1 easy questions

ΔCAD is isosceles with CA = CD (given)
∠ADC = ∠CAD = 50° (base ∠s of isos. Δ)

∠ACD + ∠ADC + ∠CAD = 180° (∠ sum of Δ)
∠ACD + 50° + 50° = 180°
∠ACD = 80°

AB = BC = CA (given)
ΔABC is equilateral.
Hence ∠BCA= 60°
∠BCD = ∠BCA + ∠ACD= 60° + 80° = 140°

ΔBCD is isosceles with BC = CD (given)
∠CDB = ∠CBD (base ∠sof isos. Δ)

∠CBD + ∠CDB + ∠BCD= 180° (∠sum of Δ)
∠CBD + ∠CBD + 140° = 180°
∠CBD = 20°