

### Populære avsnitt

Side 45 - Ir a straight line be divided into any two parts, four times the rectangle contained by the whole line, and one of the parts, together with the square of the other part, is equal to the square of the straight line which is made up of the whole and that part.

Side 41 - 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 54 - Ir any two points be taken in the circumference of a circle, the straight line which joins them shall fall within the circle. Let ABC be a circle, and A, B any two points in the circumference ; the straight line drawn from A to B shall fall within the circle.

Side 18 - ABD, the less to the greater, which is impossible ; therefore BE is not in the same straight line with BC.

Side 10 - From a given point to draw a straight line equal to a given straight line. Let A be the given point, and BC the given straight line: it is required to draw from the point A a straight line equal to BC.

Side 8 - Let it be granted that a straight line may be drawn from any one point to any other point.

Side 256 - Again ; the mathematical postulate, that " things which are equal to the same are equal to one another," is similar to the form of the syllogism in logic, which unites things agreeing in the middle term.

Side 129 - 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 23 - At a given point in a given straight line, to make a rectilineal angle equal to a given rectilineal angle. Let AB be the given straight line, and A...

Side 20 - ANY two angles of a triangle are together less than two right angles.