8[中]如图,在$△ ABC$中,$∠ ACB=90°$,CD是斜边AB上的高.
(1)求证:$△ ACD ∽ △ CBD$;
(2)求证:$BC^2=BD · AB$;
(3)若$AD=3$,$BD=2$,求CD的长.

(1)【证明】$\because CD ⊥ AB$,$\therefore ∠ CDA = ∠ CDB = 90°.$
$\because ∠ ACB=90°$,$\therefore ∠ ACD+∠ BCD=90°.$
又 $\because ∠ BCD+∠ B=90°$,$\therefore ∠ ACD=∠ B$,$\therefore △ ACD∽△ CBD.$
(2)【证明】$\because ∠ ACB=∠ CDB=90°$,$∠ B=∠ B$,$\therefore △ ACB∽△ CDB$,$\therefore \frac{BC}{AB}=\frac{BD}{BC}$,即 $BC^2=BD · AB.$
(3)【解】$\because △ ACD∽△ CBD$,$\therefore \frac{AD}{CD}=\frac{CD}{BD}$,$\therefore CD^2=AD · DB.$
$\because AD=3,BD=2$,$\therefore CD^2=6$,$\therefore CD=\sqrt{6}.$

9[较难]如图，$△ ABC$中，$AB=AC$，$AB ⊥ AC$，点$D$，$E$分别为$BC$，$AC$的中点，$AF ⊥ BE$于点$F$。
(1)求证：$AE^2 = BE · EF$；
(2)求$∠ AFC$的大小；
(3)若$DF=1$，求$△ ABF$的面积。

(1)【证明】$\because AF ⊥ BE,AB ⊥ AC$,$\therefore ∠ AFE=∠ AFB=∠ BAE=90°$,$\therefore ∠ BAF + ∠ EAF = 90°$,$∠ ABF + ∠ BAF=90°$,$\therefore ∠ EAF=∠ ABF$,$\therefore △ AEF∽△ BEA$,$\therefore \frac{AE}{BE}=\frac{EF}{AE}$,$\therefore AE^2 = BE · EF.$
(2)【解】如图,过点 $C$ 作 $CH ⊥ BE$,交 $BE$ 的延长线于 $H.$ $\because$ 点 $E$ 是 $AC$ 的中点,$\therefore AE=EC.$
又 $\because ∠ AFE=∠ H=90°$,$∠ AEF=∠ CEH$,$\therefore △ AEF≌△ CEH(AAS)$,$\therefore AF=CH$,$EF=EH.$
$\because △ AEF∽△ BEA$,$\therefore \frac{AB}{AE}=\frac{AF}{EF}.$ 又 $\because AB=AC=2AE$,$\therefore AF=2EF$,$\therefore CH=AF=EF+EH=FH$,$\therefore ∠ CFH=45°$,$\therefore ∠ AFC=135°.$
(3) $\because ∠ CFH = 45° = ∠ ACB$,$\therefore ∠ FBC + ∠ FCB = ∠ ACF+∠ FCB$,$∠ BFC = ∠ AFC = 135°$,$\therefore ∠ ACF = ∠ FBC$,$\therefore △ AFC∽△ CFB$,$\therefore \frac{BF}{CF}=\frac{CF}{AF}=\frac{BC}{AC}.$
$\because$ 点 $D,E$ 分别是 $BC,AC$ 的中点,$\therefore DB=CD$,$AE=EC$,$\therefore$ 易得 $\frac{BC}{AC}=\frac{BD}{CE}=\sqrt{2}$,$\therefore \frac{BD}{CE}=\frac{BF}{CF}=\sqrt{2}.$
又 $\because ∠ FBC = ∠ ACF$,$\therefore △ BDF∽△ CEF$,$\therefore \frac{DF}{EF}=\frac{BD}{CE}=\sqrt{2}$,$\therefore EF=\frac{DF}{\sqrt{2}}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$,$\therefore AF=2EF=\sqrt{2}.$
由(1)知 $∠ EAF = ∠ ABF$,$∠ AFE = ∠ AFB$,$\therefore △ AEF∽△ BAF$,$\therefore \frac{EF}{AF}=\frac{AF}{BF}$,$\therefore \frac{\frac{\sqrt{2}}{2}}{\sqrt{2}}=\frac{\sqrt{2}}{BF}$,$\therefore BF=2\sqrt{2}$,$\therefore △ ABF$ 的面积为 $\frac{1}{2}× AF× BF=\frac{1}{2}×\sqrt{2}×2\sqrt{2}=2.$

