Prove:

∠∠AOB = 30

∴∠AOP+∠BOP=∠AOB=30

∵ point P 1 is the symmetric point of point p about OA.

∴OP 1=OP,∠AOP 1=∠AOP

∴∠p 1op=∠aop+∠aop 1=2∠aop

Point P2 is the symmetric point of point p with respect to OB.

∴OP2=OP,∠BOP2=∠BOP

∴∠P2OP=∠BOP+∠BOP2=2∠BOP

∴OP 1=OP2，∠p 1op 2 =∠p 1op+∠p2op = 2(∠AOP+∠bop)= 60

∴ equilateral △P 1OP2

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