解：（1）已知：在$\triangle ABC$中，$\angle ACB = 90^{\circ}，$$\angle BAC = 30^{\circ}；$
求证：$BC=\dfrac{1}{2}AB。$如图①，以点$B$为圆心，$BA$长为半径作弧，交$BC$的延长线于点$D，$连接$AD，$则$AB = BD。$
$\because\angle ACB = 90^{\circ}，$$\angle BAC = 30^{\circ}，$$\therefore\angle B = 60^{\circ}。$
$\therefore\triangle ABD$是等边三角形。
$\therefore AB = AD。$
在$Rt\triangle ABC$和$Rt\triangle ADC$中，
$\begin{cases}AB = AD\\AC = AC\end{cases}，$
$\therefore Rt\triangle ABC\cong Rt\triangle ADC。$
$\therefore BC = DC。$$\because BC + DC = BD，$
$\therefore BC=\dfrac{1}{2}BD=\dfrac{1}{2}AB。$
（2）连接$DM。$$\because$对折长方形纸片$ABCD，$$EF$为折痕，
$\therefore AE = DE，$$\angle AEM=\angle DEM = 90^{\circ}。$
又$\because EM = EM，$$\therefore\triangle AEM\cong\triangle DEM。$
$\therefore AM = DM。$
$\because$沿$AP$折叠，使点$D$落在折痕$EF$上的点$M$处，
$\therefore AD = AM，$$\angle DAP=\angle PAM，$易得$\angle PDA=\angle PMA = 90^{\circ}。$
$\therefore AD = AM = DM。$$\therefore\triangle ADM$为等边三角形。
$\therefore\angle DAM = 60^{\circ}。$$\therefore\angle DAP=\angle PAM=\dfrac{1}{2}\angle DAM = 30^{\circ}。$
又$\because\angle PMA = 90^{\circ}，$$\therefore$由（1），得$PM=\dfrac{1}{2}PA。$
（3）如图②，过点$C$作$CD\perp BA$交$BA$的延长线于点$D。$
$\because AB = AC = 20\mathrm{cm}，$$\therefore\angle ACB=\angle ABC = 15^{\circ}。$
$\therefore\angle DAC=\angle ABC+\angle ACB = 30^{\circ}。$
$\therefore$由（1），得$CD=\dfrac{1}{2}AC = 10\mathrm{cm}。$
$\therefore\triangle ABC$的面积$=\dfrac{1}{2}AB\cdot CD=\dfrac{1}{2}\times20\times10 = 100(\mathrm{cm}^{2})。$
