The Elements of Euclid: Viz. the First Six Books, Together with the Eleventh and Twelfth |
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Page 11
1 . join FC , GB . Because AF is equal to + AG , and AB tofAC , Const . the two sides FA , AC are equal to the two GA , 6 Hyp . AB , each to each ; and they contain the angle FAG common to the two triangles AFC , AGB ; therefore the A ...
1 . join FC , GB . Because AF is equal to + AG , and AB tofAC , Const . the two sides FA , AC are equal to the two GA , 6 Hyp . AB , each to each ; and they contain the angle FAG common to the two triangles AFC , AGB ; therefore the A ...
Page 12
... one of them is greater than the other : let AB be the greater ; and from it cut * off DB equal to AC , the less , and join DC : therefore , because in the triangles DBC , ACB , DB is equal to AC , and BC com• 3 .
... one of them is greater than the other : let AB be the greater ; and from it cut * off DB equal to AC , the less , and join DC : therefore , because in the triangles DBC , ACB , DB is equal to AC , and BC com• 3 .
Page 13
... DB , that are CD terminated in B. Join CD ; then , in the case in which the vertex of each of the triangles is without the other triangle , because AC is equal t to A R + Hyp . AD , the angle ACD is equal * to the angle ADC : But ...
... DB , that are CD terminated in B. Join CD ; then , in the case in which the vertex of each of the triangles is without the other triangle , because AC is equal t to A R + Hyp . AD , the angle ACD is equal * to the angle ADC : But ...
Page 15
1 . off AE equal to AD ; join DE , and upon it describe * an equilateral triangle * 1.1 . DEF ; then join AF : the straight line AF shall bisect the A angle BAC . Because AD is equalt to AE , DA E | Const . and AF is common to the two ...
1 . off AE equal to AD ; join DE , and upon it describe * an equilateral triangle * 1.1 . DEF ; then join AF : the straight line AF shall bisect the A angle BAC . Because AD is equalt to AE , DA E | Const . and AF is common to the two ...
Page 16
Take any point D in AC , and make * CE equal to CD , and upon DE describe * the equilateral triangle DFE , and join FC . The straight line FC drawn from the given point C shall be at right F angles to the given straight line AB .
Take any point D in AC , and make * CE equal to CD , and upon DE describe * the equilateral triangle DFE , and join FC . The straight line FC drawn from the given point C shall be at right F angles to the given straight line AB .
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The Elements of Euclid: Viz. The First Six Books, Together With the Eleventh ... Euclid Euclid No preview available - 2018 |
Common terms and phrases
altitude angle ABC angle BAC base base BC BC is equal centre circle ABCD circumference common cone Const contained cylinder demonstrated described diameter divided double draw drawn equal angles equiangular equilateral equimultiples extremities fall figure fore four fourth given given straight line greater half inscribed join less Let ABC likewise magnitudes manner meet multiple opposite parallel parallelogram pass perpendicular plane polygon prism PROB produced PROP proportionals proved pyramid ratio reason rectangle contained rectilineal figure remaining angle right angles segment shown sides similar solid solid angle solid parallelopiped sphere square taken THEOR third touches triangle ABC vertex wherefore whole
Bibliographic information
| Title | The Elements of Euclid: Viz. the First Six Books, Together with the Eleventh and Twelfth |
| Author | Euclid |
| Editors | Robert Simson, Samuel Maynard |
| Translated by | Robert Simson |
| Publisher | Longman, Orme, Brown, Green, and Longmans, 1838 |
| Original from | University of Iowa |
| Digitized | 13 Jul 2015 |
| Length | 328 pages |
| Export Citation | BiBTeX EndNote RefMan |