Elements of Euclid Adapted to Modern Methods in GeometryWilliam Collins, Sons,, 1874 - 216 sider |
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Side 23
... bisect a given rectilineal angle ; that is , to divide it into two equal parts . Let BAC be the given angle ; it is required to bisect it . În AB take any point D , and make AE equal to AD . Join DE , and on it describe the equi ...
... bisect a given rectilineal angle ; that is , to divide it into two equal parts . Let BAC be the given angle ; it is required to bisect it . În AB take any point D , and make AE equal to AD . Join DE , and on it describe the equi ...
Side 24
... bisect an angle of two right angles . B A Let BAC be an angle which is two right angles , and which it is required to bisect . In AB take , as before , any point D , and make AE equal to AD . On DE describe an equi- lateral triangle DFE ...
... bisect an angle of two right angles . B A Let BAC be an angle which is two right angles , and which it is required to bisect . In AB take , as before , any point D , and make AE equal to AD . On DE describe an equi- lateral triangle DFE ...
Side 25
... bisect it . Describe ( I. 1 ) on it an equi- lateral triangle ACB , and bisect ( I. 7 ) the angle C by the straight A line CD ; AB is bisected in D. B Because AC in the one triangle is equal to CB in the other , ( Const . ) and CD ...
... bisect it . Describe ( I. 1 ) on it an equi- lateral triangle ACB , and bisect ( I. 7 ) the angle C by the straight A line CD ; AB is bisected in D. B Because AC in the one triangle is equal to CB in the other , ( Const . ) and CD ...
Side 27
... either of the interior remote angles A or B. Bisect AC in O ( I. 9 ) ; join B , O , and produce B K A BO to H , making OH equal to OB ( I. 2 ) ; join HC . Then in the two triangles AOB and COH we have BOOK I. 23 PROPS . 11 , 12 . 27.
... either of the interior remote angles A or B. Bisect AC in O ( I. 9 ) ; join B , O , and produce B K A BO to H , making OH equal to OB ( I. 2 ) ; join HC . Then in the two triangles AOB and COH we have BOOK I. 23 PROPS . 11 , 12 . 27.
Side 28
... bisected in K , AK joined and the line produced , it may be shewn in the same manner that the angle BCN is greater than the angle ABC . But BCN is equal to ACD ( I. 11 ) , therefore ACD is greater than ABC ; but it has been before ...
... bisected in K , AK joined and the line produced , it may be shewn in the same manner that the angle BCN is greater than the angle ABC . But BCN is equal to ACD ( I. 11 ) , therefore ACD is greater than ABC ; but it has been before ...
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Elements of Euclid, Adapted to Modern Methods in Geometry. By J. Bryce ... Uten tilgangsbegrensning - 1874 |
Elements of Euclid Adapted to Modern Methods in Geometry Euclid,James Bryce,David Munn (F.R.S.E.) Ingen forhåndsvisning tilgjengelig - 1874 |
Vanlige uttrykk og setninger
AC and CB altitude angle AOB BA and AC bisecting the angle centre chord circles touch circumference cloth coincide Const conv Cor.-Hence diagonal diameter divided draw equal angles equal to BC equal to twice equiangular equilateral triangle Euclid exterior angle Fcap GEOGRAPHY geometrical given circle given line given point given straight line greater half the perimeter Hence hypotenuse inscribed intersecting isosceles triangle less Let ABC LL.D meet middle point multiple opposite sides parallel to BC parallelogram perpendicular polygon produced Proposition Q. E. D. Cor Q. E. D. PROP radius ratio rectangle contained rectilineal figure reflex angle remaining angles required to prove right angles right-angled triangle schol segments shew shewn side BC square on AC tangent THEOREM triangle ABC twice the rectangle twice the square whole line
Populære avsnitt
Side 68 - The straight line joining the middle points of two sides of a triangle is parallel to the third side, and equal to half of it.
Side 77 - 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 50 - If a parallelogram and a triangle be upon the same base, and between the same parallels; the parallelogram is double of the triangle.
Side 87 - If a straight line be divided into two equal, and also into two unequal parts, the squares on the two unequal parts are together double of the square on half the line and of the square on the line between the points of section.
Side 30 - Any two sides of a triangle are together greater than the third side.
Side 204 - Tin; rectangle, contained by the diagonals of a quadrilateral inscribed in a circle, is equal to the sum of the rectangles contained by its opposite sides. Let ABCD be any quadrilateral inscribed in a circle...
Side 89 - To divide a given straight line into two parts, so that the rectangle contained by the whole and one of the parts, shall be equal to the square on the other part.
Side 98 - 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.