1. Solution: ∫△ABC is a right triangle.

∴MC⊥BC

∵BM segmentation∠ ∠CBA

And MD⊥BA

∴CM=MD (the distance from the point on the bisector of the angle is equal to both sides of the angle)

BC=BD

∴△ Perimeter of ∴△DMA =MD+DA+MA

=(MD+MA)+DA

=(MC+MA)+(BA-BD)

=3+ (5 BC)

=3+(5-4)

=4

2. Solution: Yes

Reason: ∵PM⊥AO,PN⊥OB

∴∠PMO=∠PNO=90

OM = on, OP=OP

∴△OMP is equal to△ PNO (H.L.).

∴∠MOP=∠PON

That is, OP divided by ∠AOB.