解：（1）补全图形如图①所示。
（2）$AB = 2BE - BC。$如图②，
过点$D$作$DF\perp AB$于点$F。$
$\because BM$平分$\angle ABC，$$DF\perp AB，$$DE\perp BC，$
$\therefore DE = DF。$
$\because AD = CD，$
$\therefore Rt\triangle ADF\cong Rt\triangle CDE。$
$\therefore AF = CE。$
$\because DE = DF，$$BD = BD，$
$\therefore Rt\triangle BDF\cong Rt\triangle BDE。$
$\therefore BF = BE。$
$\therefore AB = BF + AF = BE + CE = BE + BE - BC = 2BE - BC。$
（3）$AB = 2BE + BC。$
解析：如图③，过点$D$作$DF\perp AB$于点$F。$
$\because BM$平分$\angle ABP，$$DF\perp AB，$$DE\perp BC，$
$\therefore DE = DF。$
$\because AD = CD，$
$\therefore Rt\triangle ADF\cong Rt\triangle CDE。$
$\therefore AF = CE。$
$\because DE = DF，$$BD = BD，$
$\therefore Rt\triangle BDF\cong Rt\triangle BDE。$
$\therefore BF = BE。$
$\therefore AB = BF + AF = BE + CE = BE + BE + BC = 2BE + BC。$
