7. 如图,在等边三角形 $ ABC $ 中,点 $ D、E $ 分别在边 $ BC、AC $ 上,且 $ DE // AB $,过点 $ E $ 作 $ EF ⊥ DE $,交 $ BC $ 的延长线于点 $ F $.
(1)求 $ ∠ F $ 的度数.
(2)若 $ CD=2 $,求 $ DF $ 的长.

解：(1)$\because\triangle ABC$是等边三角形，$\therefore\angle A = \angle B = \angle ACB = 60^{\circ}。$
$\because DE// AB，$$\therefore\angle EDC = \angle B = 60^{\circ}，$$\angle CED = \angle A = 60^{\circ}，$
$\therefore\angle DCE = 60^{\circ}。$$\because EF\perp ED，$
$\therefore\angle DEF = 90^{\circ}，$
$\therefore\angle F = 90^{\circ}-\angle EDC = 90^{\circ}-60^{\circ}=30^{\circ}。$ (2)$\because\angle F + \angle FEC = \angle DCE = 60^{\circ}，$
$\therefore\angle FEC = \angle DCE - \angle F = 60^{\circ}-30^{\circ}=30^{\circ}，$
$\therefore\angle FEC = \angle F，$$\therefore CE = CF。$
$\because\angle EDC = \angle DCE = \angle CED = 60^{\circ}，$
$\therefore CE = CD = 2，$$\therefore CF = 2，$
$\therefore DF = CD + CF = 2 + 2 = 4。$

8. 已知在$△ ABC$中，$AB=AC$，$D$是边$AB$上的一点，$∠ BCD=∠ A$。
(1)如图1，求证:$CD=CB$。
(2)如图2，过点$B$作$BE⊥ AC$，垂足为$E$，$BE$与$CD$相交于点$F$。
①求证:$∠ BCD=2∠ CBE$；
②若$△ BDF$是等腰三角形，求$∠ A$的度数。

解：(1)证明：$\because AB = AC，$$\therefore\angle ABC = \angle ACB。$
$\because\angle BDC$是$\triangle ADC$的一个外角，$\therefore\angle BDC = \angle A + \angle ACD。$
$\because\angle ACB = \angle BCD + \angle ACD，$$\angle BCD = \angle A，$
$\therefore\angle BDC = \angle ACB，$
$\therefore\angle ABC = \angle BDC，$$\therefore CD = CB。$ (2)①证明：$\because BE\perp AC，$$\therefore\angle BEC = 90^{\circ}，$
$\therefore\angle CBE + \angle ACB = 90^{\circ}，$
设$\angle CBE = \alpha，$则$\angle ACB = 90^{\circ}-\alpha，$
$\therefore\angle BDC = \angle ABC = \angle ACB = 90^{\circ}-\alpha，$
$\therefore\angle BCD = 180^{\circ}-\angle BDC - \angle ABC = 180^{\circ}-(90^{\circ}-\alpha)-(90^{\circ}-\alpha)=2\alpha，$
$\therefore\angle BCD = 2\angle CBE。$
②$\because\angle BFD$是$\triangle CBF$的一个外角，$\therefore\angle BFD = \angle CBE + \angle BCD = \alpha + 2\alpha = 3\alpha。$
分三种情况：(ⅰ)当$BD = BF$时，$\angle BFD = \angle BDC = 3\alpha，$
又$\because\angle BDC = \angle ABC = \angle ACB = 90^{\circ}-\alpha，$
$\therefore 90^{\circ}-\alpha = 3\alpha，$$\therefore\alpha = 22.5^{\circ}，$
$\therefore\angle A = \angle BCD = 2\alpha = 45^{\circ}；$
(ⅱ)当$DB = DF$时，$\angle DBE = \angle BFD = 3\alpha，$
又$\because\angle DBE = \angle ABC - \angle CBE = 90^{\circ}-\alpha-\alpha = 90^{\circ}-2\alpha，$$\therefore 90^{\circ}-2\alpha = 3\alpha，$$\therefore\alpha = 18^{\circ}，$
$\therefore\angle A = \angle BCD = 2\alpha = 36^{\circ}；$
(ⅲ)当$FB = FD$时，$\angle DBE = \angle BDF，$
又$\because\angle BDF = \angle ABC＞\angle DBF，$
$\therefore$不存在$FB = FD。$
综上所述，若$\triangle BDF$是等腰三角形，则$\angle A$的度数为$45^{\circ}$或$36^{\circ}。$

1. 如图,在$△ ABC$中,$AB=AC$,D 是边 AB 上的一点,过点 D 作$DE ⊥ BC$于点 E,延长 ED和$CA$,交于点 F.
(1)求证:$△ ADF$是等腰三角形.
(2)若$∠ F=30°$,$BD=4$,$EC=6$,求 AC 的长.

(1)证明：$\because AB = AC，$$\therefore\angle B = \angle C。$
$\because FE\perp BC，$
$\therefore\angle F + \angle C = 90^{\circ}，$$\angle B + \angle BDE = 90^{\circ}，$
$\therefore\angle F = \angle BDE。$
又$\because\angle BDE = \angle FDA，$
$\therefore\angle F = \angle FDA，$$\therefore AF = AD，$
$\therefore\triangle ADF$是等腰三角形。 (2)$\because DE\perp BC，$$\therefore\angle DEB = 90^{\circ}。$
$\because\angle F = 30^{\circ}，$$\therefore\angle BDE = 30^{\circ}，$$\angle B = 60^{\circ}。$
$\because BD = 4，$$\therefore BE = \frac{1}{2}BD = 2。$
$\because AB = AC，$$\therefore\triangle ABC$是等边三角形，$\therefore AC = BC = BE + EC = 2 + 6 = 8。$