4.(乌鲁木齐模拟)如图,AB是$\odot O$的直径.点C在$\odot O$上,点D是OB的中点,过点D作AB的垂线交AC的延长线于点F,过点C作$\odot O$的切线交FD于点E.
(1)求证:$CE=EF$.
(2)若$\cos A=\dfrac{3}{5},EF=\dfrac{5}{2}$,求$\odot O$的半径.

4.(1) 证明: 连接 $OC$, 如图所示. $\because AB$ 是$\odot O$ 的直径,
$\therefore OC = OA, \therefore ∠ A = ∠ OCA. \because CE$ 是$\odot O$ 的切线, $OC$ 是$\odot O$ 的半径, $\therefore OC ⊥ CE, \therefore ∠ OCE = 90°, \therefore ∠ ECF + ∠ OCA = 180° - ∠ OCE = 90°$,
$\therefore ∠ ECF + ∠ A = 90°. \because FD ⊥ AB, \therefore △ FDA$ 是直角三角形. 在 $\mathrm{Rt}△ FDA$ 中, $∠ F + ∠ A = 90°$,
$\therefore ∠ ECF = ∠ F, \therefore CE = EF.$
(2) 解: 连接 $BC$ 交 $FD$ 于点 $H$, 如图所示. 由(1)的结论可知$∠ ECF = ∠ F, CE = EF. \because EF = \frac{5}{2}, \therefore CE = EF = \frac{5}{2}. \because AB$ 是$\odot O$ 的直径, $\therefore ∠ ACB = 90°, \therefore ∠ FCH = 180° - ∠ ACB = 90°, \therefore △ ABC$ 和 $△ FCH$ 都是直角三角形, $\therefore ∠ F + ∠ EHC = 90°.$ 又$\because ∠ FCH = ∠ ECF + ∠ ECH = 90°, ∠ ECF = ∠ F, \therefore ∠ EHC = ∠ ECH$,
$\therefore HE = CE = \frac{5}{2}, \therefore FH = HE + EF = 5. \because$ 点 $D$ 是 $OB$ 的中点, 设 $OD = BD = a, \therefore OB = OD + BD = 2a, \therefore OA = OB = 2a, \therefore AD = OA + OD = 3a.$
$\because FD ⊥ AB, \therefore ∠ BDH = ∠ FDA = 90°, \therefore △ FDA$ 和 $△ BDH$ 都是直角三角形. 在 $\mathrm{Rt}△ FDA$ 中, $∠ F + ∠ A = 90°, \cos A = \frac{AD}{AF} = \frac{3}{5}, \therefore \frac{3a}{AF} = \frac{3}{5}, \therefore AF = 5a.$ 由勾股定理, 得 $DF = \sqrt{AF^2 - AD^2} = \sqrt{(5a)^2 - (3a)^2} = 4a.$ 在 $\mathrm{Rt}△ ABC$ 中, $∠ B + ∠ A = 90°, \therefore ∠ B = ∠ F.$ 在 $\mathrm{Rt}△ BDH$ 中, $\tan B = \frac{DH}{BD} = \frac{DH}{a}.$ 在 $\mathrm{Rt}△ FDA$ 中, $\tan F = \frac{AD}{FD} = \frac{3a}{4a} = \frac{3}{4}$,
$\therefore \frac{DH}{a} = \frac{3}{4}, \therefore DH = \frac{3a}{4}, \therefore FH = FD - DH = 4a - \frac{3a}{4} = \frac{13a}{4}.$ 又$\because FH = 5, \therefore \frac{13a}{4} = 5,$ 解得 $a = \frac{20}{13}$,
$\therefore OA = 2a = \frac{40}{13},$ 即$\odot O$ 的半径为 $\frac{40}{13}$.

5.(吐鲁番二模)如图,在菱形$ABCD$中,$DH⊥ AB$于点$H$,以$DH$为直径的$\odot O$分别交$AD$,$BD$于点$E$,$F$,连接$EF$.
(1)求证:①$CD$是$\odot O$的切线.
②$△ DEF ∽ △ DBA$.

(2)若$AB=5$,$DB=6$,求$\sin ∠ DFE$的值.

