Geometrical problems deducible from the first six books of Euclid, arranged and solved. To which is added, an appendix containing the elements of plane trigonometryJ. Smith, Printer to the University; and sold by Nicholson & Son, Deighton & Sons, Thorpe, and Newby, Cambridge; and J. Mawman, and F. C. & J. Rivington, London., 1819 - 377 sider |
Fra bogen
Resultater 1-5 af 100
Side 7
... AB : BC in the given ratio . Join AC , and from P draw PDE parallel to AC . PDE is the line required . A D P E For since DE is parallel to AC , ( Eucl . vi . 2. ) DB : BE :: AB : BC , i . e . in the given ratio . ( 12. ) If from a given ...
... AB : BC in the given ratio . Join AC , and from P draw PDE parallel to AC . PDE is the line required . A D P E For since DE is parallel to AC , ( Eucl . vi . 2. ) DB : BE :: AB : BC , i . e . in the given ratio . ( 12. ) If from a given ...
Side 11
... AB is to be divided . Join CB , and draw DG , EH , FI & c . parallel to BC ; and therefore parallel to each other ; and draw DK parallel to AB . B C Then because GD is parallel to HE one of the sides . of the triangle AHE , AG : GH ...
... AB is to be divided . Join CB , and draw DG , EH , FI & c . parallel to BC ; and therefore parallel to each other ; and draw DK parallel to AB . B C Then because GD is parallel to HE one of the sides . of the triangle AHE , AG : GH ...
Side 12
... BC is parallel to FD , the angle BCG is equal to GDF and the vertically opposite angles at G are equal ; therefore the triangles DGF , BGC are similar , and BC BG :: FD : FG But FE being parallel to BC , ( Eucl . vi . 2. ) AB .. ex æquali , ...
... BC is parallel to FD , the angle BCG is equal to GDF and the vertically opposite angles at G are equal ; therefore the triangles DGF , BGC are similar , and BC BG :: FD : FG But FE being parallel to BC , ( Eucl . vi . 2. ) AB .. ex æquali , ...
Side 16
... parallel , and AD = DB , .. AI + BG = 2 DH = FC . ( 25. ) From a given point in one of two straight lines given in ... BC be the lines given in position , and A the D E H K L B I given point . Draw AD perpendicular to AB , and meet- ing BC ...
... parallel , and AD = DB , .. AI + BG = 2 DH = FC . ( 25. ) From a given point in one of two straight lines given in ... BC be the lines given in position , and A the D E H K L B I given point . Draw AD perpendicular to AB , and meet- ing BC ...
Side 17
... AB be the given line , and BAC the given angle . From B draw BD to AC , such that it may be to AB in the given ratio * ; produce it till BE the other given line . Through E draw EC parallel to AB , meeting AC in C. Join BC , and draw DF ...
... AB be the given line , and BAC the given angle . From B draw BD to AC , such that it may be to AB in the given ratio * ; produce it till BE the other given line . Through E draw EC parallel to AB , meeting AC in C. Join BC , and draw DF ...
Indhold
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12 | |
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23 | |
25 | |
30 | |
140 | |
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170 | |
36 | |
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Almindelige termer og sætninger
ABCD angle ABC angles at F base centre chord circle ABC circles cut circles touch circumference describe a circle divided draw a line Draw the diameter drawn parallel duplicate ratio equal angles equiangular Eucl extremities given angle given circle given in position given line given point given ratio given square given straight line given triangle intercepted isosceles triangle Join AE Join BD Let AB Let ABC let fall line given line joining line required lines be drawn lines drawn mean proportional opposite side parallel to BC parallelogram pendicular perpen point of bisection point of contact point of intersection point required radius rectangle rectangle contained right angles segments semicircle shewn tangent touches the circle trapezium triangle ABC