Pls solve this sir
Dear Student,
given: ABCD is a square and OB = BP
in the triangle OBP;
∠BOP = ∠OPB
∠OBP = 90/2 = 45 deg [by the symmetry]
∠BOP = (180-45) / 2 = 135 / 2 = 67.5 deg
∠BOC = 90 deg [diagonals of a square bisect at right angle]
∠POC = 90 - 67.5 = 22.5 deg
again BDC is an isosceles triangle with BCD=90 deg.
and BDC=DBC=45 deg.=2*22.5(POC).
∠BDC = 2 ∠POC
∠BOP = 67.5
67.5 = 3 *22.5
∠BOP = 3∠ POC
Regards
Regards