1.(乌鲁木齐模拟)如图,四边形ABCD和四边形AEFG均为正方形,连接CF,DG,求$\frac{DG}{CF}$的值.

1. 解:如图,连接 AC 和 AF,则$\frac{AD}{AC}=\frac{AG}{AF}=\frac{\sqrt{2}}{2}$.
$\because ∠ DAG=45°-∠ GAC,∠ CAF=45°-∠ GAC$,
$\therefore ∠ DAG=∠ CAF,\therefore △ DAG ∽ △ CAF,\therefore \frac{DG}{CF}=\frac{AD}{AC}=\frac{\sqrt{2}}{2}$.

2.(2024·新疆中考)如图,在$\odot O$中,AB是$\odot O$的直径,弦CD交AB于点E,$\overset{\frown}{AD}=\overset{\frown}{BD}$.
(1)求证:$△ ACD ∽ △ ECB$.
(2)若$AC=3,BC=1$,求CE的长.

2. (1) 证明: $\because \overset{\frown}{AD}=\overset{\frown}{BD},\therefore ∠ ACD=∠ BCE$.
$\because ∠ ADC=∠ EBC,\therefore △ ACD ∽ △ ECB$.
(2) 解:如图,过点 B 作$BH ⊥ CD$于点 H.$\because AB$是$\odot O$的直径,$\therefore ∠ ACB=∠ ADB=90°,\therefore ∠ ACD=∠ BCD=45°$. 在$\mathrm{Rt }△ ACB$ 中, $AB=\sqrt{BC^2+AC^2}=\sqrt{1^2+3^2}=\sqrt{10}$. $\because ∠ ACD=∠ BCD=45°,\therefore ∠ ABD=∠ BAD=45°,\therefore △ ABD$为等腰直角三角形,$\therefore BD=\frac{\sqrt{2}}{2}AB=\frac{\sqrt{2}}{2}×\sqrt{10}=\sqrt{5}$. 在$\mathrm{Rt}△ BCH$中,$\because ∠ BCH=45°,\therefore CH=BH=\frac{\sqrt{2}}{2}BC=\frac{\sqrt{2}}{2}$. 在 $\mathrm{Rt }△ BDH$ 中, $DH=\sqrt{BD^2-BH^2}=\sqrt{(\sqrt{5})^2-(\frac{\sqrt{2}}{2})^2}=\frac{3\sqrt{2}}{2},\therefore CD=CH+DH=\frac{\sqrt{2}}{2}+\frac{3\sqrt{2}}{2}=2\sqrt{2}$. 由(1)知$△ ACD ∽ △ ECB,\therefore CA:CE=CD:CB$,即 $3:CE=2\sqrt{2}:1$,解得$CE=\frac{3\sqrt{2}}{4}$,即 CE 的长为$\frac{3\sqrt{2}}{4}$.

1.(和田模拟)综合与实践
【实践操作】如图①,在$△ ABC$中,$AB=AC$,$∠ BAC=α$.点$E$是$△ ABC$外一点,连接$BE$,将线段$EB$绕点$E$逆时针旋转,旋转角为$α$,得到线段$ED$,连接$AE$,$BD$,$CD$.

【探究发现】(1)求证:$△ ABE ∽ △ CBD$.
【性质应用】(2)如图②,点$E$为正方形$ABCD$内一点,连接$BE$,将线段$EB$绕点$E$逆时针旋转$90°$,得到线段$EF$,连接$BF$,$AE$,$DF$,求出$AE$与$DF$之间的数量关系.

1.(1) 证明:$\because AB=AC,BE=DE,∠ BAC=∠ BED=α,\therefore ∠ ABC=∠ EBD=\frac{1}{2}(180°-α),△ BAC∽△ BED,\therefore \frac{BE}{BA}=\frac{BD}{BC},\therefore \frac{BE}{BD}=\frac{BA}{BC}$. 又$\because ∠ ABC=∠ EBD,\therefore ∠ ABE=∠ CBD,\therefore △ ABE∽△ CBD$.
(2) 解:如图,连接$BD$.$\because$ 四边形$ABCD$为正方形,$\therefore AB=AD,∠ BAD=90°,\therefore \frac{BD}{AB}=\sqrt{2}$.$\because EB=EF,∠ BEF=90°,\therefore \frac{BF}{BE}=\sqrt{2},∠ ABD=∠ EBF=45°,\therefore \frac{BF}{BE}=\frac{BD}{AB},∠ ABE=∠ DBF=45°-∠ EBD,\therefore △ ABE∽△ DBF,\therefore \frac{AE}{DF}=\frac{AB}{DB}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$.