8. 新趋势 开放性试题 如图，过四边形ABCD的四个顶点分别作对角线AC、BD的平行线，若所围成的四边形EFGH是矩形，则原四边形ABCD需满足的条件是__________.(只需写出一个符合要求的条件)

$AC\perp BD$（答案不唯一）

9. (2024·莆田期末)如图，P是矩形ABCD的对角线AC上一点，过点P作EF//BC，分别交AB、CD于点E、F，连接PB、PD. 若AE = 2，PF = 8，则图中阴影部分的面积为__________.
　

16 解析：如图，作$PM\perp AD$于点$M$，$PN\perp BC$于点$N$，则四边形$AEPM$、四边形$DFPM$、四边形$CFPN$、四边形$BEPN$都是矩形，
$\therefore S_{\triangle ADC}=S_{\triangle ABC}$，$S_{\triangle AMP}=S_{\triangle AEP}$，$S_{\triangle PBE}=S_{\triangle PBN}$，$S_{\triangle PFD}=S_{\triangle PDM}$，$S_{\triangle PFC}=S_{\triangle PCN}$，$\therefore S_{\triangle DFP}=S_{\triangle PBE}=\frac{1}{2}\times2\times8 = 8$，$\therefore S_{阴影}=8 + 8 = 16$.

10. (2024·包头月考)如图，在△ABC中，AB = 6，AC = 8，BC = 10，P为边BC上一动点(且点P不与点B、C重合)，PE⊥AB于点E，PF⊥AC于点F，M为EF的中点. 设AM的长为x，则x的取值范围是__________.
　　　

$2.4\leqslant x\lt4$ 解析：连接$AP$.$\because AB = 6$，$AC = 8$，$BC = 10$，$\therefore AB^{2}+AC^{2}=36 + 64 = 100$.$\because BC^{2}=100$，$\therefore AB^{2}+AC^{2}=BC^{2}$，$\therefore\angle BAC = 90^{\circ}$.$\because PE\perp AB$，$PF\perp AC$，$\therefore\angle AEP=\angle AFP=\angle BAC = 90^{\circ}$，$\therefore$四边形$AEPF$是矩形，$\therefore AP = EF$.$\because\angle BAC = 90^{\circ}$，$M$为$EF$的中点，$\therefore AM=\frac{1}{2}EF=\frac{1}{2}AP$. 当$AP\perp BC$时，$AP$的值最小.$\because S_{\triangle BAC}=\frac{1}{2}\times6\times8=\frac{1}{2}\times10\times AP$，解得$AP = 4.8$，$\therefore AP\geqslant4.8$，$\therefore 2AM\geqslant4.8$，$\therefore AM\geqslant2.4$（即$x\geqslant2.4$）. 当$P$和$C$重合时，$AM = 4$.$\because$点$P$和点$B$、$C$不重合，$\therefore AP\lt8$，$\therefore x\lt4$，$\therefore x$的取值范围是$2.4\leqslant x\lt4$.

11. (大庆中考)如图，在矩形ABCD中，AB = 3，BC = 4，M、N在对角线AC上，且AM = CN，E、F分别是AD、BC的中点.
(1)求证：△ABM≌△CDN；
(2)点G是对角线AC上的点，∠EGF = 90°，求AG的长.
　　　　　　　　　　　　　

（1）$\because$四边形$ABCD$是矩形，$\therefore AB// CD$，$\therefore\angle MAB=\angle NCD$. 在$\triangle ABM$和$\triangle CDN$中，$\begin{cases}AB = CD\\\angle MAB=\angle NCD\\AM = CN\end{cases}$，$\therefore\triangle ABM\cong\triangle CDN(SAS)$.
（2）连接$EF$，交$AC$于点$O$.$\because$四边形$ABCD$是矩形，$\therefore AD = BC$，$\angle ABC = 90^{\circ}$，$\therefore AC=\sqrt{AB^{2}+BC^{2}} = 5$.$\because E$、$F$分别是$AD$、$BC$的中点，$\therefore AE = BF$，$\therefore$四边形$ABFE$是矩形，$\therefore EF = AB = 3$. 在$\triangle AEO$和$\triangle CFO$中，$\begin{cases}\angle EOA=\angle FOC\\\angle EAO=\angle FCO\\AE = CF\end{cases}$，$\therefore\triangle AEO\cong\triangle CFO(AAS)$，$\therefore EO = FO$，$AO = CO=\frac{5}{2}$，$\therefore O$为$EF$、$AC$的中点.$\because\angle EGF = 90^{\circ}$，$\therefore OG=\frac{1}{2}EF=\frac{3}{2}$，$\therefore AG = OA - OG = 1$或$AG = OA+OG = 4$，$\therefore AG$的长为$1$或$4$.

12. 已知平面上四点A(0,0)、B(4,0)、C(4,2)、D(0,2)，直线y = mx - m + 2将四边形ABCD分成面积相等的两部分，则m的值为__________.

-1 解析：$\because$点$A(0,0)$、$B(4,0)$、$C(4,2)$、$D(0,2)$，$\therefore$四边形$ABCD$为矩形.$\because$直线$y = mx - m + 2$将四边形$ABCD$分成面积相等的两部分，$\therefore$直线$y = mx - m + 2$过矩形$ABCD$的对角线的交点. 而矩形$ABCD$的对角线的交点坐标为$(2,1)$，$\therefore 2m - m + 2 = 1$，$\therefore m = - 1$.

