LET ∠DBP = ∠PBC = a°And ALSO ∠ECP + ∠PCB = b°∠ABC = 180° - 2a° and ∠ACB = 180° - 2b°In ΔABC,180° - 2a° + 72° + 180° - 2b° = 180°a° + b° = 126°In ΔPBC,a° + b° + x° = 180°x° = 180° - 126° = 54°∠BPC = 90° - (1/2) × ∠AHere, ∠A = 72° ∴ ∠BPC = 90° - (1/2) × 72° ⇒ ∠BPC = 90° - 36° = 54°