10[2025河北保定期中,较难]某班数学课题学习小组对矩形内两条互相垂直的线段与矩形两邻边的数量关系进行探究,下面是他们的探究过程,请你按要求完成相应的证明和计算.
【初步发现】(1)如图(1),在矩形ABCD中,如果DE交AB于点E,CF交AD于点F,且$DE ⊥ CF$,那么$\frac{DE}{CF}\_\_\_\_\_\_\frac{AD}{CD}$(填“＞”“＜”或“=”).
【深入探究】(2)如图(2),在矩形ABCD中,$MN ⊥ GH$,MN分别交AD,BC于点M,N,GH分别交AB,CD于点H,G.求证:$\frac{GH}{MN}=\frac{AD}{CD}$.
【尝试应用】(3)在(1)的条件下,如图(3),在矩形ABCD中,点H,G分别在边AB,DC上,点M,N分别在边AD,BC上,连结MN,GH,且$MN ⊥ GH$,若$\frac{GH}{MN}=\frac{7}{5}$,求$\frac{DE}{CF}$的值.
【联系拓展】(4)如图(4),在四边形ABCD中,若$AB=BC=6$,$AD=CD=8$,$∠ BAD=90°$,$DE ⊥ CF$,直接写出$\frac{DE}{CF}$的值.

(1)【解】$\because$ 四边形 $ABCD$ 为矩形,$\therefore ∠ A=∠ ADC=90°$,$\therefore ∠ FCD+∠ DFC=90°.$
$\because ED ⊥ CF$,$\therefore ∠ ADE+∠ DFC=90°$,$\therefore ∠ ADE=∠ DCF$,$\therefore △ ADE∽△ DCF$,$\therefore \frac{DE}{CF}=\frac{AD}{CD}.$
故答案为 $=.$
(2)【证明】过 $M$ 作 $MP ⊥ BC$ 于 $P$,过 $H$ 作 $HQ ⊥ CD$ 于 $Q$,交 $MN$ 于 $O$,如图(1).
易得 $AD// HQ// BC$,$HQ=AD$,$MP=CD$,$\therefore ∠ HON=∠ MNP.$
$\because MN ⊥ GH$,$MP ⊥ BC$,$\therefore ∠ QHG+∠ HON=90°$,$∠ MNP+∠ NMP=90°$,$\therefore ∠ QHG=∠ NMP.$
$\because MP ⊥ BC$,$HQ ⊥ CD$,$\therefore ∠ HQG=∠ MPN=90°$,$\therefore △ HQG∽△ MPN$,$\therefore \frac{GH}{MN}=\frac{HQ}{MP}=\frac{AD}{CD}.$
(3)【解】由(1)可知,$\frac{DE}{CF}=\frac{AD}{CD}$,由(2)可知,$\frac{HG}{MN}=\frac{AD}{CD}$,$\therefore \frac{DE}{CF}=\frac{GH}{MN}=\frac{7}{5}.$
(4)【解】$\frac{DE}{CF}=\frac{25}{24}.$ 过点 $C$ 作 $CM ⊥ AB$ 交 $AB$ 的延长线于 $M$,过点 $D$ 作 $DN ⊥ MC$ 交 $MC$ 的延长线于 $N$,连结 $BD$,如图(2),则易知四边形 $AMND$ 是矩形,$\therefore AM=DN$,$AD=MN.$
$\because AB = 6, AD = 8, ∠ BAD = 90°, \therefore BD = \sqrt{AB^2+AD^2}=10.$
又 $\because BC=6,CD=8$,$\therefore BC^2+CD^2=BD^2$,$\therefore BC ⊥ CD$,$\therefore ∠ MCB+∠ NCD=90°.$
又 $\because ∠ MCB+∠ MBC=90°$,$\therefore ∠ MBC=∠ DCN.$
$\because CM ⊥ AB$,$DN ⊥ MN$,$\therefore ∠ M=∠ N=90°$,$\therefore △ CMB∽△ DNC$,$\therefore \frac{BM}{CN}=\frac{CM}{DN}=\frac{BC}{CD}=\frac{3}{4}.$
设 $BM=x$,则 $AM=6+x=DN$,$\therefore MC=\frac{3}{4}(6+x)$,$CN=\frac{4}{3}x.$
$\because CM+CN=MN=AD=8$,$\therefore \frac{3}{4}(6+x)+\frac{4}{3}x=8$,解得 $x=\frac{42}{25}$,$\therefore DN=\frac{192}{25}.$
由探究过程知,$\frac{DE}{CF}=\frac{AD}{DN}=\frac{8}{\frac{192}{25}}=\frac{25}{24}.$
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