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 6
... Let AB , AC be the given lines , meet- ing in A. From P the given point draw PD parallel to AC one of the lines , and make DE DA . Join EP and produce it to F , then will EF be bisected in P .. = A B E D P FC For since DP is parallel to ...
... Let AB , AC be the given lines , meet- ing in A. From P the given point draw PD parallel to AC one of the lines , and make DE DA . Join EP and produce it to F , then will EF be bisected in P .. = A B E D P FC For since DP is parallel to ...
Side 8
... Let ABC be the given angle , DE parallel to AB , and P the given point . From P draw PC parallel B с A D to DE or AB , and take BE : CF in the given ratio . Join FP and produce it to A ; APF is the line required . For since DE and CP ...
... Let ABC be the given angle , DE parallel to AB , and P the given point . From P draw PC parallel B с A D to DE or AB , and take BE : CF in the given ratio . Join FP and produce it to A ; APF is the line required . For since DE and CP ...
Side 9
... Let AB AC :: AC : AD , and from A let 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 ...
... Let AB AC :: AC : AD , and from A let 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 ...
Side 10
... Let AB be the given straight line . On it describe an equilateral triangle ABC ; bisect the angles CAB , CBA by the lines AD , BD meeting in D , and draw DE , DF parallel to CA and CB respectively . AB will be trisected in E and F ...
... Let AB be the given straight line . On it describe an equilateral triangle ABC ; bisect the angles CAB , CBA by the lines AD , BD meeting in D , and draw DE , DF parallel to CA and CB respectively . AB will be trisected in E and F ...
Side 11
... Let AB be the given straight line . Let AC be any other indefinite straight line making any angle with AB , and in it take any point D , and take as many lines DE , EF , FC & c . each equal to AD as the number of parts into which AB is ...
... Let AB be the given straight line . Let AC be any other indefinite straight line making any angle with AB , and in it take any point D , and take as many lines DE , EF , FC & c . each equal to AD as the number of parts into which AB is ...
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