The Elements of Euclid: Viz. the First Six Books, Together with the Eleventh and Twelfth |
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Page 9
... and join FC , GB . Because AF is equal to + AG , and AB to | AC , the + Constr . two sides FA , AC are equal to the two GA , AB , each $ Hyp . to each : and they contain the angle FAG common to ** 3. 1 . 1 А В , C * 4. 1 .
... and join FC , GB . Because AF is equal to + AG , and AB to | AC , the + Constr . two sides FA , AC are equal to the two GA , AB , each $ Hyp . to each : and they contain the angle FAG common to ** 3. 1 . 1 А В , C * 4. 1 .
Page 10
... 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 common to ...
... 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 common to ...
Page 13
1 . equal to AD ; join DE , and upon it describe * an equila- * 1. 1 . teral triangle DEF ; then join AF : the straight line AF shall bisect the angle BAC . Because AD is equal + to AE , and AF is common to the two triangles DAF ...
1 . equal to AD ; join DE , and upon it describe * an equila- * 1. 1 . teral triangle DEF ; then join AF : the straight line AF shall bisect the angle BAC . Because AD is equal + to AE , and AF is common to the two triangles DAF ...
Page 14
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 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 angles to the given straight line AB .
Page 15
1 . and join CH . The straight line CH , drawn from the given point C , shall be perpendicular to the given straight line AB . Join CF , CG : and because FH is equal + to HG , and + Constr . HC common to the two triangles FHC ...
1 . and join CH . The straight line CH , drawn from the given point C , shall be perpendicular to the given straight line AB . Join CF , CG : and because FH is equal + to HG , and + Constr . HC common to the two triangles FHC ...
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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
added altitude angle ABC angle BAC base Book centre circle circle ABCD circumference common cone contained cylinder definition demonstrated described diameter difference divided double draw drawn equal equal angles equiangular equimultiples Euclid excess figure fore four fourth given angle given in position given in species given magnitude given ratio given straight line greater Greek half join less likewise logarithm magnitude manner meet multiple opposite parallel parallelogram pass perpendicular plane prism produced PROP proportionals proposition proved pyramid radius reason rectangle rectilineal figure remaining right angles segment shewn sides similar sine solid sphere square square of AC taken THEOR third triangle ABC wherefore whole