13. (2024·南昌模拟)【问题发现】(1)如图①，矩形ABCD中，AB = 9，BC = 12，点P是矩形ABCD内一点，过点P作EF⊥AD，分别交AD、BC于点E、F，PE = 4，AE = 3. 则：
①PA = __________，PB = __________，PC = __________，PD = __________；
②PA² + PC²与PB² + PD²的关系是__________.
【类比探究】(2)如图②，点P是矩形ABCD外一点，过点P作EF⊥AD，分别交AD、BC的反向延长线于点E、F，②中结论还成立吗?若成立，请说明理由.
【拓展延伸】(3)如图③，在Rt△ABC中，∠BAC = 90°，P是Rt△ABC外一点，PA = 1，PB = 3，PC = $\sqrt{3}$，则BC的最小值为__________.
　　　

（1）①$5$ $\sqrt{34}$ $\sqrt{106}$ $\sqrt{97}$ 解析：如题图①.$\because$四边形$ABCD$是矩形，$AB = 9$，$BC = 12$，$\therefore AD = BC = 12$，$\angle BAD=\angle ABC=\angle ADC=\angle BCD = 90^{\circ}$.$\because$过点$P$作$EF\perp AD$，分别交$AD$、$BC$于点$E$、$F$，$\therefore\angle AEF=\angle DEF = 90^{\circ}$，$\therefore$四边形$ABFE$和四边形$DCFE$都是矩形，$\therefore EF = AB = 9$，$BF = AE = 3$，$\therefore CF = DE = AD - AE = 12 - 3 = 9$.$\because PE = 4$，$\therefore PA=\sqrt{AE^{2}+PE^{2}}=\sqrt{3^{2}+4^{2}} = 5$，$PD=\sqrt{DE^{2}+PE^{2}}=\sqrt{9^{2}+4^{2}}=\sqrt{97}$.$\because PF = EF - PE = 9 - 4 = 5$，$\therefore PB=\sqrt{BF^{2}+PF^{2}}=\sqrt{3^{2}+5^{2}}=\sqrt{34}$，$PC=\sqrt{CF^{2}+PF^{2}}=\sqrt{9^{2}+5^{2}}=\sqrt{106}$.
②$PA^{2}+PC^{2}=PB^{2}+PD^{2}$
（2）成立，理由：如题图②.$\because$四边形$ABCD$是矩形，$\therefore\angle BAD=\angle ABC=\angle ADC=\angle BCD = 90^{\circ}$，$\therefore\angle EAB=\angle FBA = 90^{\circ}$.$\because$过点$P$作$EF\perp AD$，分别交$AD$、$BC$的反向延长线于点$E$、$F$，$\therefore\angle E = 90^{\circ}$，$\therefore$四边形$ABFE$和四边形$DCFE$都是矩形，$\therefore AE = BF$，$DE = CF$.$\because PD^{2}=DE^{2}+PE^{2}=CF^{2}+PE^{2}$，$PA^{2}=AE^{2}+PE^{2}=BF^{2}+PE^{2}$，$\therefore PD^{2}-PA^{2}=CF^{2}-BF^{2}$.$\because PC^{2}=CF^{2}+PF^{2}$，$PB^{2}=BF^{2}+PF^{2}$，$\therefore PC^{2}-PB^{2}=CF^{2}-BF^{2}$，$\therefore PC^{2}-PB^{2}=PD^{2}-PA^{2}$，$\therefore PA^{2}+PC^{2}=PB^{2}+PD^{2}$.
（3）$\sqrt{11}-1$ 解析：如图，作$PM\perp CA$交$CA$的延长线于点$M$，则$\angle PMC = 90^{\circ}$，$\therefore PC^{2}=PM^{2}+CM^{2}$，$PA^{2}=PM^{2}+AM^{2}$，作$BN\perp PM$交$PM$的延长线于点$N$，作$CT\perp NB$交$NB$的延长线于点$T$，连接$AT$、$PT$.$\because\angle BAC = 90^{\circ}$，$\therefore\angle BAM = 90^{\circ}$.$\because\angle AMN=\angle N=\angle CTN = 90^{\circ}$，$\therefore$四边形$ABNM$和四边形$CTNM$都是矩形，$\therefore TN = CM$，$BN = AM$，$\therefore PT^{2}=PN^{2}+TN^{2}=PN^{2}+CM^{2}$，$PB^{2}=PN^{2}+BN^{2}=PN^{2}+AM^{2}$，$\therefore PT^{2}-PC^{2}=PN^{2}-PM^{2}$，$PB^{2}-PA^{2}=PN^{2}-PM^{2}$，$\therefore PT^{2}-PC^{2}=PB^{2}-PA^{2}$.$\because PA = 1$，$PB = 3$，$PC=\sqrt{3}$，$\therefore PT=\sqrt{PB^{2}+PC^{2}-PA^{2}}=\sqrt{3^{2}+(\sqrt{3})^{2}-1^{2}}=\sqrt{11}$.$\because TA + PA\geqslant PT$，$\therefore TA + 1\geqslant\sqrt{11}$，$\therefore TA\geqslant\sqrt{11}-1$.$\because AB// MN$，$\therefore\angle ABT=\angle N = 90^{\circ}$，$\therefore$四边形$ABTC$是矩形，$\therefore TA = BC$，$\therefore BC\geqslant\sqrt{11}-1$，$\therefore BC$的最小值为$\sqrt{11}-1$.