Solution: (first picture)

1, extending BN AC to d,

∵AN⊥BN

∴∠ANB=∠AND=90

And ∵ evenly distributed ∠BAC.

∴∠BAN=∠DAN,AN=AN

∴△ABD≌△DAN

∴BN=DN,AD=AB= 10

∴CD= 16- 10=6

∴N is the midpoint of BD

∵M is the midpoint of BC

∴MN is the center line of △ △CBD.

∴MN=CD÷2=3

2. Extend EP intersection AB at H point, DP intersection AC at G point and FP intersection BC at L point. ..

Then EH∑BC, the quadrilateral HPBD is a parallelogram,

Similarly, the quadrilateral EPLC is a parallelogram.

So △PDL is an equilateral triangle.

∴PH=PF=BD，PD=DL，PE=LC

∴PE+PF+PD=BD+DL+LC=BC=a

Solution: (the second picture)

1, (1) connecting DE, EF

∠∠EBC =∠DBE = 60

∴∠EBC-∠EBA=∠DBE-∠EBA

That is ∠DBE=∠ABC

DB = AB，BE=BC。

∴△DBE is all equal to△△△ ABC.

∴DE=AC=AF

∴ Quadrilateral ADEF is a parallelogram

(2) When AB=AC, the quadrilateral ADEF is rhombic;

When ∠ BAC = 150, the quadrilateral ADEF is a rectangle.

(3) It doesn't always exist. For example, when ∠ BAC = 60, there is no parallelogram ADEF.

2、ED = BE

∴∠EBD=∠EDB

* AD is parallel to BC.

∴∠EDB=∠DBC

∴∠EBD=∠CBD

That is, the angle bisector with BD ∠EBC.

Extend GP to BC in H.

∵PF⊥BE,PH⊥BC

∴PF=PH

∴PF+PH=AB

The above answers are based on the answers given by Kohong _ Jiyy, and the inadequacies have been corrected, so please pay attention. . .