Euclid's Elements of geometry, the first four books, by R. Potts. Corrected and improved1864 |
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Resultat 1-5 av 29
Side 9
... triangle ABC to the triangle DEF ; and the other angles to which the equal sides are opposite shall be equal , each to each , viz . the angle ABC to the angle DEF , and the angle ACB to the angle DFE . A D B C E F For , if the triangle ABC ...
... triangle ABC to the triangle DEF ; and the other angles to which the equal sides are opposite shall be equal , each to each , viz . the angle ABC to the angle DEF , and the angle ACB to the angle DFE . A D B C E F For , if the triangle ABC ...
Side 10
... angle ACF to the angle ABG , and the angle AFC to the angle AGB . And because the whole AF is equal to the whole AG ... ACB , which are the angles at the base of the triangle ABC ; and it has also been proved , that the angle FBC is ...
... angle ACF to the angle ABG , and the angle AFC to the angle AGB . And because the whole AF is equal to the whole AG ... ACB , which are the angles at the base of the triangle ABC ; and it has also been proved , that the angle FBC is ...
Side 11
... ABC , because DB is equal to AC , and BC is common to both triangles , the two sides DB , BC ' are equal to the two sides AC , CB , each to each ; and the angle DBC is equal to the angle ACB ; ( hyp . ) therefore the base DC is equal to ...
... ABC , because DB is equal to AC , and BC is common to both triangles , the two sides DB , BC ' are equal to the two sides AC , CB , each to each ; and the angle DBC is equal to the angle ACB ; ( hyp . ) therefore the base DC is equal to ...
Side 13
... angle . It is required to bisect it . A D E B C In AB take any point D ... ACB by the straight line CD meeting AB in the point D. ( 1. 9. ) Then AB ... angle ACD is equal to BCD ; ( constr . ) therefore the base AD is equal to the base BD ...
... angle . It is required to bisect it . A D E B C In AB take any point D ... ACB by the straight line CD meeting AB in the point D. ( 1. 9. ) Then AB ... angle ACD is equal to BCD ; ( constr . ) therefore the base AD is equal to the base BD ...
Side 18
... ACB ; therefore the angles ACD , ACB are greater than the angles ABC , ACB ; but the angles ACD , ACB are equal to two right angles ; ( I. 13. ) therefore the angles ABC ... angle ABC shall be greater than the angle 18 EUCLID'S ELEMENTS .
... ACB ; therefore the angles ACD , ACB are greater than the angles ABC , ACB ; but the angles ACD , ACB are equal to two right angles ; ( I. 13. ) therefore the angles ABC ... angle ABC shall be greater than the angle 18 EUCLID'S ELEMENTS .
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Euclid's Elements of Geometry, the First Four Books, by R. Potts. Corrected ... Euclides Ingen forhåndsvisning tilgjengelig - 2016 |
Euclid's Elements of geometry, the first four books, by R. Potts. Corrected ... Euclides Ingen forhåndsvisning tilgjengelig - 1864 |
Vanlige uttrykk og setninger
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
Populære avsnitt
Side 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.
Side 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.
Side 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.
Side 54 - If two triangles have two sides of the one equal to two sides of the...
Side 5 - LET it be granted that a straight line may be drawn from any one point to any other point.
Side 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.
Side 3 - A diameter of a circle is a straight line drawn through the centre, and terminated both ways by the circumference.
Side 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...
Side 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.
Side 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.