BC=√3BD, |AD|=1AC.AD = (AD+DC). AD = |AD|^2 + DC.AD = 1 + (BC - BD). AD = 1 + (√3BD - BD ).AD = 1 + (√3-1)BD.AD = 1 + (√3-1)|BD||AD| cos ∠BDA = 1 + (√3-1) |AD|^2 = √3 #