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Math activity class seven times
(1) Solution: No; Lack of conditions to prove the congruence of triangles,

∵ Only OP=OP and PM=PN cannot judge △ OPM △ OPN;

∴ It cannot be judged that OP is the bisector of ∞∠AOB;

(2) Scheme (2) is feasible,

Proof: at △OPM and △OPN

∵OM=ONOP=OPPM=PN,

∴△OPM≌△OPN(SSS),

∴∠ AOP =∠∠ BOP (the angles corresponding to congruent triangles are equal),

∴OP is the bisector of ∞∠AOB.

(3) Solution: ① If OC is inside ∠AOB (Figure 1),

∫∠AOC:∠COB = 1:3,

∴ let ∠AOC=x, ∠COB=3x,

∫∠AOB = 60,

∴x+3x=60,

X = 15,

That is ∠ AOC = 15,

∫OP shares ∠AOB,

∴∠AOP=30,

∴∠cop=∠aod-∠aoc=30- 15 = 15。

② If OC is outside ∠AOB (as shown in Figure 2),

∫∠AOC:∠COB = 1:3,

∴ let ∠AOC=x, ∠COB=3x,

∫∠AOB = 60,

∴3x-x=60,

X = 30,

∴∠AOC=x=30,∠COB=3x=3×30 =90,

∫OP shares ∠AOB,

∴∠AOP=30,

∴∠COP=∠AOC+∠AOP=30 +30 =60。

The angle formed by the bisector of ∴OC and ∠AOB is15 or 60.