3. ▶一题多解 如图1,在四边形ABCD中,∠ABC=∠BCD,点E在边BC上,且AE//CD,DE//AB.作CF//AD交线段AE于点F,连接BF.
(1)求证:△ABF≌△EAD.
(2)如图2,若AB=9,CD=5,∠ECF=∠AED,求BE的长.
(3)如图3,若BF的延长线经过AD的中点M,求$\frac{BE}{EC}$的值.

解题思路:(1)先证明四边形AFCD是平行四边形,得到$AF=CD$,再利用平行线的性质及“等角对等边”推出$AB = AE,DE = CD = AF$,进而利用“SAS”判定三角形全等即可.
(2)证明$△ ABE ∽ △ BFE$,由相似三角形对应边成比例得到$BE,AE,EF$之间的数量关系,进而求出$BE$的长. (3)证明$△ ABE ∽ △ DEC$,将问题转化为求等腰三角形$ABE$和等腰三角形$DEC$的腰长之比. 过点$M$作$DE$的平行线构造相似三角形(或延长$BM,ED$构造全等三角形和相似三角形),建立比例式求解.
(1)证明:因为 $AE// CD,AD// CF,$
所以四边形$AFCD$是平行四边形,所以 $AF = CD.$
因为 $AE// CD,DE// AB,∠ABC = ∠BCD,$
所以 $∠DEC = ∠ABC = ∠BCD = ∠AEB,∠BAF = ∠AED,$
所以 $AB = AE,DE = CD = AF$,所以 $△ ABF ≌ △ EAD.$
(2)解:由(1)知 $△ ABF ≌ △ EAD,$
所以 $BF = AD.$
因为四边形$AFCD$为平行四边形,
所以 $FC = AD$,所以 $FC = FB,$
所以 $∠FBE = ∠ECF = ∠AED = ∠BAE.$
又因为 $∠AEB = ∠BEF,$
所以 $△ ABE ∽ △ BFE$,所以 $\frac{AE}{BE} = \frac{BE}{EF}$,所以 $BE^2 = AE · EF.$
又因为 $AE = AB = 9,EF = AE - AF = AE - CD = 9 - 5 = 4,$
所以 $BE = \sqrt{9 × 4} = 6.$
(3)解:由(1)知 $AB = AE,DE = CD = AF$,易证 $△ ABE ∽ △ DEC$,所以 $\frac{BE}{EC} = \frac{AE}{CD} = \frac{AB}{DE}.$
通解 如图1,过点$M$作 $MN// DE$,交$AE$于点$N$,则 $AN = \frac{1}{2}AE,MN = \frac{1}{2}DE = \frac{1}{2}CD.$
因为 $AB// DE// MN$,所以 $△ ABF ∽ △ NMF,$
所以 $\frac{AF}{FN} = \frac{AB}{MN} = \frac{2AB}{CD}$,即 $\frac{AF}{AN - AF} = \frac{2AB}{CD}.$ ①
设 $AF = a,EF = b$,则 $CD = AF = a,AB = AE = a + b,$
所以 $AN = \frac{1}{2}AE = \frac{a + b}{2},$
所以①式为$\frac{a}{\frac{a + b}{2} - a} = \frac{2(a + b)}{a},$
整理得 $b^2 = 2a^2$,即 $b = \sqrt{2}a,$
所以 $\frac{BE}{EC} = \frac{AB}{CD} = \frac{a + b}{a} = \sqrt{2} + 1.$

另解 如图2,延长$BM$,交$ED$的延长线于点$N$,由$AB// DE$,得$∠ABM = ∠N.$
又因为 $∠AMB = ∠DMN,AM = DM,$
所以 $△ ABM ≌ △ DNM$,所以 $AB = DN.$
因为 $AB// NE$,所以 $△ ABF ∽ △ ENF,$
所以 $\frac{AB}{NE} = \frac{AF}{FE}$,即 $\frac{AB}{AB + CD} = \frac{CD}{AB - CD}.$ ②
设 $AB = m,CD = 1,$
则②式即为$\frac{m}{m + 1} = \frac{1}{m - 1}$,整理得 $m^2 - 2m - 1 = 0,$
解得 $m = \sqrt{2} + 1$(负值已舍去),所以 $\frac{BE}{EC} = \frac{AB}{CD} = \sqrt{2} + 1.$

4. [2024湖北中考]在矩形ABCD中，点E，F分别在边AD，BC上，将矩形ABCD沿EF折叠，使点A的对应点P落在边CD上，点B的对应点为点G，PG交BC于点H.
(1)如图1，求证：$△ DEP ∽ △ CPH$.
(2)如图2，当P为CD的中点，$AB=2$，$AD=3$时，求GH的长.
(3)如图3，连接BG，当P，H分别为CD，BC的中点时，探究BG与AB的数量关系，并说明理由.

