5.【基础巩固】
(1)如图1,在△ABC中,D为边AB上一点,∠ACD=∠B.求证:AC² = AD·AB.
【尝试应用】
(2)如图2,在□ABCD中,E为BC上一点,F为CD延长线上一点,连接BF,EF,∠BFE=∠A.若BF=4,BE=3,求AD的长.
【拓展提高】
(3)如图3,在菱形ABCD中,E是AB上一点,F是△ABC内一点,EF//AC,AC=2EF,∠EDF=$\frac{1}{2}$∠BAD,AE=2,DF=5,求菱形ABCD的边长.

(1)证明:$\because ∠ ACD = ∠ B$,$∠ A = ∠ A$,$\therefore △ ADC ∽ △ ACB$,
$\therefore \frac{AD}{AC}=\frac{AC}{AB}$,$\therefore AC^2 = AD · AB$.
(2)解:$\because$ 四边形 $ABCD$ 是平行四边形,
$\therefore AD = BC$,$∠ A = ∠ C$.
又$\because ∠ BFE = ∠ A$,$\therefore ∠ BFE = ∠ C$.
又$\because ∠ FBE = ∠ CBF$,$\therefore △ BFE ∽ △ BCF$,
$\therefore \frac{BF}{BC}=\frac{BE}{BF}$.
又$\because BF=4$,$BE=3$,
$\therefore BC=\frac{BF^2}{BE}=\frac{16}{3}$,$\therefore AD=BC=\frac{16}{3}$.
(3)解:如图,分别延长 $EF$,$DC$ 相交于点 $G$.
$\because$ 四边形 $ABCD$ 是菱形,
$\therefore AB // DC$,$∠ BAC = \frac{1}{2}∠ BAD$.
又$\because AC // EF$,
$\therefore$ 四边形 $AEGC$ 为平行四边形,
$\therefore AC=EG$,$CG=AE=2$,$∠ EAC = ∠ G$.
$\because ∠ EDF = \frac{1}{2}∠ BAD$,$\therefore ∠ EDF = ∠ BAC$,$\therefore ∠ EDF = ∠ G$.
又$\because ∠ DEF = ∠ GED$,$\therefore △ EDF ∽ △ EGD$,
$\therefore \frac{DE}{EG}=\frac{EF}{DE}=\frac{DF}{DG}$,$\therefore DE^2 = EF · EG$.
又$\because EG=AC=2EF$,$\therefore DE^2=2EF^2$,
$\therefore DE=\sqrt{2}EF$.
又$\because \frac{DG}{DF}=\frac{DE}{EF}=\sqrt{2}$,$\therefore DG=\sqrt{2}DF=5\sqrt{2}$,
$\therefore DC=DG-CG=5\sqrt{2}-2$,
即菱形 $ABCD$ 的边长为 $5\sqrt{2}-2$.

6. (1)如图1,在四边形ABCD中,AB//CD,点P在BC上,∠B=∠APD=90°,求证:△ABP∽△PCD.
(2)探究:如图2,在四边形ABCD中,点P在边BC上,当∠B=∠C=∠APD时,求证:△ABP∽△PCD.
(3)应用:如图3,在△ABC中,P是边BC的中点,点D,E分别在边AB,AC上.若∠B=∠C=∠DPE=45°,BC=8√2,CE=6,求DE的长.

(1)证明:$\because AB // CD$,$∠ B = ∠ APD = 90°$,
$\therefore ∠ C=90° = ∠ B$,$∠ BAP + ∠ APB = 90°$,$∠ APB + ∠ CPD = 90°$,
$\therefore ∠ BAP = ∠ CPD$,$\therefore △ ABP ∽ △ PCD$.
(2)证明:$\because ∠ B = ∠ C = ∠ APD$,$∠ CPD = 180° - ∠ APD - ∠ APB$,$∠ BAP = 180° - ∠ APB - ∠ B$,
$\therefore ∠ CPD = ∠ BAP$,$\therefore △ ABP ∽ △ PCD$.
(3)解:由(2)知$△ BDP ∽ △ CPE$,$\therefore \frac{BD}{CP}=\frac{BP}{CE}$.
$\because P$ 是 $BC$ 的中点,$BC=8\sqrt{2}$,
$\therefore BP=CP=4\sqrt{2}$.
$\because CE=6$,$\therefore \frac{BD}{4\sqrt{2}}=\frac{4\sqrt{2}}{6}$,$\therefore BD=\frac{16}{3}$.
$\because ∠ B = ∠ C = 45°$,$\therefore ∠ A=90°$.
在$\mathrm{Rt}△ ABC$中,根据勾股定理,得$AB^2 + AC^2 = BC^2=(8\sqrt{2})^2$,
$\therefore AC=AB=8$,$\therefore AD=AB-BD=8-\frac{16}{3}=\frac{8}{3}$,
$AE=AC-CE=8-6=2$.
在$\mathrm{Rt}△ ADE$中,根据勾股定理,得$DE=\sqrt{AD^2 + AE^2}=\sqrt{(\frac{8}{3})^2 + 2^2}=\frac{10}{3}$.