1. (2024·宿迁三模)如图，抛物线$y = - 2x^{2}+2x + 4$交$x$轴于$A$，$B$两点(点$A$在点$B$的左边)，交$y$轴于点$C$.
(1)求$A$，$B$，$C$三点的坐标.
(2)如图①，作直线$x = m$，分别交$x$轴、线段$BC$、抛物线于$D$，$E$，$F$三点，连接$CF$.若$\triangle BDE$与$\triangle CEF$相似，求$m$的值.
(3)如图②，将抛物线在$BC$上方的图像沿$BC$折叠后与$y$轴交于点$G$，直接写出点$G$的坐标.
　　　　

（1）在$y = -2x^{2}+2x + 4$中，令$x = 0$，得$y = 4$；令$y = 0$，得$-2x^{2}+2x + 4 = 0$，解得$x = 2$或$x = -1$.$\therefore$点$A$的坐标为$(-1,0)$，点$B$的坐标为$(2,0)$，点$C$的坐标为$(0,4)$.
（2）若$\triangle BDE$与$\triangle CEF$相似，则存在$\angle CFE$或$\angle FCE$为直角.当$\angle CFE = 90^{\circ}$时，点$C$，$F$关于抛物线的对称轴对称.$\because$抛物线的对称轴为直线$x=\frac{-2}{2\times(-2)}=\frac{1}{2}$，点$C$的坐标为$(0,4)$，$\therefore$易得点$F$的坐标为$(1,4)$.$\therefore m = 1$.当$\angle FCE = 90^{\circ}$时，由$B$，$C$两点的坐标，得直线$BC$对应的函数表达式为$y = -2x + 4$，$\therefore$易得直线$CF$对应的函数表达式为$y=\frac{1}{2}x + 4$.联立$\begin{cases}y=\frac{1}{2}x + 4\\y = -2x^{2}+2x + 4\end{cases}$，解得$\begin{cases}x = 0\\y = 4\end{cases}$（不合题意，舍去）或$\begin{cases}x=\frac{3}{4}\\y=\frac{35}{8}\end{cases}$.$\therefore m=\frac{3}{4}$.综上所述，$m$的值为$\frac{3}{4}$或$1$.
（3）设点$G$的坐标为$(0,t)$.如图，过点$G$作$GG'\perp BC$交$BC$于点$T$，交抛物线于点$G'$，过点$T$作$TH\perp y$轴于点$H$.由$B$，$C$两点的坐标，得$\tan\angle OBC = 2$.易得$\angle CGT=\angle CTH=\angle OBC$，$\therefore\tan\angle CGT=\tan\angle CTH=\tan\angle OBC = 2$.设$\tan\alpha = 2$，则易得$\sin\alpha=\frac{2}{\sqrt{5}}$，$\cos\alpha=\frac{1}{\sqrt{5}}$.$\therefore CT = CG\sin\alpha=(4 - t)\sin\alpha$，$HT = CT\cos\alpha=\frac{2}{5}(4 - t)$，$CH = CT\sin\alpha=\frac{4}{5}(4 - t)$.$\therefore OH = OC - CH = 4-\frac{4}{5}(4 - t)=\frac{4}{5}t+\frac{4}{5}$.$\therefore$点$T$的坐标为$(\frac{8}{5}-\frac{2}{5}t,\frac{4}{5}t+\frac{4}{5})$.$\because T$为$GG'$的中点，$\therefore$由中点坐标公式，得$x_{G'}=2(\frac{8}{5}-\frac{2}{5}t)-0=\frac{16}{5}-\frac{4}{5}t$，$y_{G'}=2(\frac{4}{5}t+\frac{4}{5})-t=\frac{3}{5}t+\frac{8}{5}$.$\therefore$点$G'$的坐标为$(\frac{16}{5}-\frac{4}{5}t,\frac{3}{5}t+\frac{8}{5})$.将点$G'$的坐标代入$y = -2x^{2}+2x + 4$，得$\frac{3}{5}t+\frac{8}{5}=-2(\frac{16}{5}-\frac{4}{5}t)^{2}+2(\frac{16}{5}-\frac{4}{5}t)+4$，解得$t = 4$（不合题意，舍去）或$t=\frac{73}{32}$.$\therefore$点$G$的坐标为$(0,\frac{73}{32})$.

2. (2023·连云港)如图，在平面直角坐标系中，抛物线$L_{1}:y = x^{2}-2x - 3$的顶点为$P$.直线$l$过点$M(0,m)(m\geqslant - 3)$，且平行于$x$轴，与抛物线$L_{1}$交于$A$，$B$两点(点$B$在点$A$的右侧).将抛物线$L_{1}$沿直线$l$翻折得到抛物线$L_{2}$，抛物线$L_{2}$交$y$轴于点$C$，顶点为$D$.
(1)当$m = 1$时，求点$D$的坐标.
(2)连接$BC$，$CD$，$DB$.若$\triangle BCD$为直角三角形，求此时$L_{2}$对应的函数表达式.
(3)在(2)的条件下，若$\triangle BCD$的面积为$3$，$E$，$F$两点分别在边$BC$，$CD$上运动，且$EF = CD$，以$EF$为一边作正方形$EFGH$，连接$CG$，求$CG$长的最小值，并简要说明理由.
　　　　　　　　　　　　　　

