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(1)解：在$Rt\triangle CDB$中，由勾股定理，得$CD^{2}=BC^{2}-BD^{2}=25^{2}-15^{2}=400，$$\therefore CD = 20$米，$\therefore CE = CD + DE=20 + 1.6 = 21.6$米。
答：风筝的垂直高度$CE$为$21.6$米。
(2)解：如图，由题意得，$CM = 12$米，$\therefore DM = 8$米，
$\therefore BM^{2}=DM^{2}+BD^{2}=8^{2}+15^{2}=289，$$\therefore BM = 17$米，
$\therefore BC - BM=25 - 17 = 8$米。
答：小明应该往回收线$8$米。
$ (1)证明： $
$\because CD\perp AB，$$BE\perp AC$
$\therefore\angle BDH=\angle BEA=\angle CDA = 90^{\circ}。$
$\because\angle ABC = 45^{\circ}$
$\therefore\angle BCD = 180^{\circ}-90^{\circ}-45^{\circ}=45^{\circ}=\angle ABC$
$\therefore DB = DC。$
$\because\angle BDH=\angle BEA=\angle CDA = 90^{\circ}$
$\therefore\angle A+\angle ACD = 90^{\circ}，\angle A+\angle HBD = 90^{\circ}$
$\therefore\angle HBD=\angle ACD$
在$\triangle DBH$和$\triangle DCA$中，
$\begin{cases}\angle BDH=\angle CDA\\BD = CD\\\angle HBD=\angle ACD\end{cases}$
$\therefore\triangle DBH\cong\triangle DCA(ASA)，$$\therefore BH = AC。$
(2)证明：连接$CG，$由
(1)知，$DB = CD。$
$\because F$为$BC$的中点，$\therefore DF$垂直平分$BC，$$\therefore BG = CG。$
$\because\angle ABE=\angle CBE，$$BE\perp AC，$$\therefore EC = EA。$
在$Rt\triangle CGE$中，由勾股定理，得$CG^{2}-GE^{2}=CE^{2}，$$\therefore BG^{2}-GE^{2}=EA^{2}。$
解:(2)过点$D$作$DE\perp AB$交$AB$于点$E，$过点$D$作$DF\perp AC$交$AC$的延长线于点$F。$
$ 由 $
(1)知$BD = DC，$$\angle BAD=\angle CAD，$$\therefore DE = DF。$
在$Rt\triangle BDE$与$Rt\triangle CDF$中，$\begin{cases}BD = CD\\DE = DF\end{cases}$
$\therefore Rt\triangle BDE\cong Rt\triangle CDF(HL)，$$\therefore BE = CF。$
在$Rt\triangle ADE$与$Rt\triangle ADF$中，$\begin{cases}AD = AD\\DE = DF\end{cases}$
$\therefore Rt\triangle ADE\cong Rt\triangle ADF(HL)，$$\therefore AE = AF，$
$BE = AB - AE = AB-(AC + CF)，$即$BE = AB - AC - BE，$
$\therefore BE=\frac{AB - AC}{2}。$
$\because AB = 15，$$AC = 9，$$\therefore BE=\frac{15 - 9}{2}=3。$
解: (1)根据题意，$AB\perp BD，$$ED\perp BD，$$AB = 5，$$DE = 1，$$BD = 8，$设$CD = x，$则$BC = BD - CD = 8 - x，$利用勾股定理，得$AC=\sqrt{AB^{2}+BC^{2}}=\sqrt{5^{2}+(8 - x)^{2}}=\sqrt{25+(8 - x)^{2}}，$$CE=\sqrt{CD^{2}+DE^{2}}=\sqrt{x^{2}+1}，$$\therefore AC + CE=\sqrt{25+(8 - x)^{2}}+\sqrt{x^{2}+1}。$
(2)根据题意，连接$AE，$则$AC + CE\geq AE，$当$A，$$C，$$E$三点共线时，$AC + CE$的值最小。
(3)根据$\sqrt{x^{2}+4}+\sqrt{(12 - x)^{2}+9}，$构造$AB = 3，$$DE = 2，$$BD = 12，$$CD = x。$如图所示，当$A，$$C，$$E$三点共线时，$AC + CE$最小，延长$ED$到点$F，$过点$A$作$AF\perp DF$于点$F，$易得$AF = BD = 12，$$AB = DF = 3，$$EF = ED + DF = 5。$故$AE=\sqrt{AF^{2}+EF^{2}} = 13。$
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