The Elements of Euclid, Libros 1-6;Libro 11J.M. Dent & Sons, Limited, 1933 - 298 páginas |
Dentro del libro
Resultados 1-3 de 30
Página 65
... double of the square on CA. But the square on AE is equal to the squares on EC , CA. [ I . 47 . Therefore the square on AE is double of the square on AC . Again , because GF is equal to FE , the square on GF is equal to the square on FE ...
... double of the square on CA. But the square on AE is equal to the squares on EC , CA. [ I . 47 . Therefore the square on AE is double of the square on AC . Again , because GF is equal to FE , the square on GF is equal to the square on FE ...
Página 93
... double of the angle EAB . For the same reason the angle FEC is double of the angle EAC . Therefore the whole angle BEC is double of the whole angle BAC . Next , let the centre of the circle be without the angle BAC . Then it may be ...
... double of the angle EAB . For the same reason the angle FEC is double of the angle EAC . Therefore the whole angle BEC is double of the whole angle BAC . Next , let the centre of the circle be without the angle BAC . Then it may be ...
Página 127
... double of the angle CFK , and the angle BRC is double of the angle CKF . [ I. 4 . For the same reason the angle CFD is double of the angle CFL , and the angle CLD is double of the angle CLF . And because the arc BC is equal to the arc ...
... double of the angle CFK , and the angle BRC is double of the angle CKF . [ I. 4 . For the same reason the angle CFD is double of the angle CFL , and the angle CLD is double of the angle CLF . And because the arc BC is equal to the arc ...
Otras ediciones - Ver todas
Términos y frases comunes
ABCD AC is equal adjacent angles angle ABC angle ACB angle BAC angle DEF angle EDF angles equal Axiom base BC bisected centre circle ABC circumference Constr Construction Corollary Definition 15 demonstration diameter double draw equal angles equal to F equiangular equimultiples Euclid Euclid's Elements exterior angle fourth given circle given straight line gnomon greater ratio greater than F Hypothesis less Let ABC Let the straight meet multiple opposite angle parallel to BC parallelogram perpendicular polygon produced proportionals PROPOSITION 13 Q.E.D. PROPOSITION rectangle contained remaining angle right angles segment shewn sides similar and similarly Simson square described square on AC straight line &c straight line AB THEOREM tiples touches the circle triangle ABC triangle DEF twice the rectangle Wherefore