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 2
... join EB , DB . Then AD = DB from the first part , and AE is equal to AD , DE , that is , to BD , DE , and is therefore greater than BE , ( Eucl . i . 20. ) ; therefore , & c . ( 3. ) Through a given point to draw a straight line which ...
... join EB , DB . Then AD = DB from the first part , and AE is equal to AD , DE , that is , to BD , DE , and is therefore greater than BE , ( Eucl . i . 20. ) ; therefore , & c . ( 3. ) Through a given point to draw a straight line which ...
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 ... BD , and through C draw ECF perpendicular to BD . ECF is the line required . Draw AE parallel to BD , and ...
... join AE . Then AE- AF ( i . 2. ) . And besides AE no other line can be drawn equal to AF . For , if possible , let ... BD , and through C draw ECF perpendicular to BD . ECF is the line required . Draw AE parallel to BD , and ...
Side 15
... BD C A parts in Cand into two unequal parts in D , and produced to E , so that BE : EA :: BD : DA ; then will AD ... Join AB and bisect it in D. Join CD , from which cut off DE equal to a third part of it . E is the point required ...
... BD C A parts in Cand into two unequal parts in D , and produced to E , so that BE : EA :: BD : DA ; then will AD ... Join AB and bisect it in D. Join CD , from which cut off DE equal to a third part of it . E is the point required ...
Side 17
... 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 so that it may = DE , and draw BG , GH respectively parallel ...
... 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 so that it may = DE , and draw BG , GH respectively parallel ...
Side 18
Miles Bland. in C. Take BD : AC in the given ratio , and from B draw BE parallel and equal to AC . Join DE and produce it to meet CF drawn at any angle from C , equal G B to the given line ; draw FG parallel to EB , and from G draw GH ...
Miles Bland. in C. Take BD : AC in the given ratio , and from B draw BE parallel and equal to AC . Join DE and produce it to meet CF drawn at any angle from C , equal G B to the given line ; draw FG parallel to EB , and from G draw GH ...
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 | |
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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