Class 1, Grade 8 students in a school divide a corner with a square in math activity class, and design the following scheme.

Solution: (1) Scheme (Ⅰ) is not feasible. There is no condition to prove triangle congruence.

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

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

Scheme (2) is feasible.

Proof: at △OPM and △OPN

OM=ONPM=PNOP=OP

∴△OPM≌△OPN(SSS)，

∴∠AOP=∠BOP (the angles corresponding to congruent triangles are equal) (5 points);

∴OP is the bisector of ∞∠AOB.

(2) When ∠AOB is a right angle, Scheme (Ⅰ) is feasible.

The sum of the internal angles of the quadrilateral is 360. If PM⊥OA, PN⊥OB, ∠ OMP = ∠ ONP = 90, ∠ MPN = 90,

∴∠AOB = 90° (vertical definition),

∵PM⊥OA, PN⊥OB, and PM=PN,

∴OP is the bisector of ∞∠AOB (angular bisector property);

When ∠AOB is not right angle, this scheme is also feasible.

∵PM⊥OA,PN⊥OB，

△ OPM and△ △OPN are right triangles.

In OPM and OPN.

PM = PNOP = OP

∴△OPM≌△OPN(HL)，

∴∠AOP=∠BOP (the angles corresponding to congruent triangles are equal) (5 points);

∴OP is the bisector of ∞∠AOB.