The Elements of Euclid, Libros 1-6;Libro 11 |
Dentro del libro
Resultados 1-3 de 86
Página 103
But FAC , the exterior angle of the triangle ABC , is equal to the two angles ABC , ACB ; [ I. 32 . therefore the angle BAC is equal to the angle FAC , ( Ax . 1 . and therefore each of them is a right angle . [ I. Def . 10 .
But FAC , the exterior angle of the triangle ABC , is equal to the two angles ABC , ACB ; [ I. 32 . therefore the angle BAC is equal to the angle FAC , ( Ax . 1 . and therefore each of them is a right angle . [ I. Def . 10 .
Página 115
Let ABC be the given circle , and DEF the given triangle : it is required to inscribe in the circle ABC a ... 1 at the point A , in the straight line AH , make the angle HAC equal to the angle DEF ; [ I . 23 . and , at the point A ...
Let ABC be the given circle , and DEF the given triangle : it is required to inscribe in the circle ABC a ... 1 at the point A , in the straight line AH , make the angle HAC equal to the angle DEF ; [ I . 23 . and , at the point A ...
Página 183
And , because the angle DEF is equal to the angle GEF , and the angle GEF is equal to the angle ABC , [ Constr ... the angle ACB is equal to the angle DFE , and the angle at A is equal to the angle at D. Therefore the triangle ABC is ...
And , because the angle DEF is equal to the angle GEF , and the angle GEF is equal to the angle ABC , [ Constr ... the angle ACB is equal to the angle DFE , and the angle at A is equal to the angle at D. Therefore the triangle ABC is ...
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