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Trianguli specie & magnitudine dati tres angulos ad rectas totidem positione datas, quae non sunt omnes parallelae, singulos ad singulas ponere.
Dantur positione tres rectae infinitae AB, AC, BC, & opor∣tet triangulum DEF ita locare, ut angulus ejus D lineam AB,
angulus E lineam AC, & angulus F lineam BC tangat. Super DE, DF & EF describe tria circulorum segmenta DRE, DGF,
Agatur enim Fc ipsi aD occurrens in n. Jungantur aG,
bG, PD, QD & producatur PQ ad R. Ex constructione est angulus EaD aequalis angulo CAB, & angulus EcF aequalis an∣gulo ACB, adeo{que} triangulum anc triangulo ABC aequiangu∣lum. Ergo angulus anc seu FnD angulo ABC, adeo{que} angulo FbD aequalis est, & propterea punctum n incidit in punctum b. Porro angulus GPQ, qui dimidius est anguli ad centrum G∣PD, aequalis est angulo ad circumferentiam GaD; & angulus G∣QR, qui dimidius est complementi anguli ad centrum GQD, aequalis est angulo ad circumferentiam GbD, adeo{que} eorum com∣plementa PQG, abG aequantur, sunt{que} ideo triangula GPQ, Gab similia, & Ga est ad ab ut GP ad PQ; id est (ex construc∣tione) ut Ga ad AB. Aequantur ita{que} ab & AB & propterea triangula abc, ABC, quae modo similia esse probavimus, sunt etiam aequalia. Unde cum tangant insuper trianguli DEF angu∣li D, E, F trianguli abc latera ab, ac, bc respective, compleri potest figura ABC def figurae abc DEF similis & aequalis, at{que} eam complendo solvetur Problema. Q.E▪F.
Corol. Hinc recta duci potest cujus partes longitudine datae rectis tribus positione datis interjacebunt. Concipe Triangulum DEF, puncto D ad latus EF accedente, & lateribus DE, DF in di∣rectum positis, mutari in lineam rectam, cujus pars data DE, rec∣tis positione datis AB, AC, & pars data DF rectis positione da∣tis AB, BC interponi debet; & applicando constructionem prae∣cedentem ad hunc casum solvetur Problema.