## The geometry, by T. S. Davies. Conic sections, by Stephen Fenwick |

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Resultat 1-5 av 86

Side 3

... do not meet . POSTULATES . 1. Let it be granted that a straight line may be

drawn from any one point to any other point . 2. That a terminated straight line

may be produced to any length in a straight line . 3. And that a circle may be

... do not meet . POSTULATES . 1. Let it be granted that a straight line may be

drawn from any one point to any other point . 2. That a terminated straight line

may be produced to any length in a straight line . 3. And that a circle may be

**described**... Side 4

Let AB be the given straight line ; it is required to

upon it . ... wherefore CA , AB , BC are equal to one another ; and the triangle

ABC is therefore equilateral , and it is

Let AB be the given straight line ; it is required to

**describe**an equilateral triangleupon it . ... wherefore CA , AB , BC are equal to one another ; and the triangle

ABC is therefore equilateral , and it is

**described**upon the given straight line AB . Side 5

the straight line AB ; and upon it

and produce ( 2 Post . ) the straight lines DA , DB , to E and F ; from the centre B ,

at the distance BC

...

the straight line AB ; and upon it

**describe**( 1. 1. ) the equilateral triangle DAB ,and produce ( 2 Post . ) the straight lines DA , DB , to E and F ; from the centre B ,

at the distance BC

**describe**( 3 Post . ) the circle CGH , and from the centre D , at...

Side 9

off AE equal to AD ; join DE , and upon the side DE remote from A ,

. ) an equilateral triangle DEF ; then join AF ; the straight line AF bisects the angle

BAC . Because AD is equal ( Constr . ) to AE , and AF is common to the two ...

off AE equal to AD ; join DE , and upon the side DE remote from A ,

**describe**( 1. 1. ) an equilateral triangle DEF ; then join AF ; the straight line AF bisects the angle

BAC . Because AD is equal ( Constr . ) to AE , and AF is common to the two ...

Side 10

It is required to draw a straight line perpendicular to AB from the point C. Take

any point D upon the other side of A B , and from the centre C , at the distance CD

,

It is required to draw a straight line perpendicular to AB from the point C. Take

any point D upon the other side of A B , and from the centre C , at the distance CD

,

**describe**( 3 Post . ) the circle FDG meeting AB in F , G ; and bisect ( 10. 1. ) ...### Hva folk mener - Skriv en omtale

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The geometry, by T. S. Davies. Conic sections, by Stephen Fenwick Royal Military Academy, Woolwich Uten tilgangsbegrensning - 1853 |

### Vanlige uttrykk og setninger

ABCD axis base bisected called centre circle circumference coincide common cone construction contained curve described diameter difference dihedral angles direction distance divided double draw drawn edges ellipse equal equal angles equimultiples extremities faces figure formed four fourth given line given point greater hence horizontal inclination intersection join less likewise magnitudes manner meet method multiple opposite parallel parallel planes parallelogram pass perpendicular perspective picture plane MN plane of projection plane PQ position preceding prism problem produced profile angles Prop proportional PROPOSITION proved ratio reason rectangle remaining respectively right angles SCHOLIUM segment shown sides similar sphere square straight line surface taken tangent Theor third touch trace transverse triangle triangle ABC trihedral vertex vertical Whence Wherefore whole

### Populære avsnitt

Side 19 - That, if a straight line falling on two straight lines make the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles.

Side 35 - If a straight line be divided into two equal parts, and also into two unequal parts ; the rectangle contained by the unequal parts, together with the square on the line between the points of section, is equal to the square on half the line.

Side 4 - AB; but things which are equal to the same are equal to one another...

Side 128 - EQUIANGULAR parallelograms have to one another the ratio which is compounded of the ratios of their sides.* Let AC, CF be equiangular parallelograms, having the angle BCD equal to the angle ECG : the ratio of the parallelogram AC to the parallelogram CF, is the same with the ratio which is compounded of the ratios of their sides. Let BG, CG, be placed in a straight line ; therefore DC and CE are also in a straight line (14.

Side 8 - If two triangles have two sides of the one equal to two sides of the...

Side 36 - 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...

Side 21 - BCD, and the other angles to the other angles, (4. 1.) each to each, to which the equal sides are opposite : therefore the angle ACB is equal to the angle CBD ; and because the straight line BC meets the two straight lines AC, BD, and makes the alternate angles ACB, CBD equal to one another, AC is parallel (27. 1 .) to DB ; and it was shown to be equal to it. Therefore straight lines, &c.

Side 65 - If a straight line touch a circle, and from the point of contact a straight line be drawn cutting the circle, the angles which this line makes with the line touching the circle shall be equal to the angles which are in the alternate segments of the circle.

Side 4 - Magnitudes which coincide with one another, that is, which exactly fill the same space, are equal to one another.

Side 116 - 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.