（1）$\because y = x^{2}-2x - 3=(x - 1)^{2}-4$，$\therefore$抛物线$L_{1}$的顶点$P$的坐标为$(1,-4)$.$\because m = 1$，$\therefore$点$P$与点$D$关于直线$y = 1$对称.$\therefore$点$D$的坐标为$(1,6)$.
（2）$\because$抛物线$L_{1}$的顶点$P(1,-4)$与$L_{2}$的顶点$D$关于直线$y = m$对称，$\therefore$点$D$的坐标为$(1,2m + 4)$，抛物线$L_{2}$：$y = -(x - 1)^{2}+(2m + 4)=-x^{2}+2x+2m + 3$.$\therefore$当$x = 0$时，$y = 2m + 3$，即点$C$的坐标为$(0,2m + 3)$.
①当$\angle BCD = 90^{\circ}$时，如图①，过点$D$作$DN\perp y$轴于点$N$.$\because$点$D$的坐标为$(1,2m + 4)$，$\therefore$点$N$的坐标为$(0,2m + 4)$.$\because$点$C$的坐标为$(0,2m + 3)$，$\therefore DN = NC = 1$.$\therefore\angle DCN = 45^{\circ}$.$\because\angle BCD = 90^{\circ}$，$\therefore\angle BCM = 45^{\circ}$.$\because$直线$l// x$轴，$\therefore\angle BMC = 90^{\circ}$.$\therefore\angle CBM=\angle BCM = 45^{\circ}$.又$\because m\geqslant - 3$，$\therefore BM = CM=(2m + 3)-m = m + 3$.$\therefore$点$B$的坐标为$(m + 3,m)$.$\because$点$B$在抛物线$L_{1}$：$y = x^{2}-2x - 3$上，$\therefore m=(m + 3)^{2}-2(m + 3)-3$，解得$m = 0$或$m = -3$.当$m = -3$时，点$B$的坐标为$(0,-3)$，点$C$的坐标为$(0,-3)$，此时$B$，$C$两点重合，不合题意，舍去.当$m = 0$时，符合题意.将$m = 0$代入$L_{2}$：$y = -x^{2}+2x+2m + 3$，得$L_{2}$：$y = -x^{2}+2x + 3$.
②当$\angle BDC = 90^{\circ}$时，如图②，过点$D$作$DN\perp y$轴于点$N$，过点$B$作$BT\perp ND$，交$ND$的延长线于点$T$.同理，可得$BT = DT$.$\because$点$D$的坐标为$(1,2m + 4)$，$\therefore DT = BT=(2m + 4)-m = m + 4$.$\because DN = 1$，$\therefore NT = DN + DT = 1+(m + 4)=m + 5$.$\therefore$点$B$的坐标为$(m + 5,m)$.$\because$点$B$在抛物线$L_{1}$：$y = x^{2}-2x - 3$上，$\therefore m=(m + 5)^{2}-2(m + 5)-3$，解得$m = -3$或$m = -4$.$\because m\geqslant -3$，$\therefore m = -3$，此时点$B$的坐标为$(2,-3)$，点$C$的坐标为$(0,-3)$，符合题意.将$m = -3$代入$L_{2}$：$y = -x^{2}+2x+2m + 3$，得$L_{2}$：$y = -x^{2}+2x - 3$.
③易知当$\angle DBC = 90^{\circ}$时，此种情况不存在.综上所述，$L_{2}$对应的函数表达式为$y = -x^{2}+2x + 3$或$y = -x^{2}+2x - 3$.
（3）由（2）知，当$\angle BDC = 90^{\circ}$时，$m = -3$，此时$\triangle BCD$的面积为$1$，不合题意，舍去.当$\angle BCD = 90^{\circ}$时，$m = 0$，此时$\triangle BCD$的面积为$3$，符合题意.由题意，易得$EF = FG = CD=\sqrt{2}$.如图③，取$EF$的中点$Q$，连接$CQ$，$GQ$.在$Rt\triangle CEF$中，易得$CQ=\frac{1}{2}EF=\frac{\sqrt{2}}{2}$.在$Rt\triangle FGQ$中，易得$GQ=\frac{\sqrt{10}}{2}$.易知当$Q$，$C$，$G$三点共线时，$CG$的长取得最小值，为$\frac{\sqrt{10}-\sqrt{2}}{2}$.