a)
DC || AB => ΔAMAB ≈ ΔMDC
MD/MA = DC/AB
MD/(MD+DA) = 12/18
MD/(MD+DA) = 2/3
3MD = 2MD + 2DA
3MD - 2MD = 2×6√3
MD = 12√3 cm
MA = MD + AD = 12√3 + 6√3 = 18√3 cm
- A(MAB) = MA × AB : 2 = 18√3×18/2 = 162√3 cm²
- P(MAB) = MA + AB + BM = 18√3 + 54 cm
BM = √(18√3)² + 18² = √1296 = 36 cm (Pitagora)
duc CE ⊥ AB
EB = AB - DC = 18 - 12 = 6 cm
CB = √CE²+EB² = √(6√3)²+(6)² = √108+36 = √144 = 12 cm
CB = 12 => CB = CD => ΔDCB isoscel => ∡CDB = ∡CBD
dar ∡CDB = ∡ABD (alterne interne) =>
=> ∡ABD = ∡DBC => BD bisectoarea ∡ABC
