OP' ⊥ AB
OP = PP' } = > OAP'B = romb
P(OAP'B) = 4OB } => P(OAP'B) = 4×2 cm = 8 cm
OB = OP' = 2 cm
OB = OP'
OB = BP' } => ΔBOP' echiateral => ∡B = ∡A = 60°
∡O = ∡P' = 2∡B = 2×60° = 120°
