232. SCHOLIUM I.
Uberior demonſtratio n. 558.

Demonſtravimus in congreſſu corporum elaſticorum ſummam virium ante
& poſt ictum eſſe eandem ; unde ſequitur, poſitis explicatis in n. 565. 566. AB x BN ^q + BC x BE ^q = AB x BG ^q + BC x BP ^q ; cujus & hìc geometri-
cam dabimus demonſtrationem.

Primo tendant corpora eandem partem verſus. Formentur quadrata li-
nearum BE, BG, BN, & BP; ducatur omnium diagonalis BV. Du-
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 ut ſunt ſemidifferentiæ quadratorum ita integræ differentiæ: ergo

BN ^q - BG ^q , BP ^q - BE ^q : :IN, IE, id eſt ut BC ad AB ex conſtructione. Idcirco AB x BN ^q - AB x BG ^q = BC x BP ^q - BC x BE ^q ; ideo AB x BN ^q
+ BC x BE ^q = AB x BG ^q + BC x BP ^q . quod demonſtrandum erat.

Tendant nunc corpora in partes contrarias. Formentur iterum quadrata
linearum BP, BN, BE aut B e, & BG aut B g. Propter æquales IN,
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
BP ^q - BE ^q , BN ^q - BG ^q : : AB, BC

Idcirco AB x BN ^q - AB x BG ^q = BC x BP ^q - BC x B ^Eq ; unde
deducimus AB x BN ^q + BC x BE ^q = AB x BG ^q + BC x BP ^q . Quod
demonſtrandum erat.