13.「2026辽宁大连期中，★★★」综合与实践
综合与实践课上，老师让同学们以“正方形的折叠”为主题开展数学活动.
【操作判断】
操作一:如图1,在正方形纸片ABCD的AD边上选一点E(点E不与点A,D重合),将正方形ABCD沿BE折叠,使点A落在正方形内部的点G处,得到折痕BE,延长EG交CD于点F,连接BF.
(1)求∠EBF的度数.
【深入探究】
操作二:如图2,在操作一的基础上,将△EDF沿着EF折叠,使点D落在正方形ABCD的内部,点D的对应点为点H,在折叠的过程中,同学们发现,随着点E的位置改变,点H的位置也随之改变,当点E在AD边上的某一位置时,点H恰好落在BF上,此时EH与BG交于点M.
(2)求∠AEB的度数.
(3)求证:BM=EF.
【拓展延伸】
操作三:如图3,在操作一的基础上,连接AC交BE于点P,连接PF,调整点E的位置,使∠BPF=90°.
(4)求证:BC+CF=√2 PC.

13.解析 (1)$\because$ 四边形 $ABCD$ 是正方形, $\therefore AB = BC, ∠ A = ∠ C = 90°, \because △ ABE$ 沿 $BE$ 折叠得到 $△ GBE, \therefore ∠ ABE = ∠ GBE, BG = AB, ∠ BGE = ∠ A = 90°, \therefore BG = BC, ∠ BGF = ∠ C = 90°,$ 在 $\mathrm{Rt}△ BGF$ 与 $\mathrm{Rt}△ BCF$ 中, $\begin{cases}BF = BF,\\BG = BC,\end{cases}$
$\therefore \mathrm{Rt}△ BGF ≌ \mathrm{Rt}△ BCF(\mathrm{HL}), \therefore ∠ GBF = ∠ CBF, \because ∠ ABE + ∠ EBG + ∠ GBF + ∠ CBF = ∠ ABC = 90°, \therefore 2∠ GBE + 2∠ GBF = 90°, \therefore ∠ GBE + ∠ GBF = 45°,$ 即 $∠ EBF = 45°.$
(2)$\because △ EDF$ 沿 $EF$ 折叠得到 $△ EHF$, 点 $H$ 在 $BF$ 上,
$\therefore ∠ EFD = ∠ EFB.$ 由 (1) 得 $△ BGF ≌ △ BCF, \therefore ∠ EFB = ∠ CFB, \therefore ∠ EFD = ∠ EFB = ∠ CFB, \because ∠ EFD + ∠ EFB + ∠ CFB = 180°, \therefore ∠ EFD = 60°, \because ∠ D = 90°, \therefore ∠ DEF = 90° - 60° = 30°, \therefore ∠ AEF = 180° - ∠ DEF = 180° - 30° = 150°,$
$\because △ ABE$ 沿 $BE$ 折叠得到 $△ GBE, \therefore ∠ AEB = ∠ GEB,$
$\therefore ∠ AEB = \dfrac{1}{2}∠ AEF = \dfrac{1}{2} × 150° = 75°.$
(3)证明:由(2)得 $∠ DEF = 30°, ∠ AEB = 75°, \therefore ∠ ABE = 90° - 75° = 15°, \therefore ∠ EBG = ∠ ABE = 15°, \therefore ∠ MBH = ∠ EBH - ∠ EBG = 45° - 15° = 30°, \therefore ∠ MBH = ∠ DEF, \because ∠ EBH = 45°, ∠ EHB = 90°, \therefore ∠ BEH = ∠ EBH = 45°, \therefore BH = EH, \because △ EDF$ 沿 $EF$ 折叠得到 $△ EHF, \therefore ED = EH, \therefore BH = ED,$ 又 $\because ∠ BHM = ∠ D = 90°, \therefore △ BMH ≌ △ EFD \ (\mathrm{ASA}),$
$\therefore BM = EF.$
(4)证明:如图,延长 $CB$ 至 $Q$, 使 $BQ = CF$, 连接 $PQ.$
$\because ∠ EBF = 45°, ∠ BPF = 90°, \therefore PB = PF, \because ∠ BPF = ∠ BCF = 90°, \therefore ∠ PBC + ∠ PFC = 180°, \because ∠ PBC + ∠ PBQ = 180°, \therefore ∠ PBQ = ∠ PFC, \because BQ = CF, \therefore △ PBQ ≌ △ PFC(\mathrm{SAS}), \therefore ∠ Q = ∠ PCF, \because$ 四边形 $ABCD$ 是正方形,
$\therefore ∠ ACF = ∠ ACB = 45°, \therefore ∠ Q = ∠ ACB = 45°, \therefore ∠ QPC = 90°, PQ = PC, \therefore QC = \sqrt{2}PC, \because QC = BQ + BC = CF + BC,$
$\therefore BC + CF = \sqrt{2}PC.$