1. [2026衡阳期末]世界陶瓷看中国，釉下五彩看醴陵. 如图，小宜将两根木条AC，BD在点O处固定，测量一件醴陵花瓶内径AB的宽度. 若$\frac{AC}{OC}=\frac{BD}{OD}=3$，且量得$CD=20\ \mathrm{cm}$，则AB的宽度为cm.

连接DC,AB.
∵$\frac{AC}{OC}=\frac{BD}{OD}=3$,$\therefore \frac{AC}{OC}-\frac{OC}{OC}=\frac{BD}{OD}-\frac{OD}{OD}=2$,$\therefore \frac{AO}{CO}=\frac{BO}{DO}=2$. $\because ∠ AOB = ∠ COD$,$\therefore △ AOB ∽ △ COD$,$\therefore \frac{AB}{CD}=\frac{AO}{CO}=2$. $\because CD = 20\ \mathrm{cm}$,$\therefore AB=20 × 2=40(\mathrm{cm})$.

2. [2026 合肥庐阳区期中] 如图 1, 在 $△ ABC$ 中, $∠ ACB = 90°$, $CO$ 平分 $∠ ACB$, 延长 $CO$ 至点 $D$, 使 $∠ DAB = 45°$, 连接 $DB$.
(1)①求证: $△ AOD ∽ △ COB$;
②试探究 $DB$ 与 $AB$ 的数量关系, 并说明理由.
(2)如图 2, 过点 $O$ 作 $OE ⊥ AD$ 于点 $E$, $\frac{BC}{AC} = \frac{3}{4}$, 求 $\frac{OE}{BD}$ 的值.
拍批+错题本

(1)①证明:$\because ∠ ACB=90°$,$CO$ 平分$∠ ACB$,
$\therefore ∠ ACD = ∠ BCD =45°$.
$\because ∠ DAB=45°$,$\therefore ∠ DAO = ∠ BCO$.
又$\because ∠ AOD = ∠ COB$,
$\therefore △ AOD ∽ △ COB$.
②解:$DB=\frac{\sqrt{2}}{2}AB$. 理由如下:
$\because △ AOD ∽ △ COB$,
$\therefore \frac{OA}{OC}=\frac{OD}{OB}$,$\therefore \frac{OA}{OD}=\frac{OC}{OB}$.
又$\because ∠ AOC = ∠ DOB$,$\therefore △ AOC ∽ △ DOB$,
$\therefore ∠ DBO = ∠ ACO =45°$.
又$\because ∠ DAB=45°$,
$\therefore △ ABD$ 为等腰直角三角形,
$\therefore AB^2 = AD^2 + BD^2 = 2BD^2$,
$\therefore BD=\frac{\sqrt{2}}{2}AB$.
(2)解:$\because ∠ ACB=90°$,$CO$ 平分$∠ ACB$,
$\therefore$ 点 $O$ 到 $CA$ 和 $CB$ 的距离相等(角平分线的性质定理),
$\therefore S_{△ BOC}:S_{△ AOC}=BC:AC=3:4$.
又$\because S_{△ BOC}:S_{△ AOC}=OB:OA$,
$\therefore OB:OA=3:4$,$\therefore OA:AB=4:7$.
$\because OE ⊥ AD$,$BD ⊥ AD$,$\therefore OE // BD$,$\therefore △ AOE ∽ △ ABD$,
$\therefore \frac{OE}{BD}=\frac{OA}{AB}=\frac{4}{7}$.

3. [2024湖南中考]如图,在△ABC中,D,E分别为边AB,AC的中点.下列结论中,错误的是 ()

A.$ DE // BC $
B.$ △ ADE ∽ △ ABC $
C.$ BC = 2DE $
D.$ S_{△ ADE} = \frac{1}{2} S_{△ ABC} $

$\because D,E$分别是边$AB,AC$的中点,$\therefore DE // BC$,$DE = \frac{1}{2}BC$(依据:中位线定理),即$BC=2DE$,$\therefore △ ADE ∽ △ ABC$,
$\therefore \frac{S_{△ ADE}}{S_{△ ABC}}=(\frac{DE}{BC})^2=\frac{1}{4}$,则$S_{△ ADE}=\frac{1}{4}S_{△ ABC}$.

4. 一题多解 [2026 六安联考] 如图,在Rt△ABC中,∠ACB = 90°,AC = 6,BC =12,点D在边BC上,点E在线段AD上,EF⊥AC于点F,EG⊥EF,交AB于点G.若EF=EG,则CD的长为.

通解 如图1,过点D作$DM ⊥ BC$交AB于点M,易证$DC=DM$,设$CD=x$,则$DM=x$,由$DM // AC$,得$△ BDM ∽ △ BCA$,
$\therefore \frac{BD}{BC}=\frac{DM}{AC}$,即$\frac{12-x}{12}=\frac{x}{6}$,解得$x=4$.
另解 如图2,延长$FE$,交$AB$于点$M$,易得$\mathrm{Rt}△ EMG ∽ \mathrm{Rt}△ CBA$,$\therefore EM:EG=CB:CA=12:6=2:1$. 又$\because EF=EG$,$\therefore EM:EF=2:1$. 由题中条件易得$△ AEM ∽ △ ADB$,$△ AEF ∽ △ ADC$,
$\therefore EM:BD=AE:AD$,$EF:CD=AE:AD$,$\therefore EM:BD=EF:CD$,
$\therefore BD:CD=EM:EF=2:1$,$\therefore CD=\frac{1}{3}BC=\frac{1}{3} × 12=4$.