Let CM in M be perpendicular to BD and PN in N be perpendicular to CM, and prove that all triangles MN=PC are equal to triangle QPC (PN is parallel to BD, be = BC, PC is the public side), PQ is equal to CN, cm = Mn+NC, Mn = PC, NC = PQ, PQ+PR = cm, and triangle MBC is an isosceles right triangle, cm =