Euclid's Elements of geometry, the first four books, by R. Potts. Corrected and improved1864 |
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Αποτελέσματα 1 - 5 από τα 99.
Σελίδα 7
... proved that AC is equal to AB ; therefore AC , BC are each of them equal to AB ; but things which are equal to the same thing are equal to one another ; therefore AC is equal to BC ; ( ax . 1. ) wherefore AB , BC , CA are equal to one ...
... proved that AC is equal to AB ; therefore AC , BC are each of them equal to AB ; but things which are equal to the same thing are equal to one another ; therefore AC is equal to BC ; ( ax . 1. ) wherefore AB , BC , CA are equal to one ...
Σελίδα 10
... proved to be equal to GB ; hence , because the two sides BF , FC are equal to the two CG , GB , each to each ; and the angle BFC has been proved to be equal to the angle CGB , also the base BC is common to the two triangles BFC , CGB ...
... proved to be equal to GB ; hence , because the two sides BF , FC are equal to the two CG , GB , each to each ; and the angle BFC has been proved to be equal to the angle CGB , also the base BC is common to the two triangles BFC , CGB ...
Σελίδα 11
... proved greater than the angle BCD , hence the angle BDC'is both equal to , and greater than the angle BCD ; which is impossible . Secondly . Let the vertex D of the triangle ADB fall within the triangle ACB . D E F B Produce AC to E ...
... proved greater than the angle BCD , hence the angle BDC'is both equal to , and greater than the angle BCD ; which is impossible . Secondly . Let the vertex D of the triangle ADB fall within the triangle ACB . D E F B Produce AC to E ...
Σελίδα 12
... proved greater than BCD , wherefore the angle BDC is both equal to , and greater than the angle BCD ; which is impossible . Thirdly . The case in which the vertex of one triangle is upon a side of the other , needs no demonstration ...
... proved greater than BCD , wherefore the angle BDC is both equal to , and greater than the angle BCD ; which is impossible . Thirdly . The case in which the vertex of one triangle is upon a side of the other , needs no demonstration ...
Σελίδα 16
... proved equal to the same three angles ; and things which are equal to the same thing are equal to one another ; therefore the angles CBE , EBD are equal to the angles DBA , ABC ; but the angles CBE , EBD are two right angles ; therefore ...
... proved equal to the same three angles ; and things which are equal to the same thing are equal to one another ; therefore the angles CBE , EBD are equal to the angles DBA , ABC ; but the angles CBE , EBD are two right angles ; therefore ...
Άλλες εκδόσεις - Προβολή όλων
Euclid's Elements of Geometry, the First Four Books, by R. Potts. Corrected ... Euclides Δεν υπάρχει διαθέσιμη προεπισκόπηση - 2016 |
Euclid's Elements of geometry, the first four books, by R. Potts. Corrected ... Euclides Δεν υπάρχει διαθέσιμη προεπισκόπηση - 1864 |
Συχνά εμφανιζόμενοι όροι και φράσεις
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
Δημοφιλή αποσπάσματα
Σελίδα 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.
Σελίδα 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.
Σελίδα 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.
Σελίδα 54 - If two triangles have two sides of the one equal to two sides of the...
Σελίδα 5 - LET it be granted that a straight line may be drawn from any one point to any other point.
Σελίδα 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.
Σελίδα 3 - A diameter of a circle is a straight line drawn through the centre, and terminated both ways by the circumference.
Σελίδα 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...
Σελίδα 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.
Σελίδα 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.