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 (1) 证明：因为$\angle BAC = \angle ADB，$$\angle BAC = \angle CDB，$所以$\angle ADB = \angle CDB，$即$DB$平分$\angle ADC。$

 因为$BD$平分$\angle ABC，$所以$\angle ABD = \angle CBD。$

 又因为四边形$ABCD$是圆内接四边形，所以$\angle ABC + \angle ADC = 180^{\circ}，$即$\angle ABD + \angle CBD + \angle ADB + \angle CDB = 180^{\circ}，$$2(\angle ABD + \angle ADB) = 180^{\circ}，$所以$\angle ABD + \angle ADB = 90^{\circ}，$则$\angle BAD = 180^{\circ} - 90^{\circ} = 90^{\circ}。$

 (2) 解：因为$\angle BAE + \angle DAE = 90^{\circ}，$$\angle BAE = \angle ADE，$所以$\angle ADE + \angle DAE = 90^{\circ}，$则$\angle AED = 90^{\circ}。$

 因为$\angle BAD = 90^{\circ}，$所以$BD$是圆的直径，$BD$垂直平分$AC，$所以$AD = CD。$

 又因为$AC = AD，$所以$AC = AD = CD，$$\triangle ACD$是等边三角形，$\angle ADC = 60^{\circ}，$则$\angle BDC = \frac{1}{2}\angle ADC = 30^{\circ}。$

 因为$CF// AD，$所以$\angle F + \angle BAD = 180^{\circ}，$$\angle F = 90^{\circ}。$

 因为四边形$ABCD$是圆内接四边形，所以$\angle ADC + \angle ABC = 180^{\circ}，$又因为$\angle FBC + \angle ABC = 180^{\circ}，$所以$\angle FBC = \angle ADC = 60^{\circ}，$则$\angle FCB = 30^{\circ}，$$BC = 2BF = 4。$

 因为$BD$是圆的直径，$\angle BCD = 90^{\circ}，$$\angle BDC = 30^{\circ}，$所以$BD = 2BC = 8，$此圆的半径是$4。$

 (1) 证明：连接$OD。$

 因为$OD = OA，$所以$\angle ODA = \angle A = 30^{\circ}。$

 则$\angle DOB = \angle ODA + \angle A = 60^{\circ}。$

 所以$\angle ODB = 180^{\circ} - \angle DOB - \angle B = 180^{\circ} - 60^{\circ} - 30^{\circ} = 90^{\circ}，$即$OD\perp BD。$

 又因为$OD$是$\odot O$的半径，所以$BD$是$\odot O$的切线。

 (2) 解：因为$OD\perp BD，$$\angle B = 30^{\circ}，$所以$OB = 2OD。$

 又因为$AB = 3，$$OA = OD，$$AB = OA + OB = 3OD = 3，$所以$OD = 1，$$OB = 2。$

 则$BD = \sqrt{OB^{2} - OD^{2}} = \sqrt{2^{2} - 1^{2}} = \sqrt{3}。$

 所以涂色部分的面积$S = S_{\triangle BDO} - S_{扇形 DOC} = \frac{1}{2}\times1\times\sqrt{3} - \frac{60\pi\times1^{2}}{360} = \frac{\sqrt{3}}{2} - \frac{1}{6}\pi。$

解: (1) $\odot M$与$x$轴相切。

 理由：连接$CM。$

 因为$AC$平分$\angle OAM，$所以$\angle OAC = \angle CAM。$

 又因为$AM = MC，$所以$\angle CAM = \angle ACM，$则$\angle OAC = \angle ACM，$所以$OA// MC。$

 因为$OA\perp x$轴，所以$MC\perp x$轴。

 又因为$CM$是$\odot M$的半径，所以$\odot M$与$x$轴相切。

 (2) 过点$M$作$MN\perp y$轴于点$N，$则$AN = BN = \frac{1}{2}AB。$

 因为$OA\perp x$轴，$MC\perp x$轴，$MN\perp y$轴，所以$\angle MCO = \angle AOC = \angle MNA = 90^{\circ}，$四边形$MNOC$是矩形，$MN = OC，$$MC = ON = 5。$

 设$AO = m，$则$MN = OC = 6 - m，$$AN = 5 - m。$

 在$Rt\triangle ANM$中，由勾股定理$AM^{2}=AN^{2}+MN^{2}，$可得$5^{2}=(5 - m)^{2}+(6 - m)^{2}，$

 即$25 = 25 - 10m + m^{2} + 36 - 12m + m^{2}，$

 $2m^{2}-22m + 36 = 0，$$m^{2}-11m + 18 = 0，$

 $(m - 2)(m - 9) = 0，$解得$m_{1}=2，$$m_{2}=9$（不合题意，舍去）。

 所以$AN = 3，$$AB = 6。$

 (3) 连接$AD$交$CM$于点$E。$

 因为$BD$是$\odot M$的直径，所以$\angle BAD = 90^{\circ}，$$AD// x$轴，$AD\perp MC，$四边形$OAEC$为矩形，$AE = DE，$$AE = OC。$

由

 (2)可得$MN = OC = 4，$$OA = 2，$所以点$C$的坐标为$(4,0)，$$AD = 2AE = 2OC = 8，$点$D$的坐标为$(8,-2)。$

 设直线$CD$对应的函数解析式为$y = kx + b，$则$\begin{cases}4k + b = 0\\8k + b = -2\end{cases}，$

 两式相减得：$4k=-2，$解得$k = -\frac{1}{2}，$

 把$k = -\frac{1}{2}$代入$4k + b = 0，$得$-2 + b = 0，$$b = 2。$

 所以直线$CD$对应的函数解析式为$y = -\frac{1}{2}x + 2。$