11. 如图,在$△ ABC$中,已知$∠ BAC=45°$,$AD⊥ BC$于点$D$.小明同学灵活运用轴对称知识将图形进行翻折变换:分别以直线$AB,AC$为对称轴,画出$△ ABD$,$△ ACD$的轴对称图形,$D$的对称点分别为$E,F$,延长$EB,FC$相交于点$G$.请按照小明同学的思路,探究并解答下列问题:
(1)求证:四边形$AEGF$是正方形；
(2)若$AD=6$,$BD=2$,则$DC=$.

11. 解：(1)根据题意得,$△ ABD≌△ ABE,△ ACD≌△ ACF$,
$\therefore AD=AE,∠ DAB=∠ EAB,AD=AF,∠ DAC=∠ FAC$,
$\because ∠ BAC=45°$,
$\therefore ∠ EAF=∠ DAB+∠ DAC+∠ EAB+∠ FAC=∠ BAC+∠ BAC=90°$.
$\because AD\bot BC$,
$\therefore ∠ ADB=∠ ADC=90°$,
$\therefore ∠ E=∠ ADB=90°,∠ F=∠ ADC=90°$,
$\therefore$四边形$AEGF$是矩形.
$\because AD=AE,AD=AF$,
$\therefore AE=AF$,
$\therefore$矩形$AEGF$是正方形.
(2)3

12. 在正方形$ABCD$中,点$E$为射线$AC$上一点,连接$DE$,过点$E$作$EF⊥ DE$交射线$BC$于点$F$,以$DE,EF$为邻边作矩形$DEFG$,连接$CG$.
【发现问题】(1)如图①,当点$E$在线段$AC$上时.
①求证:四边形$DEFG$是正方形；
②猜想$CG$与$AE$之间的数量关系:.

12. 解：(1)①证明：过点$E$作$EP\bot CD$于$P,EQ\bot BC$于$Q$,如图①.
$\therefore ∠ EPD=∠ EQC=90°$,
$\because$四边形$ABCD$是正方形,
$\therefore ∠ DCA=∠ BCA=45°$,
$\therefore EQ=EP,∠ QEC=∠ PEC=45°$,
$\therefore ∠ PEQ=90°$,
$\therefore ∠ QEF+∠ PEF=90°$.
又四边形$DEFG$是矩形,
$\therefore ∠ DEF=90°$,
$\therefore ∠ DEP+∠ PEF=90°$,
$\therefore ∠ DEP=∠ QEF$,
$\therefore △ EQF≌△ EPD(\mathrm{ASA})$,
$\therefore EF=ED$,
$\therefore$矩形$DEFG$是正方形.
②$AE=CG$.
(2)过点$E$作$EP\bot CD$于$P,EQ\bot BC$于$Q$,如图②.
由(1)知,矩形$DEFG$是正方形,
$\therefore DE=DG,∠ EDG=90°$,
$\because ∠ ADE+∠ EDC=90°,∠ CDG+∠ EDC=90°$,
$\therefore ∠ ADE=∠ CDG$,
$\therefore △ ADE≌△ CDG(\mathrm{SAS})$,
$\therefore ∠ DCG=∠ DAC=45°$.
(3)过点$E$作$EP\bot CD$于点$P,EQ\bot BC$于点$Q$,如图③.
由(1)知,四边形$DEFG$是正方形,
$\therefore DE=DG,∠ EDG=90°$,
$\therefore ∠ ADC+∠ CDE=∠ EDG+∠ CDE$,
即$∠ ADE=∠ CDG$.
$\therefore △ ADE≌△ CDG(\mathrm{SAS})$,
$\therefore AE=CG,∠ DCG=∠ DAC=45°$,
$\therefore ∠ ACG=90°$,
即$∠ GCE=90°$,
在$\mathrm{Rt}△ ADC$中,$AD=DC=4$,
$\therefore AC=\sqrt{AD^{2}+DC^{2}}=\sqrt{4^{2}+4^{2}}=4\sqrt{2}$,
$\therefore AG=AC+CE=4\sqrt{2}+\sqrt{2}=5\sqrt{2}$,
$\therefore CG=5\sqrt{2}$,
在$\mathrm{Rt}△ GCE$中,
$\therefore EG=\sqrt{CE^{2}+CG^{2}}=\sqrt{(\sqrt{2})^{2}+(5\sqrt{2})^{2}}=2\sqrt{13}$.

【拓展运用】(3)如图③,当点$E$在线段$AC$的延长线上时,若正方形$ABCD$的边长为4,$CE=\sqrt{2}$,求$GE$的长.

12. 解：(3)过点$E$作$EP\bot CD$于点$P,EQ\bot BC$于点$Q$,如图③.
由(1)知,四边形$DEFG$是正方形,
$\therefore DE=DG,∠ EDG=90°$,
$\therefore ∠ ADC+∠ CDE=∠ EDG+∠ CDE$,
即$∠ ADE=∠ CDG$.
$\therefore △ ADE≌△ CDG(\mathrm{SAS})$,
$\therefore AE=CG,∠ DCG=∠ DAC=45°$,
$\therefore ∠ ACG=90°$,
即$∠ GCE=90°$,
在$\mathrm{Rt}△ ADC$中,$AD=DC=4$,
$\therefore AC=\sqrt{AD^{2}+DC^{2}}=\sqrt{4^{2}+4^{2}}=4\sqrt{2}$,
$\therefore AG=AC+CE=4\sqrt{2}+\sqrt{2}=5\sqrt{2}$,
$\therefore CG=5\sqrt{2}$,
在$\mathrm{Rt}△ GCE$中,
$\therefore EG=\sqrt{CE^{2}+CG^{2}}=\sqrt{(\sqrt{2})^{2}+(5\sqrt{2})^{2}}=2\sqrt{13}$.