自主拓展
如图（1），以$□ ABCD$的较短边$CD$为一边作菱形$CDEF$，使点$F$落在边$AD$上，连接$BE$，交$AF$于点$G$.
（1）猜想$BG$与$EG$的数量关系，并说明理由；
（2）延长$DE$，$BA$交于点$H$，其他条件不变.
① 如图（2），若$∠ ADC=60^{\circ}$，求$\dfrac{DG}{BH}$的值；
② 如图（3），若$∠ ADC=α(0^{\circ}＜α＜90^{\circ})$，直接写出$\dfrac{DG}{BH}$的值.（用含$α$的三角函数表示）
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答案：(1) $BG = EG$，理由如下：$\because$四边形$ABCD$是平行四边形，$\therefore AB// CD$，$AB = CD$. $\because$四边形$CDEF$是菱形，$\therefore CD// EF$，$CD = EF$. $\therefore AB// EF$，$AB = EF$. $\therefore∠ ABG=∠ FEG$. 又$\because∠ AGB=∠ FGE$，$\therefore△ ABG≌△ FEG$（AAS）. $\therefore BG = EG$.
(2) ① 方法1：如图(1)，过点$G$作$GM// BH$，交$DH$于点$M$，
$\therefore∠ EMG=∠ EHA$. $\because∠ GEM=∠ BEH$，$\therefore△ GME∼△ BHE$. $\therefore\frac{GM}{BH}=\frac{GE}{BE}$. 由(1)知$BG = EG$. $\therefore EG=\frac{1}{2}BE$. $\therefore\frac{GM}{BH}=\frac{GE}{BE}=\frac{1}{2}$. $\because$四边形$CDEF$为菱形，$\therefore∠ ADC=∠ EDF = 60^{\circ}$. $\because$四边形$ABCD$是平行四边形，$\therefore AB// CD$. $\therefore∠ CDF=∠ HAD = 60^{\circ}$. $\because GM// AH$，$\therefore∠ MGD=∠ HAD = 60^{\circ}$. $\therefore∠ GMD = 180^{\circ}-∠ MGD-∠ MDG = 60^{\circ}$，即$∠ GMD=∠ MGD=∠ MDG = 60^{\circ}$. $\therefore△ MGD$是等边三角形. $\therefore DG = MG$. $\therefore\frac{DG}{BH}=\frac{MG}{BH}=\frac{1}{2}$.
方法2：如图(2)，延长$ED$，$BC$交于点$M$，
$\because$四边形$CDEF$为菱形，$\therefore∠ EDF=∠ CDF = 60^{\circ}$. $\because$四边形$ABCD$为平行四边形，$\therefore∠ ABC=∠ ADC = 60^{\circ}$，$AD// BC$. $\therefore∠ EDF=∠ M = 60^{\circ}$，$∠ H = 180^{\circ}-∠ HBM-∠ M = 180^{\circ}-60^{\circ}=60^{\circ}$，即$∠ HBM=∠ M=∠ H = 60^{\circ}$. $\therefore△ HBM$为等边三角形. $\therefore HB = MB$. $\because AD// BC$，$\therefore∠ EGD=∠ EBM$，$∠ EDG=∠ M$. $\therefore△ EDG∼△ EMB$，$\therefore\frac{DG}{MB}=\frac{EG}{EB}$. 由(1)知$BG = EG$. $\therefore EG=\frac{1}{2}BE$. $\therefore\frac{DG}{MB}=\frac{GE}{BE}=\frac{1}{2}$. $\because HB = MB$，$\therefore\frac{DG}{BH}=\frac{DG}{MB}=\frac{1}{2}$.
② $\cosα$. 如图(3)，连接$EC$交$DF$于$O$，
$\because$四边形$CFED$是菱形，$\therefore EC⊥ AD$，$FD = 2FO$. 设$FG = a$，$AB = b$，则$AG = a$，$EF = ED = CD = b$. $Rt△ EFO$中，$\cosα=\frac{OF}{EF}$，$\therefore OF = b\cosα$，$\therefore DG = a + 2b\cosα$. 过$H$作$HM⊥ AD$于$M$，$\because∠ ADC=∠ HAD=∠ ADH=α$，$\therefore AH = HD$，$\therefore AM=\frac{1}{2}AD=\frac{1}{2}(2a + 2b\cosα)=a + b\cosα$，$Rt△ AHM$中，$\cosα=\frac{AM}{AH}$，$\therefore AH=\frac{a + b\cosα}{\cosα}$，$\therefore\frac{DG}{BH}=\frac{a + 2b\cosα}{b+\frac{a + b\cosα}{\cosα}}=\cosα$.
