439400Archimedis
Educamus igitur E G parallelam ipſi
507[Figure 507] A B, & iungamus D B, D G: & quia duo
anguli D E G, D G E ſunt æquales, erit
angulus G D C duplus anguli D E G,
& quia angulus B D C æqualis eſt angu-
lo B C D, & angulus C E G æqualis eſt
angulo A C E, erit angulus G D C du-
plus anguli C D B, & totus angulus B
D G triplus anguli B D C, & arcus B G
æqualis arcui A E, triplus eſt arcus B F,
& hoc eſt, quod voluimus.
507[Figure 507] A B, & iungamus D B, D G: & quia duo
anguli D E G, D G E ſunt æquales, erit
angulus G D C duplus anguli D E G,
& quia angulus B D C æqualis eſt angu-
lo B C D, & angulus C E G æqualis eſt
angulo A C E, erit angulus G D C du-
plus anguli C D B, & totus angulus B
D G triplus anguli B D C, & arcus B G
æqualis arcui A E, triplus eſt arcus B F,
& hoc eſt, quod voluimus.
SCHOLIVM ALMOCHTASSO.
HAEc quidem propoſitio elegantiſſima eſt, quæ ſi problematicè reſolui poſ-
ſet via plana, reperta iam eßet tripartitio cuiuſlibet anguli.
ſet via plana, reperta iam eßet tripartitio cuiuſlibet anguli.
Breuius tamen demonſtratio
509[Figure 509] perfici poteſt hac ratione. Iuncta
recta E B, quia in triangulo Iſo-
ſcele B D C duo anguli C, & C
D B æquales ſunt, eſtque pariter
externus angulus B D C duplus an-
guli D E B in triangulo Iſoſcelio
D E B, ergo angulus C duplus eſt
anguli B E C, & propterea illi an-
guli ſimul ſumpti, ſeu externus an-
gulus A B E triplus erit anguli B
E F, & circunferentia A E tripla ipſius B F.
509[Figure 509] perfici poteſt hac ratione. Iuncta
recta E B, quia in triangulo Iſo-
ſcele B D C duo anguli C, & C
D B æquales ſunt, eſtque pariter
externus angulus B D C duplus an-
guli D E B in triangulo Iſoſcelio
D E B, ergo angulus C duplus eſt
anguli B E C, & propterea illi an-
guli ſimul ſumpti, ſeu externus an-
gulus A B E triplus erit anguli B
E F, & circunferentia A E tripla ipſius B F.