11. 如图8，在$△ ABC$中，$AB = AC$.
（1）尺规作图：求作$AB$的垂直平分线$DE$，分别交$AB$，$BC$于点$D$，$E$；（保留作图痕迹，不写作法）
（2）在（1）的条件下，连接$AE$，若$AC = EC$，求$∠ C$的度数.

11. 解：(1) 如答图1，DE即为所求.
(2) $\because AB=AC, \therefore ∠ C = ∠ B$.
又$\because DE$垂直平分$AB$,
$\therefore AE=BE. \therefore ∠ B = ∠ BAE$.
$\because AC=EC, \therefore ∠ CAE = ∠ CEA$.
$\because ∠ CEA = ∠ B + ∠ BAE = 2∠ B = 2∠ C, \therefore ∠ CAE = 2∠ C$.
$\because ∠ C + ∠ CAE + ∠ CEA = 180°$,
$\therefore ∠ C + 2∠ C + 2∠ C = 180°. \therefore ∠ C = 36°$.

12. 如图9①，在△ABC中，AB=AC，D是边AB上一个动点，DF⊥BC于点F，交CA延长线于点E.
(1) 直接判断△ADE的形状，并说明理由.
(2) 如图9②，当点D在BA的延长线上时，其他条件不变，(1)中的结论是否还成立？请说明理由.

12. 解：(1) $△ ADE$是等腰三角形. 理由：
$\because AB=AC, \therefore ∠ B = ∠ C$.
$\because DF ⊥ BC$,
$\therefore ∠ BDF + ∠ B = 90°, ∠ C + ∠ E = 90°. \therefore ∠ E = ∠ BDF$.
$\because ∠ BDF = ∠ EDA, \therefore ∠ E = ∠ EDA. \therefore AE = AD$.
$\therefore △ ADE$是等腰三角形.
(2) (1)中的结论还成立. 理由：
$\because AB=AC, \therefore ∠ B = ∠ C$.
$\because DF ⊥ BC, \therefore ∠ BDF + ∠ B = 90°, ∠ C + ∠ FEC = 90°$.
$\therefore ∠ FEC = ∠ BDF$.
$\because ∠ FEC = ∠ AED, \therefore ∠ ADE = ∠ AED. \therefore AE = AD$.
$\therefore △ ADE$是等腰三角形.

13. 如图10，在$△ ADB$中，$∠ ADB = 60°$，$DC$平分$∠ ADB$，交$AB$于点$C$，且$DC ⊥ AB$，过点$C$作$CE // DA$交$DB$于点$E$，连接$AE$。
（1）求证：$△ ADB$是等边三角形；
（2）求证：$AE ⊥ DB$。

13. 证明：(1) $\because DC$平分$∠ ADB, \therefore ∠ ADC = ∠ BDC$.
$\because ∠ ADB = 60°, \therefore ∠ ADC = ∠ BDC = 30°$.
$\because DC ⊥ AB, \therefore ∠ DCB = ∠ DCA = 90°$.
$\therefore ∠ B = ∠ DAB = 90° - 30° = 60°$.
$\therefore ∠ ADB = ∠ B = ∠ DAB = 60°. \therefore △ ADB$是等边三角形.
(2) $\because CE // DA$,
$\therefore ∠ BEC = ∠ ADB = 60°, ∠ BCE = ∠ DAB = 60°$.
$\therefore ∠ CEB = ∠ CBE = ∠ ECB = 60°$.
$\therefore △ CEB$是等边三角形. $\therefore CE = BE = CB$.
$\because ∠ BDC = 30°, ∠ DCB = 90°$,
$\therefore BC = \frac{1}{2}BD. \therefore BE = \frac{1}{2}BD$.
$\therefore E$是$BD$的中点，即$AE$是边$BD$的中线.
$\because △ ADB$是等边三角形, $\therefore AE ⊥ BD$.