247160PHYSICES ELEMENTA
SCHOLIUM I.
Uberior demonſtratio n. 558.
Demonſtravimus in congreſſu corporum elaſticorum ſummam virium ante
11TAB. XX.
fig. 12.& poſt ictum eſſe eandem ; unde ſequitur, poſitis explicatis in n. 565. 566. AB x BNq + BC x BEq = AB x BGq + BC x BPq; cujus & hìc geometri-
22470. cam dabimus demonſtrationem.
11TAB. XX.
fig. 12.& poſt ictum eſſe eandem ; unde ſequitur, poſitis explicatis in n. 565. 566. AB x BNq + BC x BEq = AB x BGq + BC x BPq; cujus & hìc geometri-
22470. cam dabimus demonſtrationem.
Primo tendant corpora eandem partem verſus.
Formentur quadrata li-
33587. nearum BE, BG, BN, & BP; ducatur omnium diagonalis BV. Du-
44TAB. XX.
fig. 18. catur IS parallela ad PV; & per S, punctum, in quo diagonalem ſecat,
ducatur XSK, parallela PB; continuentur GR & EQ in Z & K; quia
IN & IG ſunt æquales, ut & IP & IE, triangula YST, RSZ ſunt æ-
qualia, etiam triangula SXV, SKQ. Idcirca Trapezium GRTN æ-
quale eſt rectangulo GZYN, & trapezium EQVP æquale rectangulo
EKXP.
33587. nearum BE, BG, BN, & BP; ducatur omnium diagonalis BV. Du-
44TAB. XX.
fig. 18. catur IS parallela ad PV; & per S, punctum, in quo diagonalem ſecat,
ducatur XSK, parallela PB; continuentur GR & EQ in Z & K; quia
IN & IG ſunt æquales, ut & IP & IE, triangula YST, RSZ ſunt æ-
qualia, etiam triangula SXV, SKQ. Idcirca Trapezium GRTN æ-
quale eſt rectangulo GZYN, & trapezium EQVP æquale rectangulo
EKXP.
Semidifferentia quadratorum linearum BN, BG eſt trapezium GRTN,
id eſt rectangulum GZYN. Eodem modo ſemidifterentia quadratorum linea-
rum BP, BE eſt rectangulum EKXP; Sed rectangula hæc, propter communem
altitudinem IS, ſunt ut baſes , aut ut baſium ſemiſſes IN, IE; etiam 551. El. VI. ſunt ſemidifferentiæ quadratorum ita integræ differentiæ: ergo
id eſt rectangulum GZYN. Eodem modo ſemidifterentia quadratorum linea-
rum BP, BE eſt rectangulum EKXP; Sed rectangula hæc, propter communem
altitudinem IS, ſunt ut baſes , aut ut baſium ſemiſſes IN, IE; etiam 551. El. VI. ſunt ſemidifferentiæ quadratorum ita integræ differentiæ: ergo
BNq - BGq, BPq - BEq:
:IN, IE, id eſt ut BC ad AB ex conſtructione.
Idcirco AB x BNq - AB x BGq = BC x BPq - BC x BEq; ideo AB x BNq
+ BC x BEq = AB x BGq + BC x BPq. quod demonſtrandum erat.
Idcirco AB x BNq - AB x BGq = BC x BPq - BC x BEq; ideo AB x BNq
+ BC x BEq = AB x BGq + BC x BPq. quod demonſtrandum erat.
Tendant nunc corpora in partes contrarias.
Formentur iterum quadrata
66588. linearum BP, BN, BE aut B e, & BG aut B g. Propter æquales IN,
77TAB. XX.
fig. 29. IG, & IP, IE, æquales ſunt NP, EG, aut e g; addamus utrim-
que e N, erunt æquales e P, g N. Differentia quadratorum BV & BQ,
id eſt quadratorum linearum BP, BE, eſt rectangulum, cujus baſis eſt PV,
& e Q, id eſt PE, & altitudo e P; differentia quadratorum BT, BR,
id eſt quadratorum linearum BN, B g aut BG, eſt rectangulum, cujus ba-
ſis eſt NT, & g R, id eſt NG, & altitudo g N; propter æquales alti-
tudines rectangula hæc ſunt ut baſes PE, NG, aut ut harum ſemiſſes IE,
IN, quæ ſuntut AB, BC; ergo
BPq - BEq, BNq - BGq: : AB, BC
66588. linearum BP, BN, BE aut B e, & BG aut B g. Propter æquales IN,
77TAB. XX.
fig. 29. IG, & IP, IE, æquales ſunt NP, EG, aut e g; addamus utrim-
que e N, erunt æquales e P, g N. Differentia quadratorum BV & BQ,
id eſt quadratorum linearum BP, BE, eſt rectangulum, cujus baſis eſt PV,
& e Q, id eſt PE, & altitudo e P; differentia quadratorum BT, BR,
id eſt quadratorum linearum BN, B g aut BG, eſt rectangulum, cujus ba-
ſis eſt NT, & g R, id eſt NG, & altitudo g N; propter æquales alti-
tudines rectangula hæc ſunt ut baſes PE, NG, aut ut harum ſemiſſes IE,
IN, quæ ſuntut AB, BC; ergo
BPq - BEq, BNq - BGq: : AB, BC
Idcirco AB x BNq - AB x BGq = BC x BPq - BC x BEq;
unde
deducimus AB x BNq + BC x BEq = AB x BGq + BC x BPq. Quod
demonſtrandum erat.
deducimus AB x BNq + BC x BEq = AB x BGq + BC x BPq. Quod
demonſtrandum erat.
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