## 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 9

... the greater , cut off AG equal ( 1. 3. ) to AF , the less , and join FC , GB .

Because AF is equal to AG , and AB to AC ( Hyp . ) , the two sides FA , AC are

equal to the two GA , A B , each to each ; and they contain the angle FAG

... the greater , cut off AG equal ( 1. 3. ) to AF , the less , and join FC , GB .

Because AF is equal to AG , and AB to AC ( Hyp . ) , the two sides FA , AC are

equal to the two GA , A B , each to each ; and they contain the angle FAG

**common**to ... Side 12

... to AD ; join DE , and upon it , on the side towards B and c , describe ( 1 . 1 . ) an

equilateral triangle DEF ; then join AF ; the straight line AF bisects the triangle

BAC . Because AD is equal to A E , and AF is

... to AD ; join DE , and upon it , on the side towards B and c , describe ( 1 . 1 . ) an

equilateral triangle DEF ; then join AF ; the straight line AF bisects the triangle

BAC . Because AD is equal to A E , and AF is

**common**to the two triangles DAF ... Side 13

Because DC is equal to CE , and FC

the two sides DC , CF , are equal to the two EC , CF , each to each ; and the base

DF is equal to the base EF ; therefore the angle dcf is equal ( 1. 8. ) to the ...

Because DC is equal to CE , and FC

**common**to the two triangles DCF , ECF ; ADthe two sides DC , CF , are equal to the two EC , CF , each to each ; and the base

DF is equal to the base EF ; therefore the angle dcf is equal ( 1. 8. ) to the ...

Side 59

than the third , BE , EF are greater than BF ; but AE is equal to EB ; therefore AE ,

EF , A that is AF , is greater than by : Again , because BE is equal to CE , and FE

than the third , BE , EF are greater than BF ; but AE is equal to EB ; therefore AE ,

EF , A that is AF , is greater than by : Again , because BE is equal to CE , and FE

**common**to the triangles BEF , CEF , the two sides BE , EF are equal to the two ... Side 86

For , by construction , the angle DAE is equal to the angle DAF , and the angles at

D are right angles , and moreover Ad is

ADF , therefore ( E. 1. 26. ) the angle AED is equal to the angle AFD . 4. From two

...

For , by construction , the angle DAE is equal to the angle DAF , and the angles at

D are right angles , and moreover Ad is

**common**to both the triangles ADE andADF , therefore ( E. 1. 26. ) the angle AED is equal to the angle AFD . 4. From two

...

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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.