In △ABC, point P is the midpoint of BC. (1) as shown in figure (1), verification: AP.

BP = PC

∴ Quadrilateral ABHC is a parallelogram,

∴AB=HC

At △ ah, ah.

∴2ap<； AB+AC,

In △ABC, point P is the midpoint of BC. (1) as shown in figure (1), verification: AP.

BP = PC

∴ Quadrilateral ABHC is a parallelogram,

∴AB=HC

At △ ah, ah.

∴2ap<； AB+AC, which is AP.

Proof: cross B as BH∨AE, cross DE as H, and connect CH and AH.

∴∠ 1=∠BAC=60

DB = AC，AB=CE，

∴AD=AE

∴△AED is an equilateral triangle,

∴∠D=∠ 1=∠2=60

∴△BDH is an equilateral triangle

∴BD=DH=BH=AC

∴ quadrilateral ABHC is a parallelogram.

Point p is the midpoint of BC,

∴AH and BC are divided equally at point P, that is, AH=2AP.

At △ADH and △EDB,

∴△ADH≌△EDB

∴ah=be=2ap； ② Proof: There are two situations:

I. when AB=AC,

∴AB=AC=DB=CE

∴ BC = Germany; Two. When AB≠AC, BD and BC are taken as a set of adjacent sides to make a parallelogram BDGC (as shown in the figure).

∴DB=GC=AC,∠BAC=∠ 1,BC=DC

AB = CE

∴△ABC≌△CEC

∴BC=EC=DG

At △DGE, DG+ge >；; Delaware

∴2bc>； Germany, namely BC & gtDE, to sum up, BC≥ DE.