Connecting BD, it is known that ABCD is a diamond with an angle of a = 60, and ABD and BCD are equilateral triangles.

BP, British Columbia, DQ. ? Do you know △DQC∽△BCP? Get DQ/DC=BC/BP, because DC=BC=BD, get dq/BD = BD/BP; ? In addition, ∠ QDB = ∠ DBP = 120, so △QDB∠△DBP, ∠PDB=∠BQD, ∠ PRQ = ∠ RDQ+∠ BQ.