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. . .