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第32页

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 (1)证明：$\because DF\perp AB，$$\angle ACB = 90^{\circ}，$

 $\therefore \triangle ADF，$$\triangle ACE$是直角三角形.

 在$Rt\triangle ACE$和$Rt\triangle ADF$中，

 $\begin{cases}AE = AF \\AC = AD\end{cases}$

 $\therefore Rt\triangle ACE\cong Rt\triangle ADF(HL)，$

 $\therefore CE = DF.$

 (2)证明：$\because EF\perp BC，$$DF\perp AB，$

 $\therefore \angle CEF = 90^{\circ}，$$\angle BDF = 90^{\circ}，$

 $\therefore \angle BFD + \angle ABF = 90^{\circ}.$

 又$\because \angle ACB = \angle ACF + \angle ECF = 90^{\circ}，$$\angle ACF = \angle ABF，$

 $\therefore \angle BFD = \angle ECF.$

 在$\triangle FCE$和$\triangle BFD$中，

 $\begin{cases}\angle ECF = \angle DFB \\\angle CEF = \angle FDB = 90^{\circ} \\EF = FD\end{cases}$

 $\therefore \triangle FCE\cong\triangle BFD(ASA)，$

 $\therefore FC = FB.$

 又$\because EF\perp BC，$

 $\therefore$点$F$在$BC$的垂直平分线上.

 (1)证明：$\because MN\perp AB，$

 $\therefore \angle ACP = \angle BCP = 90^{\circ}.$

 在$\triangle ACP$和$\triangle BCP$中，

 $\begin{cases}AC = BC \\\angle ACP = \angle BCP \\PC = PC\end{cases}$

 $\therefore \triangle ACP\cong\triangle BCP(SAS)，$

 $\therefore PA = PB.$

 (2)①解：$\because MN$垂直平分$AB，$

 $\therefore MB = MA.$

 $\because \triangle MBC$的周长是$18\ cm，$

 $\therefore MB + MC + BC = MA + MC + BC = AC + BC = 18\ cm.$

 $\because AC = AB = 10\ cm，$

 $\therefore BC = 8\ cm.$

 ②解：存在.

 $\because MN$垂直平分$AB，$

 $\therefore PA = PB，$

 $\therefore \triangle PBC$的周长$= PB + PC + BC = PA + PC + BC，$

 $\therefore$当$A，$$P，$$C$三点共线，即点$P$与点$M$重合时，$PA + PC$的值最小，即此时$\triangle PBC$的周长最小，最小值为$PA + PC + 8 = AC + 8 = 18\ cm.$