Euclid's Elements of geometry, the first four books, by R. Potts. Corrected and improved1864 |
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Side 12
... , DF ; wherefore likewise the angle BAC coincides with the angle ÉDF , and is equal to it . ( ax . 8. ) Therefore if two triangles have two sides , & c . Q.E.D. PROPOSITION IX . PROBLEM . To bisect a given rectilineal 12 EUCLID'S ELEMENTS .
... , DF ; wherefore likewise the angle BAC coincides with the angle ÉDF , and is equal to it . ( ax . 8. ) Therefore if two triangles have two sides , & c . Q.E.D. PROPOSITION IX . PROBLEM . To bisect a given rectilineal 12 EUCLID'S ELEMENTS .
Side 13
... bisect it . A D E B C In AB take any point D ; from AC cut off AE equal to AD , ( 1. 3. ) and join DE ; on the side ... bisected by the straight line AF . Q.E.F. PROPOSITION X. PROBLEM . To bisect a given finite straight line , that is ...
... bisect it . A D E B C In AB take any point D ; from AC cut off AE equal to AD , ( 1. 3. ) and join DE ; on the side ... bisected by the straight line AF . Q.E.F. PROPOSITION X. PROBLEM . To bisect a given finite straight line , that is ...
Side 15
... 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 is equal to HG , ( constr . ) and HC is common to ...
... 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 is equal to HG , ( constr . ) and HC is common to ...
Side 18
... bisected , and AC be pro- duced to G ; it may be demonstrated that the angle BCG , that is , the angle ACD , ( I. 15. ) is greater than the angle ABC . Therefore , if one side of a triangle , & c . Q.E.D. PROPOSITION XVII . THEOREM ...
... bisected , and AC be pro- duced to G ; it may be demonstrated that the angle BCG , that is , the angle ACD , ( I. 15. ) is greater than the angle ABC . Therefore , if one side of a triangle , & c . Q.E.D. PROPOSITION XVII . THEOREM ...
Side 31
... bisect it . B D Because AB is parallel to CD , and BC meets them , therefore the angle ABC is equal to the alternate angle BCD . ( 1. 29. ) And because AC is parallel to BD , and BC meets them , therefore the angle ACB is equal to the ...
... bisect it . B D Because AB is parallel to CD , and BC meets them , therefore the angle ABC is equal to the alternate angle BCD . ( 1. 29. ) And because AC is parallel to BD , and BC meets them , therefore the angle ACB is equal to the ...
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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.