The Elements of Euclid, Libros 1-6;Libro 11 |
Dentro del libro
Resultados 1-3 de 91
Página 127
1 1 Aud because FB is equal to FC , and FK is common to the two triangles BFK , CFK ; the two sides BF , FK are equal to the two sides CF , FK , each to each ; and the base BK was shewn equal to the base CK ; therefore the angle BFK is ...
1 1 Aud because FB is equal to FC , and FK is common to the two triangles BFK , CFK ; the two sides BF , FK are equal to the two sides CF , FK , each to each ; and the base BK was shewn equal to the base CK ; therefore the angle BFK is ...
Página 150
And it was shewn that FG is not less than K , and EF is greater than D ; [ Construction . therefore the whole EG is greater than K and D together . But K and D together are equal to L ; [ Construction , therefore EG is greater than L.
And it was shewn that FG is not less than K , and EF is greater than D ; [ Construction . therefore the whole EG is greater than K and D together . But K and D together are equal to L ; [ Construction , therefore EG is greater than L.
Página 225
And because EA is equal to EB , [ Construction and EF is common and at right angles to them , [ Hypothesis . therefore the base AF is equal to the base BF . [ I. 4 . For the same reason CF is equal to DF . And since it has been shewn ...
And because EA is equal to EB , [ Construction and EF is common and at right angles to them , [ Hypothesis . therefore the base AF is equal to the base BF . [ I. 4 . For the same reason CF is equal to DF . And since it has been shewn ...
Comentarios de la gente - Escribir un comentario
No encontramos ningún comentario en los lugares habituales.
Otras ediciones - Ver todas
Términos y frases comunes
ABCD angle ABC angle ACB angle BAC Axiom base BC is equal bisected Book centre circle circle ABC circumference common Construction contained Corollary Definition demonstration described diameter divided double draw drawn edition Elements equal equal angles equiangular equilateral equimultiples Euclid exterior angle extremities fall figure four fourth given straight line greater half Hypothesis impossible join less Let ABC magnitudes manner meet multiple namely parallel parallelogram pass perpendicular plane polygon PROBLEM produced proportionals Q.E.D. PROPOSITION ratio reason rectangle rectangle contained rectilineal figure right angles segment shewn sides similar Simson solid square straight line &c suppose Take taken THEOREM third touches the circle triangle ABC Wherefore whole