12. ★★★【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,CD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.

12. (1) = 【点拨】$\because$ 四边形$ABCD$是矩形，$\therefore ∠ A=∠ ADC=90°$。$\therefore ∠ FCD+∠ DFC=90°$。$\because ED⊥ CF$，$\therefore ∠ ADE+∠ DFC=90°$。$\therefore ∠ ADE=∠ DCF$。$\therefore △ ADE ∽ △ DCF$。$\therefore \frac{DE}{CF}=\frac{AD}{CD}$。
(2)【证明】过点$M$作$MP⊥ BC$于$P$，过点$H$作$HQ⊥ CD$于$Q$，交$MN$于$O$，如图①。易得$AD// HQ// BC$，$HQ=AD$，$MP=CD$，$\therefore ∠ HON=∠ MNP$。$\because MN⊥ GH$，$MP⊥ BC$，$\therefore ∠ QHG+∠ HON=90°$，$∠ MNP+∠ NMP=90°$。$\therefore ∠ QHG=∠ NMP$。又$\because ∠ HQG=∠ MPN=90°$，$\therefore △ HQG ∽ △ MPN$。$\therefore \frac{GH}{MN}=\frac{HQ}{MP}=\frac{AD}{CD}$。
(3)【解】由(1)可知$\frac{DE}{CF}=\frac{AD}{CD}$，由(2)可知$\frac{GH}{MN}=\frac{AD}{CD}$，$\therefore \frac{DE}{CF}=\frac{GH}{MN}=\frac{7}{5}$。
(4)【解】$\frac{DE}{CF}=\frac{25}{24}$。【点拨】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，$\therefore AM=DN$，$AD=MN$。$\because AB=6$，$AD=8$，$∠ BAD=90°$，$\therefore BD=10$。又$\because BC=6$，$CD=8$，$\therefore BC^2+CD^2=BD^2$。$\therefore BC⊥ CD$。$\therefore ∠ MCB+∠ NCD=90°$。又$\because ∠ MCB+∠ MBC=90°$，$\therefore ∠ MBC=∠ DCN$。又$\because ∠ CMB=∠ DNC=90°$，$\therefore △ CMB ∽ △ DNC$。$\therefore \frac{BM}{CN}=\frac{CM}{DN}=\frac{BC}{CD}=\frac{3}{4}$。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(1)易知，$\frac{DE}{CF}=\frac{AD}{DN}=\frac{8}{\frac{192}{25}}=\frac{25}{24}$。