19. [2024浙江中考] 如图,在圆内接四边形ABCD中,AD ＜ AC,∠ADC ＜ ∠BAD,延长AD至点E,使AE = AC,延长BA至点F,连接EF,使∠AFE = ∠ADC.
(1)若∠AFE = 60°,CD为直径,求∠ABD的度数.
(2)求证:①EF // BC;②EF = BD.

(1)解:$\because CD$为直径,$\therefore ∠ CAD = 90°.$
$\because ∠ ADC = ∠ AFE = 60°$,$\therefore ∠ ACD = 90° - 60° = 30°$,
$\therefore ∠ ABD = ∠ ACD = 30°.$
(2)证明:①如图,延长$AB$,
$\because$ 四边形$ABCD$是圆内接四边形,
$\therefore ∠ CBM = 180° - ∠ ABC = ∠ ADC.$
又$\because ∠ AFE = ∠ ADC$,
$\therefore ∠ AFE = ∠ CBM$,$\therefore EF // BC.$

②如图,过点$D$作$DG // BC$交$\odot O$于点$G$,连接$AG$,$CG.$
$\because DG // BC$,$\therefore ∠ CDG = ∠ BCD$,
$\therefore \overset{\frown}{CG} = \overset{\frown}{BD}$,$\therefore CG = BD.$
$\because$ 四边形$ACGD$是圆内接四边形,
$\therefore ∠ GDE = 180° - ∠ ADG = ∠ ACG.$
$\because EF // BC$,$DG // BC$,$\therefore EF // DG$,
$\therefore ∠ DEF = ∠ GDE$,$\therefore ∠ DEF = ∠ ACG.$
$\because ∠ AFE = ∠ ADC$,$∠ ADC = ∠ AGC$,
$\therefore ∠ AFE = ∠ AGC.$
在$△ AEF$和$△ ACG$中,$\begin{cases} ∠ AEF = ∠ ACG, \\ ∠ AFE = ∠ AGC, \\ AE = AC, \end{cases}$
$\therefore △ AEF ≌ △ ACG(\mathrm{AAS})$,
$\therefore EF = CG$,$\therefore EF = BD.$

20. 推理能力 如图1，点A,B,C,D均在$\odot O$上，$AB=2$，$CD=1$，$AD ⊥ BD$于点D，直线AD,BC相交于点E.
(1)求$∠ E$的度数.
(2)如果点C,D在$\odot O$上运动，且保持弦CD的长度不变，那么直线AD,BC相交所成的锐角的度数是否改变？试就以下两种情况进行探究，并说明理由.
①如图2，弦AB与弦CD交于点F.
②如图3，弦CD在弦AB下方.

(1)解:$\because AD ⊥ BD$,$\therefore ∠ ADB = 90°$,
$\therefore$ 弦$AB$是$\odot O$的直径.
如图1,连接$OD$,$OC$,则$OD = OC = 1$,
$\because CD = 1$,$\therefore △ DOC$是等边三角形,
$\therefore ∠ DOC = 60°$,$\therefore ∠ DBC = 30°.$
$\because ∠ EDB = 90°$,$\therefore ∠ E = 60°.$
(2)直线$AD$,$BC$相交所成的锐角的度数不改变. 理由如下：
①如图2,连接$OD$,$OC$. 由(1)知$∠ DOC = 60°$.
$\because ∠ CDB = \frac{1}{2}∠ BOC$,$∠ DCB = \frac{1}{2}∠ DOB$,$∠ DBE = ∠ CDB + ∠ DCB$,
$\therefore ∠ DBE = \frac{1}{2}∠ BOC + \frac{1}{2}∠ DOB = \frac{1}{2}∠ DOC = 30°.$
$\because ∠ EDB = 90°$,$\therefore ∠ E = 60°.$
②如图3,连接$OD$,$OC$.
由(1)知$∠ DOC = 60°$,$\therefore ∠ DBC = 30°.$
$\because ∠ EDB = 90°$,$\therefore ∠ BED = 60°.$

如图1,BD是$\odot O$的直径,点A,C在圆上,连接AB,BC,AC,CD,BD与AC交于点E.

(1)若$∠ CBD = 20°$,则$∠ BAC =$ $°$.
(2)连接OA,若$OA // CD$,且$∠ BDC = 50°$,则$∠ OBA =$ $°$.
(3)若$CD = OB$,则$∠ BAC =$ $°$.
(4)当$BD ⊥ AC$时,如图2.
①若$AC = 8$,$CD = 2\sqrt{5}$,则直径BD的长为.
②若$AB = 4$,$CD = 2$,则直径BD的长为.
③将题目中“BD是$\odot O$的直径”改为“BD是$\odot O$的弦”,若$AB = 4$,$CD = 2$,则$\odot O$的直径长为.

(1)70
$BD$是$\odot O$的直径$\to ∠ BCD = 90° \to ∠ BDC = 90° - ∠ CBD = 70° \to ∠ BAC = ∠ BDC = 70°.$
(2)25
$OA // CD \to ∠ AOD = ∠ BDC = 50° \to ∠ OBA = \frac{1}{2}∠ AOD = 25°.$
(3)60
连接$OC. CD = OB = OC = OD \to △ OCD$是等边三角形$\to ∠ BDC = 60° \to ∠ BAC = ∠ BDC = 60°.$
(4)①10
$\because BD ⊥ AC$,$AC = 8$,$\therefore AE = CE = 4$,在$\mathrm{Rt}△ CDE$中,$CD = 2\sqrt{5}$,
$\therefore DE = \sqrt{CD^2 - CE^2} = 2$,$\therefore OE = OD - DE = OD - 2$. 连接$OC$,则$OC = OD$,在$\mathrm{Rt}△ COE$中,$OC^2 = OE^2 + CE^2$,$\therefore OC^2 = (OC - 2)^2 + 4^2$,$\therefore OC = 5$,$\therefore BD = 2OC = 10.$
②$2\sqrt{5}$
$\because BD$是$\odot O$的直径,$\therefore ∠ BCD = 90°. \because BD ⊥ AC$,$\therefore BD$垂直平分$AC$,$\therefore BC = AB = 4$. 在$\mathrm{Rt}△ BCD$中,$BD = \sqrt{BC^2 + CD^2} = \sqrt{4^2 + 2^2} = 2\sqrt{5}.$
③$2\sqrt{5}$
如图,连接$BO$并延长交$\odot O$于点$F$,连接$AF. \because BF$为$\odot O$的直径,$\therefore ∠ BAF = 90°$,即$∠ BAC + ∠ CAF = 90°. \because BD ⊥ AC$,
$\therefore ∠ DCE + ∠ CDB = 90°. \because ∠ BAC = ∠ CDB$,$\therefore ∠ CAF = ∠ DCE$,$\therefore \overset{\frown}{CF} = \overset{\frown}{AD}$,$\therefore \overset{\frown}{CD} = \overset{\frown}{AF}$,$\therefore AF = CD = 2$. 在$\mathrm{Rt}△ ABF$中,$BF = \sqrt{AB^2 + AF^2} = \sqrt{4^2 + 2^2} = 2\sqrt{5}.$