The Synoptical Euclid; Being the First Four Books of Euclid's Elements of Geometry, from the Edition of Dr. Robert Simson ... With ExercisesCharles Henry Law, 1854 - 120 páginas |
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Página 10
... take any point F , and from AE , the greater , cut off AG equal ( I. 3. ) to AF , the less , and join FC , GB . Because AF is equal ( Constr . ) to 4G , and AB ( Hypothesis ) to AC , the two sides FA , AC are equal to the two GA , AB ...
... take any point F , and from AE , the greater , cut off AG equal ( I. 3. ) to AF , the less , and join FC , GB . Because AF is equal ( Constr . ) to 4G , and AB ( Hypothesis ) to AC , the two sides FA , AC are equal to the two GA , AB ...
Página 13
... Take any point Din AB , and from AC cut ( I. 3. ) off AE equal to AD ; join DE , and upon it , on the side remote from 4 , describe ( I. 1. ) an equilateral triangle DEF ; then join AF ; the straight line AF bisects the triangle BAC ...
... Take any point Din AB , and from AC cut ( I. 3. ) off AE equal to AD ; join DE , and upon it , on the side remote from 4 , describe ( I. 1. ) an equilateral triangle DEF ; then join AF ; the straight line AF bisects the triangle BAC ...
Página 14
... right angles to AB . Take any point D in AC , and ( I. 3. ) make CE equal to CD , and upon DE describe ( I. 1. ) the equilateral triangle DFE , and join FC ; the straight line FC drawn from the given point C ' 14 EUCLID'S ELEMENTS .
... right angles to AB . Take any point D in AC , and ( I. 3. ) make CE equal to CD , and upon DE describe ( I. 1. ) the equilateral triangle DFE , and join FC ; the straight line FC drawn from the given point C ' 14 EUCLID'S ELEMENTS .
Página 16
... Take any point D upon the other side of AB , and from the centre C , at the distance CD , describe ( Post . 3. ) the circle EGF meeting AB in F , G ; and bisect ( I. 10. ) FG in H , and join CH ; the straight line CH , drawn from the ...
... Take any point D upon the other side of AB , and from the centre C , at the distance CD , describe ( Post . 3. ) the circle EGF meeting AB in F , G ; and bisect ( I. 10. ) FG in H , and join CH ; the straight line CH , drawn from the ...
Página 18
... Take away the common angle ABC , and ( Ax . 3. ) 3 . The remaining angle ABE is equal to the remaining angle ABD , the less to the greater , which is impossible ; therefore 4. BE is not in the same straight line with BC . And in like ...
... Take away the common angle ABC , and ( Ax . 3. ) 3 . The remaining angle ABE is equal to the remaining angle ABD , the less to the greater , which is impossible ; therefore 4. BE is not in the same straight line with BC . And in like ...
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Términos y frases comunes
AB is equal AC is equal adjacent angles angle ABC angle ACB angle AGH angle BAC angle BCD angle EDF angle equal base BC bisected circle ABC circumference diameter double draw equal angles equal Constr equal Hyp equal straight lines equal to BC equilateral and equiangular EUCLID'S ELEMENTS exterior angle given circle given point given rectilineal angle given straight line given triangle gnomon greater inscribed interior and opposite less Let ABC Let the straight likewise opposite angles parallel to CD parallelogram pentagon perpendicular point F produced Q.E.D. PROP rectangle AE rectangle contained remaining angle required to describe right angles semicircle side BC square of AC straight line AB straight line AC straight line drawn touches the circle triangle ABC twice the rectangle