Bonjour,
2)a)
(AB;AF) = π/6 (La droite (AF) est la médiatrice de (AB;AC))
(AF;AD) = (AF;AC) + (AC;AE) + (AE;AD) = π/6 + π/3 + π = 9π/6 = 3π/2
(CB;CE) = (CB;CF) + (CF;CE) = π/3 + π = 4π/3
b)
(CF;BD) = (CF;CA) = 4π/3
(AD;BC) = (AD;AE) = π
(EB;CD) = (EB;EF) + (EF;CD) = π/6 + 3π/2 = 10π/6 = 5π/3