(29)guli aequlateri angulus continet 60 gr.(120) patet, quâ ratione solo circino &regulâ fieri queat angulus 60 graduum.124. PROBL. In extremo puncto re-ctae AB (quae defectu spatii produci ne-queat) perpendicularem ducere.SOLUTIO I. Super AB fac triangu-lum aequilaterum ABL (122) II. Late-re BL producto fac OL=AL. III. Du-cta ex O in A recta OA est perpendicu-laris.DEMONST. Quia quilibet aequila-teri angulus = 60 gr. (120) erit LAB †LBA= 120 gr. : sed OLA=LABLBA (112) ergo OLA=120 gr. (22)jam quia OL=AL ex constr., erit AOLaequicrurum (115) undeangulus AOL-OAL (116) sed AOL † OAL †OLA= 180 gr. (103) ex his ergo si subtra-hatur OLA = 120 gr., relinquenturAOL+OAL=60 gr., adeóque OAL= 30 gr., cui si addis LAB= 60 gr.(120) erit OAL+LAB=OAB=90gr., qui proinde rectus est (43) & hincOA perpendicularis (58) q. e. d.125. THEOR. Cujusvis trianguli illeangulus major est, qui opponitur majorilateri.DEMONST. Sit OB=AB: dico,erit etiam angulus OAB » AOB, Namex OB abscindatur pars BL=AB: du-ctâ AL erit ABL aequicrurum (115) &angulus

16.