7. [2026长春期中]如图,在正方形ABCD中,F是BC边上一点,连接AF,以AF为对角线作正方形AEFG,边FG与正方形ABCD的对角线AC相交于点H,连接DG.以下四个结论:①$∠ EAB = ∠ GAD$;②$△ AFC ∽ △ AGD$;③$2AE^2 = AG · AC$;④$DG ⊥ AC$.其中正确的个数为 ()


A.1
B.2
C.3
D.4

$\because$ 四边形 $AEFG$ 和四边形 $ABCD$ 均为正方形,$AF,AC$ 为对角线,$\therefore ∠ EAG = ∠ BAD = 90°$,$∠ FAG = ∠ AFG = ∠ DAC = ∠ ACB = 45°$,$AC=\sqrt{2}AD$,$AF=\sqrt{2}AG$. $\because ∠ EAB = ∠ EAG - ∠ BAG$,$∠ GAD = ∠ BAD - ∠ BAG$,$\therefore ∠ EAB = ∠ GAD$,故①正确. $\because AC=\sqrt{2}AD$,$AF=\sqrt{2}AG$,$\therefore \frac{AC}{AD}=\sqrt{2}$,$\frac{AF}{AG}=\sqrt{2}$,$\therefore \frac{AC}{AD}=\frac{AF}{AG}$. $\because ∠ FAG = ∠ DAC = 45°$,$\therefore ∠ FAG - ∠ GAC = ∠ DAC - ∠ GAC$,$\therefore ∠ CAF = ∠ DAG$,$\therefore △ AFC ∽ △ AGD$,故②正确. $\because ∠ AFH = ∠ ACF = 45°$,$∠ FAH = ∠ CAF$,$\therefore △ HAF ∽ △ FAC$,$\therefore \frac{AF}{AC}=\frac{AH}{AF}$,即$AF^2 = AC · AH$. 又$\because AF=\sqrt{2}AE$,$\therefore 2AE^2 = AH · AC$. 在$\mathrm{Rt}△ AGH$中,$∠ AGH=90°$,$\therefore AH＞AG$,$\therefore 2AE^2 ≠ AG · AC$,故③错误. 延长 $DG$ 交 $AC$ 于点 $N$,$\because △ AFC ∽ △ AGD$,$\therefore ∠ ADG = ∠ ACF = 45°$,$\therefore ∠ AND = 180° - ∠ ADG - ∠ DAC = 90°$,即$DG ⊥ AC$,故④正确. 综上可知,正确的个数为3.

8. 问题背景 如图1,已知△ABC∽△ADE,求证:△ABD∽△ACE.
尝试应用 如图2,在△ABC和△ADE中,∠BAC=∠DAE=90°,∠ABC=∠ADE=30°,AC与DE相交于点F.点D在BC边上,$\frac{AD}{BD}=\sqrt{3}$,求$\frac{DF}{CF}$的值.
拓展创新 如图3,D是△ABC内一点,∠BAD=∠CBD=30°,∠BDC=90°,AB=4,AC=$2\sqrt{3}$,直接写出AD的长.

问题背景 证明:$\because △ ABC ∽ △ ADE$,
$\therefore \frac{AB}{AD}=\frac{AC}{AE}$,$∠ BAC = ∠ DAE$,$\therefore \frac{AB}{AC}=\frac{AD}{AE}$,$∠ BAC - ∠ DAC = ∠ DAE - ∠ DAC$,$\therefore ∠ BAD = ∠ CAE$,
$\therefore △ ABD ∽ △ ACE$.
尝试应用 解:连接 $CE$,设 $BD=t$,
则 $AD=\sqrt{3}BD=\sqrt{3}t$.
根据题意,易得$△ ABC ∽ △ ADE$,$\therefore \frac{AB}{AD}=\frac{AC}{AE}$,$\therefore \frac{AB}{AC}=\frac{AD}{AE}$.
又$\because ∠ BAC = ∠ DAE$,$\therefore ∠ BAC - ∠ DAC = ∠ DAE - ∠ DAC$,
$\therefore ∠ BAD = ∠ CAE$,$\therefore △ ABD ∽ △ ACE$,
$\therefore \frac{CE}{BD}=\frac{AC}{AB}$,$∠ ACE = ∠ ABC$.
在$\mathrm{Rt}△ ABC$中,$∠ ABC=30°$,$\therefore BC=2AC$,
由勾股定理,得$AB=\sqrt{3}AC$,$\therefore \frac{AC}{AB}=\frac{\sqrt{3}}{3}$,
$\therefore CE=\frac{\sqrt{3}}{3}BD=\frac{\sqrt{3}}{3}t$,$\therefore \frac{AD}{CE}=\frac{\sqrt{3}t}{\frac{\sqrt{3}}{3}t}=3$.
$\because ∠ ADE = ∠ ABC$,$∠ ABC = ∠ ACE$,$\therefore ∠ ACE = ∠ ADE$.
又$\because ∠ AFD = ∠ EFC$,
$\therefore △ ADF ∽ △ ECF$,$\therefore \frac{DF}{CF}=\frac{AD}{CE}=3$.
拓展创新 解:$AD$ 的长为$\sqrt{5}$.
如图,过点 $D$ 作 $AD$ 的垂线交 $AB$ 于点 $M$,连接 $CM$.
根据题意,易证$△ ADB ∽ △ MDC$,
$\therefore ∠ DMC = ∠ DAB = 30°$,$\frac{AB}{CM}=\frac{AD}{MD}$,
在$\mathrm{Rt}△ AMD$中,$∠ MAD=30°$,$\therefore AM=2MD$,
由勾股定理,得$AD=\sqrt{3}MD$,$\therefore \frac{AD}{MD}=\sqrt{3}$,
$\therefore CM=\frac{AB}{\sqrt{3}}=\frac{4\sqrt{3}}{3}$,$∠ AMC = ∠ AMD + ∠ DMC = ∠ AMD + ∠ DAB = 90°$.
在$\mathrm{Rt}△ AMC$中,$AM=\sqrt{AC^2 - CM^2}=\frac{2\sqrt{15}}{3}$,
$\therefore MD=\frac{\sqrt{15}}{3}$,$\therefore AD=\sqrt{5}$.