Conjecture1 : 
Conjecturer la position du point P à l'intérieur du triangle ABC rendant la somme des distances AP, BP et CP minimale
(On pourra affiner en pilotant le point P avec les flèches du clavier)

A = (2, 5)
C = (11.362156575088145, -1.2752104107208244)
B = (-1.637843424911849, -1.2752104107208244)
poly1, c, a, b = Polygon(A, C, B)
P = (3.5763848709314665, 0.44952258088792596)
d = Segment(P, A)
e = Segment(P, B)
f = Segment(P, C)
AE = Distance(A, P)
EC = Distance(P, B)
EB = Distance(P, C)
g: AE + EC + EB
α = Angle(A, P, B)
β = Angle(B, P, C)
γ = Angle(C, P, A)
texte1: "AP + BP + CP =  " + (LaTeX[g]) + ""
texte3: "\widehat{CPB} = " + (LaTeX[β]) + ""
texte2: "\widehat{APB} = " + (LaTeX[α]) + ""
texte4: "\widehat{CPA} = " + (LaTeX[γ]) + ""