$y=ax+b$与反比例函数$y=\frac{k}{x}$的图象相交于$A(1,4),B(4,m)$两点,与$x$轴交于点$C$,点$D$与点$A$关于点$O$对称,连接$AD$.
(1)求一次函数和反比例函数的表达式.
(2)点$P$在$x$轴的负半轴上,且$△ AOC$与$△ POD$相似,求点$P$的坐标.

19. (1)解:把$A(1,4)$代入反比例函数$y=\dfrac{k}{x}$,则$k=4$,则反比例函数表达式为$y=\dfrac{4}{x}$,把$B(4,m)$代入$y=\dfrac{4}{x}$,则$m=1$,$\therefore B(4,1)$,再把$A(1,4),B(4,1)$代入$y=ax+b$,则$\begin{cases}a+b=4,\\4a+b=1,\end{cases}$解得$\begin{cases}a=-1,\\b=5,\end{cases}$则一次函数的表达式为$y=-x+5.$ (2)解:令$y=-x+5=0$,则$x=5$,$\therefore C(5,0).$ $\because$ 点$D$与点$A$关于点$O$对称,且$A(1,4)$,$\therefore D(-1,-4)$,设点$P(x,0)(x＜0)$,$\because O(0,0)$,$\therefore PO=-x.$ 又$\because A(1,4),C(5,0),D(-1,-4)$,$\therefore AO=\sqrt{(1-0)^2+(4-0)^2}=\sqrt{17}$,$OD=\sqrt{(-1-0)^2+(-4-0)^2}=\sqrt{17}$,$OC=5.$ $\because △ AOC$与$△ POD$相似,$∠ AOC=∠ POD$,$\therefore$ 分两种情况:$\dfrac{AO}{PO}=\dfrac{OC}{OD}$或$\dfrac{AO}{OD}=\dfrac{OC}{PO}$,当$\dfrac{AO}{PO}=\dfrac{OC}{OD}$时,即$\dfrac{\sqrt{17}}{-x}=\dfrac{5}{\sqrt{17}}$,解得$x=-\dfrac{17}{5}$,此时点$P$的坐标是$(-\dfrac{17}{5},0)$,当$\dfrac{AO}{OD}=\dfrac{OC}{PO}$时,即$\dfrac{\sqrt{17}}{\sqrt{17}}=\dfrac{5}{-x}$,解得$x=-5$,此时点$P$的坐标是$(-5,0).$ 综上:当点$P$在$x$轴的负半轴上,且$△ AOC$与$△ POD$相似,点$P$的坐标为$(-\dfrac{17}{5},0)$或$(-5,0).$

20. (13分)(2026·南充中考)在菱形ABCD中,对角线AC,BD相交于点O,AP=AB.
(1)【初步感知】如图①,点P在线段AC上,若OP=2CP,AC=6,求AB的长.
(2)【深入探究】如图②,点P在线段CD上,若CP=DP,设AB长为x,AC长为y,求y关于x的函数表达式.
(3)【拓展运用】如图③,点P在线段CD上,将△ADP沿直线AP折叠,若点D落在BC边上,求$\frac{DP}{DC}$的值.

20. (1)解:$\because$ 四边形$ABCD$是菱形,$AC=6$,$\therefore OC=OA=3. \because OP=2CP$,$\therefore OP=2$,$\therefore AB=AP=OP+OA=5.$ (2)解:过点$P$作$PE⊥ AC$于点$E$,如图①,$\because$ 四边形$ABCD$是菱形,$\therefore AC⊥ BD$,$\therefore PE// BD$,$\therefore △ CPE∽ △ CDO$,$\therefore \dfrac{CE}{OC}=\dfrac{CP}{CD}. \because CP=DP$,$\therefore \dfrac{CE}{OC}=\dfrac{CP}{CD}=\dfrac{1}{2}. \because AB=x$,$AC=y$,$\therefore CP=\dfrac{1}{2}x$,$OC=\dfrac{1}{2}y$,$CE=\dfrac{1}{2}OC=\dfrac{1}{4}y$,$\therefore AE=AC-CE=\dfrac{3}{4}y.$ 在$\mathrm{Rt}△ PCE$中,$PE^2=PC^2-CE^2=(\dfrac{x}{2})^2-(\dfrac{y}{4})^2$,在$\mathrm{Rt}△ PAE$中,$PE^2=PA^2-AE^2=x^2-(\dfrac{3y}{4})^2$,即$(\dfrac{x}{2})^2-(\dfrac{y}{4})^2=x^2-(\dfrac{3y}{4})^2$,整理得$y^2=\dfrac{3}{2}x^2$,即$y=\dfrac{\sqrt{6}}{2}x$ 或 $y=-\dfrac{\sqrt{6}}{2}x$(舍去);$\therefore y$关于$x$的函数表达式为$y=\dfrac{\sqrt{6}}{2}x(x＞0).$ (3)解:如图②,设$AP$交$BD$于点$F$,将$△ ADP$沿直线$AP$折叠,使点$D$落在$BC$边上的$D'$处,连接$PD',AD'$,$\because AP=AB$,$\therefore AD=AD'=AP=AB$,$∠ D'AP=∠ DAP.$ 又$∠ ABD'=∠ ADP$,$\therefore ∠ ADP=∠ APD=∠ APD'=∠ AD'P=∠ AD'B=∠ ABD'$,$\therefore ∠ DAP=∠ D'AP=∠ BAD'. \because AB// CD$,$\therefore ∠ APD=∠ PAB=2∠ DAP. \because ∠ PDA+∠ APD+∠ DAP=180°$,$\therefore 5∠ DAP=180°$,$\therefore ∠ DAP=36°$,$∠ APD=∠ ADP=72°$,$\therefore ∠ ADF=∠ PDF=\dfrac{1}{2}∠ ADP=36°$,$\therefore ∠ PDF=∠ DAP=∠ ADF$,$\therefore ∠ DPF=∠ DFP=72°$,$\therefore DF=AF=PD.$ 又$∠ DPF=∠ APD$,$\therefore △ PDF∽ △ PAD$,$\therefore \dfrac{DP}{AP}=\dfrac{FP}{DP}$,$\therefore DP^2=AP· FP$,设$DP=m$,$DC=n$,则$AP=n$,$FP=n-m$,$\therefore m^2=n(n-m)$,$\therefore m^2+mn-n^2=0$,$\therefore (\dfrac{m}{n})^2 + \dfrac{m}{n} -1=0$,取正数解得$\dfrac{m}{n}=\dfrac{\sqrt{5}-1}{2}$,$\therefore \dfrac{DP}{DC}=\dfrac{m}{n}=\dfrac{\sqrt{5}-1}{2}.$