5.(1) 证明: ①$\because$ 四边形 $ABCD$ 是菱形, $\therefore AB // CD.$
$\because DH ⊥ AB, \therefore ∠ CDH = ∠ DHA = 90°, \therefore CD ⊥ OD. \because OD$ 为$\odot O$ 的半径, $\therefore CD$ 是$\odot O$ 的切线.
②如图, 连接 $HF, \therefore ∠ DEF = ∠ DHF. \because DH$ 为$\odot O$ 的直径, $\therefore ∠ DFH = 90°, \therefore ∠ DHF = 90° - ∠ BDH. \because ∠ DHB = 90°, \therefore ∠ DBA = 90° - ∠ BDH, \therefore ∠ DHF = ∠ DBA = ∠ DEF. \because ∠ EDF = ∠ BDA, \therefore △ DEF ∽ △ DBA.$
(2) 解: 如图, 连接 $AC$ 交 $BD$ 于点 $G. \because$ 四边形 $ABCD$ 为菱形, $BD = 6, \therefore AC ⊥ BD, AG = GC, DG = GB = 3.$ 在 $\mathrm{Rt}△ AGB$ 中, $AG = \sqrt{AB^2 - GB^2} = 4, \therefore AC = 2AG = 8. \because S_{\mathrm{菱形}ABCD} = \frac{1}{2} AC · BD = AB · DH, \therefore DH = \frac{\frac{1}{2} × 8 × 6}{5} = \frac{24}{5}.$ 由$△ DEF ∽ △ DBA$ 知$∠ DFE = ∠ DAH, \therefore \sin∠ DFE = \sin∠ DAH = \frac{DH}{AD} = \frac{\frac{24}{5}}{5} = \frac{24}{25}.$

6.(乌鲁木齐模拟)如图,已知AB是$\odot O$的直径,点E是$\odot O$上异于A,B的点,点F是$\overset{\frown}{EB}$的中点,连接AE,AF,BF,过点F作$FC ⊥ AE$交AE的延长线于点C,交AB的延长线于点D,$∠ ADC$的平分线DG交AF于点G,交FB于点H.
(1)求证:CD是$\odot O$的切线.
(2)求$\sin∠ FHG$的值.
(3)若$GH=4\sqrt{2},HB=2$,求$\odot O$的直径.

6.(1) 证明: 连接 $OF. \because OA = OF, \therefore ∠ OAF = ∠ OFA. \because \overset{\frown}{EF} = \overset{\frown}{FB}, \therefore ∠ CAF = ∠ FAB, \therefore ∠ CAF = ∠ AFO, \therefore OF // AC. \because AC ⊥ CD, \therefore OF ⊥ CD. \because OF$ 是$\odot O$ 的半径, $\therefore CD$ 是$\odot O$ 的切线.
(2) 解: $\because AB$ 是$\odot O$ 的直径, $\therefore ∠ AFB = 90°. \because OF ⊥ CD, \therefore ∠ OFD = ∠ AFB = 90°, \therefore ∠ AFO = ∠ DFB. \because ∠ OAF = ∠ OFA, \therefore ∠ DFB = ∠ OAF. \because DG$ 平分$∠ ADF, \therefore ∠ ADG = ∠ FDG. \because ∠ FGH = ∠ OAF + ∠ ADG, ∠ FHG = ∠ DFB + ∠ FDG, \therefore ∠ FGH = ∠ FHG = 45°, \therefore \sin∠ FHG = \frac{\sqrt{2}}{2}.$
(3) 解: 过点 $H$ 作 $HM ⊥ DF$ 于点 $M, HN ⊥ AD$ 于点 $N. \because DH$ 平分$∠ ADF$,
$\therefore HM = HN, \therefore \frac{S_{△ DHF}}{S_{△ DHB}} = \frac{FH}{HB} = \frac{\frac{1}{2} DF · HM}{\frac{1}{2} DB · HN} = \frac{DF}{DB}.$ 由(2)易得$△ FGH$ 是等腰直角三角形, $GH = 4\sqrt{2}, \therefore FH = FG = 4, \therefore \frac{DF}{DB} = \frac{4}{2} = 2.$ 设 $DB = k$, 则 $DF = 2k. \because ∠ FDB = ∠ ADF, ∠ DFB = ∠ DAF, \therefore △ DFB ∽ △ DAF, \therefore \frac{DF}{DA} = \frac{DB}{DF}, \therefore DF^2 = DB · DA, \therefore AD = 4k. \because DG$ 平分$∠ ADF$,
$\therefore ∠ FDH = ∠ ADG, \therefore △ FDH ∽ △ ADG, \therefore \frac{FH}{AG} = \frac{DF}{AD} = \frac{1}{2}, \therefore AG = 8, \therefore AF = AG + FG = 12.$
$\because ∠ AFB = 90°, FB = FH + HB = 6, \therefore AB = \sqrt{AF^2 + BF^2} = \sqrt{12^2 + 6^2} = 6\sqrt{5}, \therefore \odot O$ 的直径为 $6\sqrt{5}.$