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 60
Side 1
... shewn to be greater than AC ; there- fore AC is the least . ( 2. ) If a perpendicular be drawn bisecting a given straight line ; any point in this perpendicular is at equal A distances , and any point without the perpendicular is at.
... shewn to be greater than AC ; there- fore AC is the least . ( 2. ) If a perpendicular be drawn bisecting a given straight line ; any point in this perpendicular is at equal A distances , and any point without the perpendicular is at.
Side 4
... shewn that Ap = pC . Hence AP and BP together are equal to BC , and Ap , pB are equal to Cp , pB . Now ( Eucl . i . 20. ) BC is less than Bp , PC , and therefore AP , PB are less than Ap , pB ; therefore , & c . ( 7. ) Of all straight ...
... shewn that Ap = pC . Hence AP and BP together are equal to BC , and Ap , pB are equal to Cp , pB . Now ( Eucl . i . 20. ) BC is less than Bp , PC , and therefore AP , PB are less than Ap , pB ; therefore , & c . ( 7. ) Of all straight ...
Side 5
... shewn that AH is greater than AG . · = And from A there can only be drawn to BC two straight lines equal to each other , viz . one on each side of AD . Make DE = DF and join AE . Then AE- AF ( i . 2. ) . And besides AE no other line can ...
... shewn that AH is greater than AG . · = And from A there can only be drawn to BC two straight lines equal to each other , viz . one on each side of AD . Make DE = DF and join AE . Then AE- AF ( i . 2. ) . And besides AE no other line can ...
Side 11
... shewn that HI = IB ; and so on , if there be any other parts ; therefore AG , GH , HI , IB , & c , are all equal , and AB is divided as was required . COR . If it be required to divide the line into parts which shall have a given ratio ...
... shewn that HI = IB ; and so on , if there be any other parts ; therefore AG , GH , HI , IB , & c , are all equal , and AB is divided as was required . COR . If it be required to divide the line into parts which shall have a given ratio ...
Side 32
... shewn that every line drawn from A to BCD will be divided by the circum- ference of the circle GFH in the same ratio , i . e . GFH . will be the locus required . ( 14. ) Having given the radius of a circle ; to de- termine its centre ...
... shewn that every line drawn from A to BCD will be divided by the circum- ference of the circle GFH in the same ratio , i . e . GFH . will be the locus required . ( 14. ) Having given the radius of a circle ; to de- termine its centre ...
Indhold
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12 | |
17 | |
19 | |
23 | |
25 | |
30 | |
140 | |
146 | |
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154 | |
157 | |
158 | |
164 | |
170 | |
36 | |
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58 | |
63 | |
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91 | |
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134 | |
171 | |
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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