## Euclid's Elements of Geometry: The First Six, the Eleventh and Twelfth Books |

### From inside the book

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Page 169

I , A Right lined figure is faid to be

angle of the figure

every ...

I , A Right lined figure is faid to be

**inscribed**in a right lined figure , when everyangle of the figure

**inscribed**touches every side of the figure in which it is**inscribed**. 2. A right lined figure is said to circumscribe a right lined figure , whenevery ...

Page 172

B Wherefore a triangle is

Which was to be done . * This problem may be otherwise constructed thus : Let

A B C be the given circle , whose centre is H , and d e f a given triangle to which a

...

B Wherefore a triangle is

**inscribed**in a circle , equiangular to a given triangle .Which was to be done . * This problem may be otherwise constructed thus : Let

A B C be the given circle , whose centre is H , and d e f a given triangle to which a

...

Page 187

... and is

equiangular hexagon is

Corollary . From hence it is manifest that the fide of an hexagon * is equal to the

femidiameter ...

... and is

**inscribed**in the circle A B C D E F. Therefore an equilateral andequiangular hexagon is

**inscribed**in a given circle . Which was to be done .Corollary . From hence it is manifest that the fide of an hexagon * is equal to the

femidiameter ...

Page 188

It is required to

AB C D. Let A c be the side of an equilateral triangle

, and A B the side of an equilateral pentagon

It is required to

**inscribe**an equilateral and equiangular quindecagon in the circleAB C D. Let A c be the side of an equilateral triangle

**inscribed**in the circle ABC D, and A B the side of an equilateral pentagon

**inscribed**in it ; then if the circle ... Page 202

A Let ABCDEFGH be a regular octagon

Biseat [ by 10. 1. ) the side A B in the point K , and draw the semidiameter L i ,

and join A I , I B. Then because K A is equal to K B , the angles AK I , BKI [ by 3. 3.

) ...

A Let ABCDEFGH be a regular octagon

**inscribed**in a circle , whose centre is L.Biseat [ by 10. 1. ) the side A B in the point K , and draw the semidiameter L i ,

and join A I , I B. Then because K A is equal to K B , the angles AK I , BKI [ by 3. 3.

) ...

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### Common terms and phrases

A B C ABCD added alſo altitude baſe becauſe centre circle circumference common cone cylinder definition demonſtrated deſcribed diameter difference divided double draw drawn equal equal angles equiangular equimultiples Euclid exceeds fall fame fides figure firſt folid fore four fourth given right line greater half inſcribed join leſs magnitudes manner meet multiple oppoſite parallel parallelogram perpendicular plane polygon priſms PROP proportional propoſition proved pyramid ratio rectangle remaining angle right angles right line A B right lined figure ſame ſay ſecond ſegment ſhall ſides ſimilar ſince ſolid ſome ſphere ſquare ſtand ſum taken THEOR theſe third thoſe thro touch triangle triangle ABC twice vertex Wherefore whole whoſe baſe

### Popular passages

Page 247 - 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.

Page 30 - If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, viz. either the sides adjacent to the equal...

Page 248 - But it was proved that the angle AGB is equal to the angle at F ; therefore the angle at F is greater than a right angle : But by the hypothesis, it is less than a right angle ; which is absurd.

Page 18 - When a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right, and the straight line standing on the other is called a perpendicular to that on which it stands.

Page 32 - Let the straight line EF, which falls upon the two straight lines AB, CD, make the alternate angles AEF, EFD equal to one another; AB is parallel to CD.

Page 56 - Therefore all the angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has sides.

Page 391 - KL: but the cylinder CM is equal to the cylinder EB, and the axis LN to the axis GH; therefore as the cylinder EB to...

Page 110 - If any two points be taken in the circumference of a circle, the straight line which joins them shall fall within the circle.

Page 130 - When you have proved that the three angles of every triangle are equal to two right angles...

Page 183 - FK : in the same manner it may be demonstrated, that FL, FM, FG are each of them equal to FH, or FK : therefore the five straight lines FG, FH, FK, FL, FM are equal to one another : wherefore the circle described from the centre F, at the distance of...