Elements of Geometry: Containing the First Six Books of Euclid, with a Supplement on the Quadrature of the Circle, and the Geometry of Solids; to which are Added, Elements of Plane and Spherical TrigonometryB. & S. Collins; W. E. Dean, printer, 1836 - 311 páginas |
Dentro del libro
Resultados 6-10 de 67
Página 33
... fore the angle DAF is equal ( 5. 1. ) to the angle EAF wherefore the given rectilineal angle BAC is bisected by the straight line AF . : SCHOLIUM . D 01 ! E Jons F B C By the same construction , each of the halves BAF , CAF , may be ...
... fore the angle DAF is equal ( 5. 1. ) to the angle EAF wherefore the given rectilineal angle BAC is bisected by the straight line AF . : SCHOLIUM . D 01 ! E Jons F B C By the same construction , each of the halves BAF , CAF , may be ...
Página 39
... fore the parallelogram FECG is equal to the triangle ABC , and it has one of its angles CEF equal to the given angle D ; Wherefore there has been described a parallelogram FECG equal to a given triangle ABC , having one of its angles ...
... fore the parallelogram FECG is equal to the triangle ABC , and it has one of its angles CEF equal to the given angle D ; Wherefore there has been described a parallelogram FECG equal to a given triangle ABC , having one of its angles ...
Página 41
... fore equivalent . Draw , now , the diagonal CA and BG parallel to it , meeting EA produced : join CG , and the polygon ABCF will be re- duced to an equivalent triangle ; and thus the pentagon ABCDE B C D G A E will be reduced to an ...
... fore equivalent . Draw , now , the diagonal CA and BG parallel to it , meeting EA produced : join CG , and the polygon ABCF will be re- duced to an equivalent triangle ; and thus the pentagon ABCDE B C D G A E will be reduced to an ...
Página 45
... fore AB.AC + AB.BC = AB2 . SCHOLIUM . A D C с B FE This property is evident from Algebra : let AB be denoted by a , and the segments AC , CB , by b and d , respectively ; then , a = b + d ; therefore , multiplying both members of this ...
... fore AB.AC + AB.BC = AB2 . SCHOLIUM . A D C с B FE This property is evident from Algebra : let AB be denoted by a , and the segments AC , CB , by b and d , respectively ; then , a = b + d ; therefore , multiplying both members of this ...
Página 46
... fore AB.BC = AC.CB + BC2 . A C a + be F D Са SCHOLIUM . B E In this proposition let AB be denoted by a , and the segments AC and CB , by b and c ; then a = b + c : therefore , multiplying both members of this equality by c , we shall ...
... fore AB.BC = AC.CB + BC2 . A C a + be F D Са SCHOLIUM . B E In this proposition let AB be denoted by a , and the segments AC and CB , by b and c ; then a = b + c : therefore , multiplying both members of this equality by c , we shall ...
Otras ediciones - Ver todas
Términos y frases comunes
ABC is equal ABCD adjacent angles altitude angle ABC angle ACB angle BAC angles equal base BC bisected centre chord circle ABC circumference cosine cylinder demonstrated diameter divided equal and similar equal angles equiangular equilateral equilateral polygon equimultiples Euclid exterior angle fore four right angles given straight line greater Hence hypotenuse inscribed join less Let ABC Let the straight magnitudes meet multiple opposite angle parallel parallelogram parallelopiped perpendicular polygon prism PROP proposition quadrilateral radius ratio rectangle contained rectilineal figure remaining angle right angled triangle SCHOLIUM segment semicircle shewn side BC sine solid angle solid parallelopiped spherical angle spherical triangle square straight line BC THEOR third touches the circle triangle ABC triangle DEF wherefore