Guichard, Claude, 1861- · Paris : Viubert et Nony · leaf 272
Fig. 169. … Fig. 170. Lun des arcs PA ou PB doit traverser le triangle; supposonsque ce soit larc PA [fig. 170). On a A + B-C = (P + Y) + (Y-a)-(!3-a) = 2y. 3° Le point P est extérieur au triangle, mais du côté opposéau sommet G par rapport au grand cercle AB [fig. 171). On a alors A-hB-G = (!i —Y)-l-(a —y)…
Geometry
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