## Euclid's Elements of geometry, the first four books, by R. Potts. Corrected and improved1864 |

### Dentro del libro

Resultados 1-5 de 100

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**produced**ever so far both ways , do not meet . A. A parallelogram is a four - sided figure , of which the opposite sides are parallel : and the diameter , or the diagonal is the straight line joining two of its opposite angles ... Página 6

... these straight lines being continually

... these straight lines being continually

**produced**, shall at length meet upon that side on which are the angles which are less than two right angles . PROPOSITION I. PROBLEM . To describe an equilateral triangle upon 6 EUCLID'S ELEMENTS . Página 9

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**produced**, the angles on the other side of the base shall be equal . Let ABC be an isosceles triangle of which the side AB is equal to AC , and let the equal sides AB , AC be**produced**to D and E. Then the angle ABC shall be equal to the ... Página 15

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**produced**if necessary , in F and G : ( post . 3. ) bisect FG in H ( 1. 10. ) , and join CH . Then the straight line CH drawn from the given point C , shall be perpendicular to the given straight line AB . Join FC , and CG . Because FH ... Página 17

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**produced**, the exterior angle is greater than either of the interior opposite angles . Let ABC be a triangle , and let the side BC be**produced**to D. Then the exterior angle ACD shall be greater than either of the interior opposite ...### Otras ediciones - Ver todas

Euclid's Elements of Geometry, the First Four Books, by R. Potts. Corrected ... Euclides Sin vista previa disponible - 2016 |

### Términos y frases comunes

ABCD AC is equal adjacent angles angle ABC angle ACB angle BAC angle equal Apply Euc axiom base BC bisecting the angle chord circle ABC circumference construction demonstrated describe a circle diagonals diameter double draw equal angles equal to twice equiangular equilateral triangle Euclid Euclid's Elements exterior angle Geometry given angle given circle given line given point given straight line gnomon greater hypotenuse inscribed intersection isosceles triangle Let ABC line AC line CD line joining lines be drawn meet the circumference opposite angles opposite sides parallel parallelogram pentagon perpendicular porism problem produced Prop proved quadrilateral figure radius rectangle contained remaining angle right angles right-angled triangle segment semicircle shew shewn side BC square on AC tangent THEOREM touches the circle trapezium triangle ABC twice the rectangle vertex vertical angle wherefore

### Pasajes populares

Página 118 - Guido, with a burnt stick in his hand, demonstrating on the smooth paving-stones of the path, that the square on the hypotenuse of a right-angled triangle is equal to the sum of the squares on the other two sides.

Página 90 - If a straight line be divided into any two parts, the squares of the whole line, and of one of the parts, are equal to twice the rectangle contained by the whole and that part, together with the square of the other part. Let the straight line AB be divided into any two parts in the point C; the squares of AB, BC are equal to twice the rectangle AB, BC, together with the square of AC.

Página 30 - ... twice as many right angles as the figure has sides ; 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.

Página 54 - If two triangles have two sides of the one equal to two sides of the...

Página 5 - LET it be granted that a straight line may be drawn from any one point to any other point.

Página 85 - If there be two straight lines, one of which is divided into any number of parts, the rectangle contained by the two straight lines is equal to the rectangles contained by the undivided line, and the several parts of the divided line.

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

Página 96 - In obtuse-angled triangles, if a perpendicular be drawn from either of the acute angles to the opposite side produced, the square on the side subtending the obtuse angle is greater than the squares on the sides containing the obtuse angle, by twice the rectangle contained by the side...

Página 41 - If the square described upon one of the sides of a triangle, be equal to the squares described upon the other two sides of it ; the angle contained by these two sides is a right angle.

Página 126 - EF, that is, AF, is greater than BF : 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 CE, EF; but the angle BEF is greater than the angle CEF ; therefore the base BF is greater (24. 1.) than the base FC ; for the same reason, CF is greater than GF. Again, because GF, FE are greater (20.