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 3
... Join AE , EB . Since AF FB , and FE is common , and the angles at F are right angles , therefore AE = EB . F с E B B ( 5. ) From two given points on the same side of a line given in position , to draw two lines which shall meet in that ...
... Join AE , EB . Since AF FB , and FE is common , and the angles at F are right angles , therefore AE = EB . F с E B B ( 5. ) From two given points on the same side of a line given in position , to draw two lines which shall meet in that ...
Side 5
... join AE . Then AE- AF ( i . 2. ) . And besides AE no other line can be drawn equal to AF . For , if possible , let AI - AF . Then be- cause AI = AF and AF = AE , therefore AI = AE , i . e . a line more remote is equal to one nearer the ...
... join AE . Then AE- AF ( i . 2. ) . And besides AE no other line can be drawn equal to AF . For , if possible , let AI - AF . Then be- cause AI = AF and AF = AE , therefore AI = AE , i . e . a line more remote is equal to one nearer the ...
Side 9
... AE be drawn equal to AC , inclined at any angle to AB ; join EB , EC , ED ; the angle BEC B is equal to the angle CED . E For since AB : AC :: AC : AD , and AE = AC ; .. AB AE :: AE : AD , i . e . the sides about the angle A are ...
... AE be drawn equal to AC , inclined at any angle to AB ; join EB , EC , ED ; the angle BEC B is equal to the angle CED . E For since AB : AC :: AC : AD , and AE = AC ; .. AB AE :: AE : AD , i . e . the sides about the angle A are ...
Side 20
... Join AF ; and draw BK parallel to AG cutting AF in L ; and draw LM parallel to KE cutting AE in M and AG in N. Then FE : LM :: GF : ( NL = ) AB · and FE : DG :: FG : AB by construction ; : . LM = DG = IA , if therefore ILO be drawn , IL ...
... Join AF ; and draw BK parallel to AG cutting AF in L ; and draw LM parallel to KE cutting AE in M and AG in N. Then FE : LM :: GF : ( NL = ) AB · and FE : DG :: FG : AB by construction ; : . LM = DG = IA , if therefore ILO be drawn , IL ...
Side 21
... Join AE , FE , and HC parallel Then ( Eucl . vi . 2. ) AI : IH :: AB : BE in the given ratio of the remainders ; and ... AE to meet CB in E making the angle AEC = the given angle to be made by the line to be drawn , with BC . In AE take ...
... Join AE , FE , and HC parallel Then ( Eucl . vi . 2. ) AI : IH :: AB : BE in the given ratio of the remainders ; and ... AE to meet CB in E making the angle AEC = the given angle to be made by the line to be drawn , with BC . In AE take ...
Indhold
1 | |
6 | |
12 | |
17 | |
19 | |
23 | |
25 | |
30 | |
140 | |
146 | |
147 | |
154 | |
157 | |
158 | |
164 | |
170 | |
36 | |
42 | |
43 | |
47 | |
52 | |
58 | |
63 | |
65 | |
69 | |
75 | |
80 | |
86 | |
87 | |
91 | |
92 | |
97 | |
103 | |
109 | |
114 | |
120 | |
126 | |
134 | |
171 | |
177 | |
179 | |
185 | |
191 | |
193 | |
199 | |
203 | |
209 | |
212 | |
218 | |
220 | |
226 | |
232 | |
237 | |
244 | |
249 | |
255 | |
273 | |
295 | |
301 | |
Andre udgaver - Se alle
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