答案：
26. 问题情境：线段$DN$，$MB$，$EC$之间的数量关系为$DN + MB = EC$.理由如下：因为四边形$ABCD$是正方形，所以$\angle ABE = \angle BCD = 90^{\circ}$，$AB = BC = CD$，$AB // CD$.如图①，过点$B$作$BR // MN$分别交$AE$，$CD$于$L$，$R$两点，则$\angle ABE = \angle BCR = 90^{\circ}$，四边形$MBRN$为平行四边形.所以$NR = MB$.因为$MN ⊥ AE$，所以$BR ⊥ AE$.所以$\angle BLE = 90^{\circ}$.所以$\angle CBR + \angle AEB = 90^{\circ}$.因为$\angle BAE + \angle AEB = 90^{\circ}$，所以$\angle CBR = \angle BAE$.所以$\triangle ABE \cong \triangle BCR$（ASA）.所以$BE = CR$.所以$DN + NR + CR = BE + EC$，即$DN + NR = EC$.所以$DN + MB = EC$.

问题探究：
（1）连接$AQ$，过点$Q$作$HI // AB$，分别交$AD$，$BC$于$H$，$I$两点，如图②.因为四边形$ABCD$是正方形，所以四边形$ABIH$为矩形.所以$HI ⊥ AD$，$HI ⊥ BC$，$HI = AB = AD$.所以$\angle AHQ = \angle DHQ = \angle QIE = 90^{\circ}$，即$\angle QEI + \angle EQI = 90^{\circ}$，$\angle HQD + \angle BDA = 90^{\circ}$.因为$BD$是正方形$ABCD$的对角线，所以$\angle BDA = 45^{\circ}$.所以$\angle HQD = 90^{\circ} - \angle BDA = 45^{\circ}$，即$\angle HQD = \angle BDA$.所以$HQ = HD$.所以$AD - HD = HI - HQ$，即$AH = QI$.因为$MN ⊥ AE$，$P$为$AE$的中点，所以$MN$是$AE$的垂直平分线，即$AQ = QE$.所以$Rt \triangle AHQ \cong Rt \triangle QIE$（HL）.所以$\angle AQH = \angle QEI$.所以$\angle AQH + \angle EQI = \angle QEI + \angle EQI = 90^{\circ}$.又$\angle AQH + \angle AQE + \angle EQI = 180^{\circ}$，所以$\angle AQE = 90^{\circ}$.所以$\triangle AQE$是等腰直角三角形.所以$\angle EAQ = \angle AEQ = 45^{\circ}$，即$\angle AEF = 45^{\circ}$.
（2）如图③，过点$P$作$GK // AD$，交$AB$，$CD$于$G$，$K$两点，取$AB$的中点为$S'$，连接$PS'$.因为四边形$ABCD$是边长为$4$的正方形，所以$AB = AD = 4$，$AD // BC$，$\angle BAD = 90^{\circ}$，$\angle KDP = \angle ABD = 45^{\circ}$.易得四边形$AGKD$是矩形，所以$AG = DK$，$\angle AGP = \angle PKN = 90^{\circ}$.又$\angle DPK + \angle KDP = 90^{\circ}$，所以$\angle DPK = 90^{\circ} - \angle KDP = 45^{\circ}$，即$PK = DK$.所以$AG = PK$.因为$MN ⊥ AE$，所以$\angle APN = 90^{\circ}$.又$\angle APN + \angle APG + \angle KPN = 180^{\circ}$，所以$\angle APG + \angle KPN = 90^{\circ}$.因为$\angle KPN + \angle PNK = 90^{\circ}$，所以$\angle APG = \angle PNK$.所以$\triangle AGP \cong \triangle PKN$（AAS）.所以$AP = PN$，即$\triangle APN$是等腰直角三角形.所以$\angle PAN = 45^{\circ}$.由翻折的性质，得$\angle P'AN = \angle PAN = 45^{\circ}$，$AP' = AP$，所以$\angle PAP' = 90^{\circ}$.因为$\angle S'AP + \angle PAD = \angle BAD = 90^{\circ}$，$\angle SAP' + \angle PAD = \angle PAP' = 90^{\circ}$，所以$\angle S'AP = \angle SAP'$.又$AB = AD$，所以$\frac{1}{2}AB = \frac{1}{2}AD$，即$AS' = AS$.所以$\triangle AS'P \cong \triangle ASP$（SAS）.所以$PS' = P'S$.因为$S'$为定点，所以当$PS' ⊥ BD$时，$PS'$的长取最小值，即$P'S$的长取最小值.当$PS' ⊥ BD$时，$\angle BS'P + \angle ABD = 90^{\circ}$，所以$\angle BS'P = \angle ABD = 45^{\circ}$，即$PS' = BP$.在$Rt \triangle S'PB$中，$BS' = \frac{1}{2}AB = 2$，由勾股定理，得$PS'^2 + BP^2 = BS'^2$，即$2PS'^2 = 4$，解得$PS' = \sqrt{2}$（负值已舍去）.则$P'S = \sqrt{2}$.所以$P'S$的长的最小值为$\sqrt{2}$.