## Elements of Geometry: Containing the First Six Books of Euclid, with a Supplement of the Quadrature of the Circle and the Geometry of Solids |

### Inni boken

Side 32

Let the straight line GHK

Let the straight line GHK

**cut**AB , EF , CD ; and because GHK**cuts**the**parallel**straight lines AB , EF , the angle AGH is ... and**BC**the given straight line ; it is required to**draw**a A straight line through the E F point A ,**parallel**to ... Side 158

ADA E B Next , let the sides AB , AC

ADA E B Next , let the sides AB , AC

**of**the triangle ABC , or these sides produced , be**cut**proportionally in the points D , E , that is , so that BD be to DA , as CE to EA ; and**join**DE ; DE is**parallel**to**BC**. Side 168

Containing the First Six Books of Euclid, with a Supplement of the Quadrature of the Circle and the Geometry of ... is to contain the

Containing the First Six Books of Euclid, with a Supplement of the Quadrature of the Circle and the Geometry of ... is to contain the

**part**which is to be A**cut off**from it ;**join BC**, and**draw**DE**parallel**to it ; then AE is the**part**reE ... Side 230

Let two planes AB ,

Let two planes AB ,

**BC cut**each other , and let B , D bie two points in the line**of**their common section . ...**draw**GH**parallel**to AD , and make HF = HA ;**join**FG , and produce it to meet CA in C . D ;**join**BD , BG , BF .### Hva folk mener - Skriv en omtale

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Elements of Geometry: Containing the First Six Books of Euclid with a ... John Playfair Uten tilgangsbegrensning - 1855 |

Elements of Geometry: Containing the First Six Books of Euclid, with a ... Euclid,John Playfair Uten tilgangsbegrensning - 1853 |

Elements of Geometry: Containing the First Six Books of Euclid, with a ... John Playfair Uten tilgangsbegrensning - 1847 |

### Vanlige uttrykk og setninger

ABC is equal ABCD altitude angle ABC angle ACB angle BAC arch base bisected Book centre circle circle ABC circumference coincide common compounded contained cylinder definition demonstrated described diameter difference divided double draw drawn equal equal angles equiangular equilateral equimultiples exterior angle extremities fall figure fore fourth given straight line greater half inscribed interior join less Let ABC magnitudes manner meet multiple opposite parallel parallelogram pass perpendicular plane polygon prism PROB produced PROP proportional proposition proved pyramid Q. E. D. PROP ratio reason rectangle contained rectilineal figure right angles segment shown sides similar solid space square taken THEOR third triangle ABC wherefore whole

### Populære avsnitt

Side 121 - 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...

Side 42 - TO a given straight line to apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.

Side 63 - Therefore, in obtuse-angled triangles, &c. QED PROP. XIII. THEOREM. In every triangle, the square of the side subtending either of the acute angles is less than the squares of the sides containing that angle, by twice the rectangle contained by either of these sides, and the straight line intercepted between the perpendicular let fall upon it from the opposite angle, and the acute angle.

Side 3 - A circle is a plane figure contained by one line, which is called the circumference, and is such that all straight lines drawn from a certain point within the figure to the circumference, are equal to one another.

Side 183 - 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 •f the ratios of their sides.

Side 3 - A diameter of a circle is a straight line drawn through the centre, and terminated both ways by the circumference.

Side 291 - All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides.

Side 160 - ... extremities of the base shall have the same ratio which the other sides of the triangle have to one...

Side 10 - ... shall be greater than the base of the other. Let ABC, DEF be two triangles, which have the two sides AB, AC, equal to the two DE, DF, each to each, viz.

Side 14 - Therefore, 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 extretnity equal to one another.