(1)证明:
∵ 四边形ABCD是矩形，
∴ $∠A = ∠D = ∠C = 90°,$
∴ $∠DEP + ∠DPE = 90°.$
由折叠的性质,得 $∠EPH = ∠A = 90°,$
∴ $∠DPE + ∠HPC = 90°,∴ ∠DEP = ∠HPC,$
∴ $△ DEP ∽ △ CPH$(提示:一线三直角模型).
(2)解:
∵ 四边形ABCD是矩形，
∴ $CD = AB = 2,∠A = ∠D = ∠C = 90°.$
∵ $P$为$CD$的中点,$∴ DP = CP = \frac{1}{2} × 2 = 1.$
由折叠的性质,得 $EP = AE,GP = AB = 2.$
设 $EP = AE = x$,则 $ED = AD - AE = 3 - x.$
在$\mathrm{Rt}△ EDP$中,$EP^2 = ED^2 + DP^2,$
即 $x^2 = (3 - x)^2 + 1$,解得 $x = \frac{5}{3},$
∴ $EP = AE = \frac{5}{3},ED = \frac{4}{3}.$
由(1)知 $△ DEP ∽ △ CPH,$
∴ $\frac{ED}{PC} = \frac{EP}{PH}$,即 $\frac{\frac{4}{3}}{1} = \frac{\frac{5}{3}}{PH},$
∴ $PH = \frac{5}{4},∴ GH = PG - PH = \frac{3}{4}.$
(3)解:$\frac{AB}{BG} = \sqrt{6}.$ 理由如下：
通解 如图1,延长$AB,PG$交于点$M$,连接$AP.$

∵ $AB// CD,∴ ∠M = ∠CPH,∠MBH = ∠C.$
∵ $H$是$BC$的中点,$∴ BH = CH,$
∴ $△ MBH ≌ △ PCH,∴ PC = BM,PH = MH.$
∵ $AE = EP,∴ ∠EAP = ∠EPA.$
又
∵ $∠EPH = ∠EAB = 90°,∴ ∠MAP = ∠MPA,∴ MA = MP.$
设 $AB = DC = 2a(a ＞ 0),BC = AD = 2b(b ＞ 0),$
则 $CH = b,BM = CP = DP = a,$
∴ $AM = 3a,PH = \sqrt{CH^2 + CP^2} = \sqrt{b^2 + a^2},$
$AP = \sqrt{AD^2 + DP^2} = \sqrt{(2b)^2 + a^2},$
∴ $MP = 2\sqrt{b^2 + a^2},∴ 3a = 2\sqrt{b^2 + a^2}$,化简,得 $5a^2 = 4b^2,$
∴ $AP = \sqrt{6}a.$
由折叠的性质,得$EF$垂直平分$AP$,$EF$垂直平分$BG$,
∴ $BG// AP,∴ △ MBG ∽ △ MAP,$
∴ $\frac{MB}{MA} = \frac{BG}{AP}$,即 $\frac{a}{3a} = \frac{BG}{\sqrt{6}a},∴ BG = \frac{\sqrt{6}a}{3},∴ \frac{AB}{BG} = \sqrt{6}.$
另解 设 $AB = DC = 2a(a ＞ 0),BC = AD = 2b(b ＞ 0)$,则 $CH = b,CP = DP = a.$
由(1)知 $△ DEP ∽ △ CPH,∴ \frac{ED}{PC} = \frac{DP}{CH}$,即 $\frac{ED}{a} = \frac{a}{b},$
∴ $ED = \frac{a^2}{b},∴ EP = AE = 2b - \frac{a^2}{b}.$
在$\mathrm{Rt}△ DEP$中,$EP^2 = ED^2 + DP^2$,即$(2b - \frac{a^2}{b})^2 = (\frac{a^2}{b})^2 + a^2,$
化简,得 $5a^2 = 4b^2,∴ EP = AE = \frac{6}{5}b,a = \frac{2\sqrt{5}}{5}b.$
如图2,连接$AP$,过点$B$作 $BN// EF$,交$AD$于点$N$,则四边形$BNEF$是平行四边形,$∴ EN = BF.$
∵ $EF ⊥ AP,∴ BN ⊥ AP.$
易证$△ BAN ∽ △ ADP$(提示:矩形中的十字模型),
∴ $\frac{AN}{DP} = \frac{AB}{AD}$,即 $\frac{AN}{a} = \frac{2a}{2b},∴ AN = \frac{a^2}{b} = \frac{4}{5}b,$
∴ $BF = NE = AE - AN = \frac{2}{5}b,BN = \sqrt{AB^2 + AN^2} = \frac{4\sqrt{6}}{5}b.$
延长$EF$交$BG$于点$Q$,则 $EQ ⊥ BG,$
∴ $∠BQF = 90° = ∠BAN,BQ = QG.$
同上可得 $BG// AP,∴ BG ⊥ BN,∴ ∠GBN = 90° = ∠ABC,$
∴ $∠QBF = ∠ABN,∴ △ BFQ ∽ △ BNA,$
∴ $\frac{BA}{BQ} = \frac{BN}{BF} = \frac{\frac{4\sqrt{6}}{5}b}{\frac{2}{5}b} = 2\sqrt{6},∴ \frac{BA}{BG} = \sqrt{6}.$