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1. [2025安庆大观区期中]已知线段$AB=4$厘米，点$P$是线段$AB$的黄金分割点，那么线段$AP=$
$(2\sqrt{5}-2)$或$(6-2\sqrt{5})$
厘米.(结果保留根号)答案
分两种情况:①当$AP＞BP$时，
∵点$P$是线段$AB$的黄金分割点，
∴$AP=\frac{\sqrt{5}-1}{2}AB=2\sqrt{5}-2$.
②当$AP＜BP$时，$AP=AB-BP=4-(2\sqrt{5}-2)=6-2\sqrt{5}$.综上，$AP$的长为$(2\sqrt{5}-2)$厘米或$(6-2\sqrt{5})$厘米.
∵点$P$是线段$AB$的黄金分割点，
∴$AP=\frac{\sqrt{5}-1}{2}AB=2\sqrt{5}-2$.
②当$AP＜BP$时，$AP=AB-BP=4-(2\sqrt{5}-2)=6-2\sqrt{5}$.综上，$AP$的长为$(2\sqrt{5}-2)$厘米或$(6-2\sqrt{5})$厘米.
2. [2025宣城段考]如图，在$□ ABCD$中，E是AD边上一点，且点E将AD边分为2:3两部分，连接BE，AC相交于点F，则$S_{△ AEF}: S_{△ CBF} =$
$4:25$或$9:25$
。答案
点$E$将$AD$边分为$2:3$两部分，不知道$AE$，$ED$的占比，故分两种情况讨论:①当$AE: ED=2: 3$时，$AE: AD=2: 5$，在$□ABCD$中，$AD=BC$，$AD// BC$，$\therefore △ AEF∽ △ CBF$，$AE: CB=AE: AD=2: 5$，$\therefore S_{△ AEF}: S_{△ CBF}=4: 25$. ②当$AE: ED=3: 2$时，同理可得$S_{△ AEF}: S_{△ CBF}=9: 25$.
3. [2026安庆期中]如图,在$△ ABC$中,$AB = AC = 6$,$BC = 8$,D是BC边上的一个动点,点E在AC上,点D在运动过程中始终保持$∠ 1 = ∠ B$.设BD的长为$x(0 ＜ x ＜ 8)$.
(1)求证:$△ DCE ∽ △ ABD$.
(2)用含$x$的代数式表示CE的长,当$CE = 2$时,求$x$的值.
(3)当$x$为何值时,$△ ADE$为等腰三角形?
(1)求证:$△ DCE ∽ △ ABD$.
(2)用含$x$的代数式表示CE的长,当$CE = 2$时,求$x$的值.
(3)当$x$为何值时,$△ ADE$为等腰三角形?
答案
(1)证明:如图,$\because ∠ ADC = ∠ 1 + ∠ 2 = ∠ B + ∠ 3, ∠ 1 = ∠ B$,
$\therefore ∠ 2 = ∠ 3$.
$\because AB = AC, \therefore ∠ C = ∠ B$,
$\therefore △ DCE ∽ △ ABD$.
(2)解:$\because BD = x, \therefore CD = BC - BD = 8 - x$.
由(1)知$△ DCE ∽ △ ABD$,
$\therefore \frac{CE}{BD} = \frac{DC}{AB}$,即$\frac{CE}{x} = \frac{8 - x}{6}, \therefore CE = -\frac{1}{6}x^2 + \frac{4}{3}x$.
$\because CE = 2, \therefore -\frac{1}{6}x^2 + \frac{4}{3}x = 2$,解得$x = 2$或$x = 6$.
(3)解:①当$DA = DE$时,$△ DCE ≌ △ ABD$,
$\therefore DC = AB = 6$,即$8 - x = 6$,解得$x = 2$.
②当$EA = ED$时,$∠ DAE = ∠ 1 = ∠ B = ∠ C$.
$\therefore △ DAC ∽ △ ABC, \therefore \frac{DC}{AC} = \frac{AC}{BC}$,即$\frac{8 - x}{6} = \frac{6}{8}$,解得$x = \frac{7}{2}$.
③当$AD = AE$时,点$D$与点$B$重合,点$E$与点$C$重合,此时$x = 0$,即此种情况不成立.
综上所述,当$x = 2$或$x = \frac{7}{2}$时,$△ ADE$为等腰三角形.
4. [2026铜陵期末]如图,在四边形ABCD中,AB // CD,对角线AC,BD交于点E,点F在边AB上,连接CF交线段BE于点G,CG^2 = GE · GD.
(1)求证:$∠ ACF = ∠ ABD$;
(2)连接EF,求证:$EF · CG = EG · CB$.
(1)求证:$∠ ACF = ∠ ABD$;
(2)连接EF,求证:$EF · CG = EG · CB$.
答案
(1)证明:$\because CG^2 = GE · GD, \therefore \frac{CG}{GE} = \frac{GD}{CG}$.
