142415
VERA
CIRCULI ET HYPERBOLÆ
QUADRATURA.
CIRCULI ET HYPERBOLÆ
QUADRATURA.
Sit circuli, ellipſeos vel hyperbolæ ſegmentum B I P
11TAB. XLIII.
Fig. 1. 2. 3. cujus centrum A: compleatur triangulum A B P, &
ſegmentum in punctis, B, P, tangentes ducantur re-
ctæ B F, P F, ſe invicem ſecantes in puncto F; pro-
ducatur (ſi opus ſit) recta A F ſegmentum interſecans in
puncto I & rectam B P in puncto Q; deinde jungantur re-
ctæ B I, P I.
11TAB. XLIII.
Fig. 1. 2. 3. cujus centrum A: compleatur triangulum A B P, &
ſegmentum in punctis, B, P, tangentes ducantur re-
ctæ B F, P F, ſe invicem ſecantes in puncto F; pro-
ducatur (ſi opus ſit) recta A F ſegmentum interſecans in
puncto I & rectam B P in puncto Q; deinde jungantur re-
ctæ B I, P I.
PROP. I. THEOREMA.
Dico trapezium B A P I eſſe medium propor-
tionale inter trapezium B A P F, &
triangulum B A P.
tionale inter trapezium B A P F, &
triangulum B A P.
Quoniam recta A Q ducitur per F concurſum duarum re-
ctarum F B, F P, ſegmentum in punctis B, P, tan-
gentium; igitur recta A Q rectam B P contactuum
puncta jungentem bifariam ſecabit in puncto Q; & proinde
triangulum A B Q eſt æquale triangulo A Q P, & trian-
gnlum F B Q triangulo F Q P; & igitur triangulum A B F
æquale eſt triangulo A P F; eſt ergo triangulum A B F di-
midium trapezii A B F P: eodem modo probatur triangu-
lum A B I eſſe dimidium trapezii A B I P; & triangulum
A B Q eſt dimidium trianguli A B P: cumque triangula
A B F, A B I, A B Q, eandem habeant altitudinem, in-
ter ſe ſunt ut baſes, ſed eorum baſes nempe A F, A I, A Q,
ſunt continuè proportionales; & igitur ipſa quoque triangu-
la ſunt continuè proportionalia; & proinde eorum dupla ni-
mirum trapezia A B F P, A B I P, & triangulum A B P
ſunt continuè proportionalia in ratione A F ad A I, quod
demonſtrare oportuit.
ctarum F B, F P, ſegmentum in punctis B, P, tan-
gentium; igitur recta A Q rectam B P contactuum
puncta jungentem bifariam ſecabit in puncto Q; & proinde
triangulum A B Q eſt æquale triangulo A Q P, & trian-
gnlum F B Q triangulo F Q P; & igitur triangulum A B F
æquale eſt triangulo A P F; eſt ergo triangulum A B F di-
midium trapezii A B F P: eodem modo probatur triangu-
lum A B I eſſe dimidium trapezii A B I P; & triangulum
A B Q eſt dimidium trianguli A B P: cumque triangula
A B F, A B I, A B Q, eandem habeant altitudinem, in-
ter ſe ſunt ut baſes, ſed eorum baſes nempe A F, A I, A Q,
ſunt continuè proportionales; & igitur ipſa quoque triangu-
la ſunt continuè proportionalia; & proinde eorum dupla ni-
mirum trapezia A B F P, A B I P, & triangulum A B P
ſunt continuè proportionalia in ratione A F ad A I, quod
demonſtrare oportuit.