Consider(AQ2+BP2)FromΔACQAQ2=AC2+CQ2⋯(1)FromΔBPCBP2=CP2+BC2⋯(2)FromΔQPCPQ2=CP2+CQ2⋯(3)FromΔABCAB2=CA2+CB2⋯(4)From(1),(2)⟹AQ2+BP2=AC2+CQ2+CP2+BC2From(3),(4)⟹AB2+PQ2=AC2+CQ2+CP2+BC2∴AQ2+BP2=AB2+PQ2Hence proved.