又$\because ∠ CGD = ∠ EGC$,
$\therefore △ GCD ∽ △ GEC, \therefore ∠ GDC = ∠ GCE$.
$\because AB// CD, \therefore ∠ ABD = ∠ BDC$,
$\therefore ∠ ACF = ∠ ABD$.
(2)$\because ∠ ABD = ∠ ACF, ∠ BGF = ∠ CGE$,
$\therefore △ BGF ∽ △ CGE, \therefore \frac{FG}{EG} = \frac{BG}{CG}$,即$\frac{FG}{BG} = \frac{EG}{CG}$.
又$\because ∠ FGE = ∠ BGC, \therefore △ FGE ∽ △ BGC$,
$\therefore \frac{EF}{CB} = \frac{EG}{CG}, \therefore EF · CG = EG · CB$.
又$\because ∠ CGD = ∠ EGC$,
$\therefore △ GCD ∽ △ GEC, \therefore ∠ GDC = ∠ GCE$.
$\because AB// CD, \therefore ∠ ABD = ∠ BDC$,
$\therefore ∠ ACF = ∠ ABD$.
(2)$\because ∠ ABD = ∠ ACF, ∠ BGF = ∠ CGE$,
$\therefore △ BGF ∽ △ CGE, \therefore \frac{FG}{EG} = \frac{BG}{CG}$,即$\frac{FG}{BG} = \frac{EG}{CG}$.
又$\because ∠ FGE = ∠ BGC, \therefore △ FGE ∽ △ BGC$,
$\therefore \frac{EF}{CB} = \frac{EG}{CG}, \therefore EF · CG = EG · CB$.
5. [2025安庆期末]如图,在$Rt△ ABC$中,$∠ ACB=90°$,D是斜边AB上的中点,E是边BC上的点,AE与CD交于点F,且$AC^2 = CE · CB$.
(1)求证:$AE ⊥ CD$;
(2)连接BF,若E是BC的中点,求证:$AE · BF = 2BE · CD$.
(1)求证:$AE ⊥ CD$;
(2)连接BF,若E是BC的中点,求证:$AE · BF = 2BE · CD$.
答案
(1)证明:$\because AC^2 = CE · CB, \therefore \frac{AC}{CE} = \frac{CB}{AC}$.
又$\because ∠ ACB = ∠ ECA = 90°$,
$\therefore △ ACB ∽ △ ECA, \therefore ∠ ABC = ∠ EAC$.
$\because D$是$AB$的中点,$\therefore CD = \frac{1}{2}AB = AD$,
$\therefore ∠ ACD = ∠ CAD$.
$\because ∠ CAD + ∠ ABC = 90°$,
$\therefore ∠ ACD + ∠ EAC = 90°$,
$\therefore ∠ AFC = 90°, \therefore AE ⊥ CD$.
(2)$\because AE ⊥ CD, \therefore ∠ EFC = 90°$,
$\therefore ∠ ACE = ∠ EFC$.
又$\because ∠ AEC = ∠ CEF, \therefore △ EAC ∽ △ ECF$,
$\therefore \frac{EC}{EA} = \frac{EF}{EC}$.
$\because E$是$BC$的中点,$\therefore CE = BE$,
$\therefore \frac{BE}{EA} = \frac{EF}{BE}$.
又$\because ∠ BEF = ∠ AEB, \therefore △ BEF ∽ △ AEB$,
$\therefore \frac{BE}{AE} = \frac{BF}{AB}, \therefore AE · BF = BE · AB$.
$\because CD = \frac{1}{2}AB, \therefore AB = 2CD$,
$\therefore AE · BF = 2BE · CD$.
又$\because ∠ ACB = ∠ ECA = 90°$,
$\therefore △ ACB ∽ △ ECA, \therefore ∠ ABC = ∠ EAC$.
$\because D$是$AB$的中点,$\therefore CD = \frac{1}{2}AB = AD$,
$\therefore ∠ ACD = ∠ CAD$.
$\because ∠ CAD + ∠ ABC = 90°$,
$\therefore ∠ ACD + ∠ EAC = 90°$,
$\therefore ∠ AFC = 90°, \therefore AE ⊥ CD$.
(2)$\because AE ⊥ CD, \therefore ∠ EFC = 90°$,
$\therefore ∠ ACE = ∠ EFC$.
又$\because ∠ AEC = ∠ CEF, \therefore △ EAC ∽ △ ECF$,
$\therefore \frac{EC}{EA} = \frac{EF}{EC}$.
$\because E$是$BC$的中点,$\therefore CE = BE$,
$\therefore \frac{BE}{EA} = \frac{EF}{BE}$.
又$\because ∠ BEF = ∠ AEB, \therefore △ BEF ∽ △ AEB$,
$\therefore \frac{BE}{AE} = \frac{BF}{AB}, \therefore AE · BF = BE · AB$.
$\because CD = \frac{1}{2}AB, \therefore AB = 2CD$,
$\therefore AE · BF = 2BE · CD$.
