如图, $BD$ 是$\odot O$的直径,$AC$是$\odot O$的弦,$AE ⊥ CD$,垂足为$E$,且$∠ EDA = ∠ ADB$.

(1)求证:$AB = AC$.
(2)求证:$AE$是$\odot O$的切线.
(3)若$\odot O$的半径为$5$,$CE = 8$,求$AD$的长.
(4)若$BD$不是$\odot O$的直径,将条件$AE ⊥ CD$换成$AE // BC$,其他条件不变.
①(2)中的结论是否成立?请给出理由.
②若$BD ⊥ AC$,$AB = 8$,$CD = 4$,则$\odot O$的半径为.
③若点$M$在$BD$上且点$M$为$△ ABC$的内心,$∠ ACB = 80°$,则$∠ MAD =$$°$.
(5)已知$∠ ACD = 30°$,$AD = 4$.
①若沿直径$BD$将$\odot O$剪开,扇形$OBD$围成一个圆锥的侧面,求该圆锥底面圆的半径;
②求$\overset{\frown}{AD}$的长和阴影部分的面积(结果保留$π$).

(1)证明:
∵ 四边形 ABCD 是⊙O 的内接四边形,
∴ ∠ABC + ∠ADC = 180°.
∵ ∠ADC + ∠ADE = 180°,
∴ ∠ADE = ∠ABC.
∵ ∠ADB = ∠ACB,∠ADB = ∠ADE,
∴ ∠ABC = ∠ACB,
∴ AB = AC.
(2)证明:如图1,连接 OA,则 OA = OD,
∴ ∠OAD = ∠ODA.
∵ ∠EDA = ∠ADB,
∴ ∠OAD = ∠EDA,
∴ OA//DE.
∵ AE⊥CE,
∴ OA⊥AE.
∵ OA 是⊙O 的半径,
∴ AE 是⊙O 的切线.

(3)解:如图2,连接 AO 并延长交 BC 于点 F.
∵ ∠E = ∠EAF = ∠ECF = 90°,
∴ 四边形 AECF 是矩形,
∴ AE = CF,CE = AF = 8,∠AFB = 90°,
∴ OF = AF - AO = 8 - 5 = 3,
易知 OF 为△BCD 的中位线,
∴ CD = 2OF = 6,CF = BF = √(OB² - OF²) = √(5² - 3²) = 4,
∴ AE = 4,ED = CE - CD = 8 - 6 = 2,
∴ AD = √(DE² + AE²) = √(2² + 4²) = 2√5.

(4)解:①(2)中的结论成立.理由如下:
如图3,连接 AO 并延长交 BC 于点 G,连接 OB,OC.
同理(1)可得 AB = AC,
∵ OB = OC,
∴ AG 垂直平分 BC,
∴ ∠AGB = 90°.
∵ AE//BC,
∴ ∠OAE = ∠AGB = 90°.
∵ OA 是⊙O 的半径,
∴ AE 是⊙O 的切线.

②2√5
如图4,设AC与BD相交于点H,连接BO并延长交⊙O于点P,连接AP.
∵ BD⊥AC,
∴ ∠CHB=90°,
∴ ∠HBC + ∠ACB = 90°.
∵ BP是⊙O的直径,
∴ ∠BAP=90°,
∴ ∠APB + ∠ABP =90°.
∵ ∠APB = ∠ACB,
∴ ∠ABP = ∠HBC,
∴ ⌢AP = ⌢DC,
∴ AP = DC = 4. 在Rt△ABP中,AB = 8,
∴ BP = √(AB² + AP²) =√(8² + 4²) =4√5,
∴ ⊙O 的半径为2√5.
③50
在△ABC中,
∵ ∠ACB = 80°,
∴ ∠BAC + ∠ABC = 100°,∠ADB =80°.
∵ 点M为△ABC的内心,
∴ AM,BM分别为∠BAC,∠ABC的平分线,
∴ ∠BAM = 1/2 ∠BAC,∠ABM = 1/2 ∠ABC,
∴ ∠BAM +∠ABM = 1/2(∠BAC + ∠ABC) = 1/2 ×100° = 50°,
∴ ∠AMD =50°.在△AMD中,∠MAD = 180° - ∠AMD - ∠ADM = 180° -50° -80° =50°.

(5)解:①如图5,连接 OA.
∵ ⌢AD = ⌢AD,
∴ ∠AOD = 2∠ACD.
∵ ∠ACD = 30°,
∴ ∠AOD = 60°.
又
∵ OA = OD,
∴ △OAD 是等边三角形,
∴ OA = AD.
∵ AD = 4,
∴ OA = 4.
l⌢_BCD = 1/2 ×2π ×4 =4π.
设围成的圆锥的底面半径为 r,则 2πr =4π.
解得 r =2,即该圆锥底面圆的半径为2.

②如图5,过点 O 作 ON⊥AD 于点 N.
∵ BD 是⊙O 的直径,
∴ ∠BAD = 90°.
∵ ∠ACD = 30°,
∴ ∠ABD = 30°,
∴ BD = 2AD = 8.
由①知⌢AD的长= (60π ×4)/180 = 4π/3.
∵ ON⊥AD,OA = OD,
∴ AN = DN = 1/2 AD = 1/2 ×4 =2,
∴ ON = √(OA² - AN²) = √(4² - 2²) = 2√3,
∴ S_△AOD = 1/2 AD · ON = 1/2 ×4 ×2√3 =4√3.
又
∵ S_扇形OAD = (60π × OA²)/360 = (60π ×4²)/360 = 8π/3,
∴ S_阴影 = S_扇形OAD - S_△AOD = 8π/3 -4√3.