1.(乌鲁木齐模拟)如图，AB是$\odot O$的直径，C，D是$\odot O$上的点，E是AB上一点，连接CE并延长交AD于点F，延长AD，CB交于点G，$\overset{\frown}{AC}=\overset{\frown}{CD}$，$∠ BCE=∠ G$，连接AC.
(1)求证：$CF⊥ AB$.
(2)若$EF=2$，$AG=14$，求$\odot O$的半径.

1.(1) 证明: $\because AB$ 为$\odot O$ 的直径, $\therefore ∠ ACB = 90°$,
$\therefore ∠ CAD + ∠ G = 90°. \because \overset{\frown}{AC}=\overset{\frown}{CD}, \therefore ∠ CAD = ∠ CBE, \therefore ∠ CBE + ∠ G = 90°. \because ∠ BCE = ∠ G$,
$\therefore ∠ CBE + ∠ BCE = 90°, \therefore ∠ CEB = 90°,$ 即 $CF ⊥ AB$.
(2) 解: 由题意, 得$∠ ACE + ∠ BCE = 90°, ∠ CAG + ∠ G = 90°. \because ∠ BCE = ∠ G, \therefore ∠ CAG = ∠ ACE$,
$\therefore AF = CF = FG = \frac{1}{2}AG = 7. \because EF = 2, \therefore CE = CF - EF = 7 - 2 = 5.$ 在 $\mathrm{Rt}△ AEF$ 中, $AE = \sqrt{AF^2 - EF^2} = \sqrt{7^2 - 2^2} = 3\sqrt{5}.$ 在 $\mathrm{Rt}△ AEC$ 中,
$AC = \sqrt{AE^2 + CE^2} = \sqrt{(3\sqrt{5})^2 + 5^2} = \sqrt{70}.$
$\because ∠ CAB = ∠ EAC, ∠ ACB = ∠ AEC = 90°$,
$\therefore △ CAB ∽ △ EAC, \therefore \frac{AB}{AC} = \frac{AC}{AE},$ 即 $\frac{AB}{\sqrt{70}} = \frac{\sqrt{70}}{3\sqrt{5}}$,
$\therefore AB = \frac{14\sqrt{5}}{3}, \therefore AO = \frac{7\sqrt{5}}{3}, \therefore \odot O$ 的半径为 $\frac{7\sqrt{5}}{3}$.

2.(乌鲁木齐模拟)如图,在四边形ABCD中,BD=CD,∠C=∠BAD,以AB为直径的⊙O经过点D,且与边CD交于点E,连接AE,BE.
(1)求证:BC为⊙O的切线.

(2)若AD=2,cos∠BAD=$\dfrac{1}{3}$,求BE的长.

2.(1) 证明: $\because BD = CD, \therefore ∠ C = ∠ DBC. \because ∠ C = ∠ BAD, \therefore ∠ BAD = ∠ DBC. \because AB$ 为$\odot O$ 的直径, $\therefore ∠ ADB = 90°, \therefore ∠ BAD + ∠ DBA = 90°$,
$\therefore ∠ DBC + ∠ DBA = 90°,$ 即 $∠ CBA = 90°, \therefore AB ⊥ BC.$ 又$\because OB$ 为$\odot O$ 的半径, $\therefore BC$ 为$\odot O$ 的切线.
(2) 解: 如图, 作 $BF ⊥ CD$ 于点 $F. \because AD = 2$,
$\cos∠ BAD = \frac{1}{3}, ∠ ADB = 90°, \therefore AB = \frac{AD}{\cos∠ BAD} = 6.$ 在 $\mathrm{Rt}△ ABD$ 中, 由勾股定理, 得 $BD = \sqrt{AB^2 - AD^2} = 4\sqrt{2}, \therefore CD = BD = 4\sqrt{2}. \because ∠ C = ∠ BAD, ∠ BEC = ∠ BAD, \therefore ∠ C = ∠ BEC, \therefore BC = BE. \because BF ⊥ CD, \therefore EF = CF = \frac{1}{2}CE.$
$\because \cos∠ BAD = \frac{1}{3}, \therefore \cos∠ BEC = \frac{EF}{BE} = \frac{1}{3}.$ 设 $EF = x,$ 则 $BC = BE = 3x, CE = 2EF = 2x. \because ∠ DBC = ∠ BAD, ∠ BEC = ∠ BAD, \therefore ∠ DBC = ∠ BEC.$
又$\because ∠ C = ∠ C, \therefore △ CBE ∽ △ CDB, \therefore \frac{BC}{CE} = \frac{CD}{BC}$,
即 $BC^2 = CE · CD, \therefore (3x)^2 = 2x × 4\sqrt{2},$ 解得 $x = \frac{8\sqrt{2}}{9}, \therefore BE = 3x = \frac{8\sqrt{2}}{3}.$

3. 如图，CD是$△ ABC$的高，以AB为直径作$\odot O$交CB的延长线于点E，连接DE，$DE=CD$.
(1)求证:$DE$是$\odot O$的切线.
(2)若$CD=4$，$BD=2$，求$△ ABC$的面积.

3.(1) 证明: 如图, 连接 $OE$, 则 $OE = OB, \therefore ∠ OEB = ∠ OBE. \because ∠ OBE = ∠ CBD, \therefore ∠ OEB = ∠ CBD.$
$\because DE = CD, \therefore ∠ DEC = ∠ DCE. \because CD$ 是$△ ABC$ 的高, $\therefore CD ⊥ AB, \therefore ∠ ADC = 90°, \therefore ∠ OED = ∠ OEB + ∠ DEC = ∠ CBD + ∠ DCE = 90°, \therefore OE ⊥ DE. \because OE$ 是$\odot O$ 的半径, $\therefore DE$ 是$\odot O$ 的切线.
(2) 解: $\because ∠ OED = 90°, \therefore OE^2 + DE^2 = OD^2. \because DE = CD = 4, BD = 2, \therefore OD = OB + 2. \because OE = OB$,
$\therefore OB^2 + 4^2 = (OB + 2)^2,$ 解得 $OB = 3, \therefore AB = 2OB = 6, \therefore S_{△ ABC} = \frac{1}{2} AB · CD = \frac{1}{2} × 6 × 4 = 12$,
$\therefore △ ABC$ 的面积为 12.