## The first three books of Euclid's Elements of geometry, with theorems and problems, by T. Tate |

### Inni boken

Resultat 1-5 av 5

Side 55

D B But if it be not in CE , let , if possible , G be the centre , and

Then , because da is equal to DB , and DG common to the two triangles ADG ,

BDG , the с two sides AD , DG are equal to the two BD , DG , each to each ; and ...

D B But if it be not in CE , let , if possible , G be the centre , and

**join**GA , GD , GB :Then , because da is equal to DB , and DG common to the two triangles ADG ,

BDG , the с two sides AD , DG are equal to the two BD , DG , each to each ; and ...

Side 62

K cumference DEF in more than two А points , viz . in B , G , F ; take the centre k

of the circle ABC , and

is taken the point K , from which E to the circumference DEF fall more than two ...

K cumference DEF in more than two А points , viz . in B , G , F ; take the centre k

of the circle ABC , and

**join**KB , KG , KF : And because within the circle DEF thereis taken the point K , from which E to the circumference DEF fall more than two ...

Side 63

If two circles touch each other externally , the straight line which

centres shall pass through the point of contact . ... and

F is the centre of the circle ABC , AF is equal to Fc : Also because G is the centre

of the ...

If two circles touch each other externally , the straight line which

**joins**theircentres shall pass through the point of contact . ... and

**join**FA , AG : And becauseF is the centre of the circle ABC , AF is equal to Fc : Also because G is the centre

of the ...

Side 95

The line

line that is drawn parallel to the base ... then PQ is bisected in the point G. If PG

be not equal to GQ , one of them is the greater : let pg be the greater ; and

DQ ...

The line

**joining**the vertex and the middle of the base of a triangle , bisects everyline that is drawn parallel to the base ... then PQ is bisected in the point G. If PG

be not equal to GQ , one of them is the greater : let pg be the greater ; and

**join**DQ ...

Side 101

the given height of the triangle ; through r draw FD parallel to AB , meeting the arc

of the circle in D ;

Ds parallel to FB , then BFDs is a parallelogram , and ( E. I. 34. ) Ds is equal to ...

the given height of the triangle ; through r draw FD parallel to AB , meeting the arc

of the circle in D ;

**join**B , D and A , D : Then ABD is the triangle required . DrawDs parallel to FB , then BFDs is a parallelogram , and ( E. I. 34. ) Ds is equal to ...

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The First Three Books of Euclid's Elements of Geometry from the Text of Dr ... Euclid,Thomas Tate Ingen forhåndsvisning tilgjengelig - 2014 |

### Vanlige uttrykk og setninger

ABCD alternate angle ABC angle ACB angle BAC angle equal base BC BC is equal bisect centre circle ABC circumference coincide common construct demonstrated describe diameter divided double draw equal angles equal to FB equilateral exterior angle extremity figure fore four given point given straight line gnomon greater impossible interior isosceles triangle join less Let ABC likewise line be drawn meet opposite angles opposite sides parallel parallelogram pass perpendicular PROB produced PROP Q. E. D. PROP rectangle contained remaining angle right angles segment semicircle shown sides squares of AC straight line AC Take taken THEOR third touch touches the circle triangle ABC twice the rectangle vertex wherefore whole

### Populære avsnitt

Side 6 - If a straight line meets 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 5 - Let it be granted that a straight line may be drawn from any one point to any other point.

Side 20 - If two triangles have two sides of the one equal to two sides of the...

Side 30 - Parallelograms upon equal bases, and between the same parallels, are equal to one another.

Side 17 - Any two angles of a triangle are together less than two right angles. Let ABC be any triangle ; any two of its angles together are less than two right angles.

Side 84 - IF from a point without a circle there be drawn two straight lines, one of which cuts the circle, and the other meets it; if the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, be equal to the square of the line which meets it, the line which meets it shall touch the circle.

Side 82 - If from any point without a circle two straight lines be drawn, one of -which cuts the circle, and the other touches it; the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, shall be equal to the square of the line which touches it.

Side 11 - UPON the same base, and on the same side of it, there cannot be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise those which are terminated in the other extremity.

Side 19 - To make a triangle of which the sides shall be equal to three given straight lines, but any two whatever of these must be greater than the third, (i.

Side 7 - From the greater of two given straight lines to cut off a part equal to the less. Let AB and C be the two given straight lines, whereof AB is the greater.