21.「2025江苏宿迁宿豫月考，★☆」（10分）如图，四边形ABCD内接于$\odot O$，$∠ 1=∠ 2$，延长BC到点E，使得$CE=AB$，连接ED。
（1）求证：$BD=ED$。
（2）若$AB=4$，$BC=6$，$∠ ABC=60°$，求四边形ABCD的面积。

21.解析 (1)证明:$\because$ 四边形ABCD内接于$\odot O, \therefore ∠ A+∠ BCD=180°, \because ∠ BCD+∠ DCE=180°, \therefore ∠ A=∠ DCE$,
$\because ∠ 1=∠ 2, \therefore \overset{\frown}{AD}=\overset{\frown}{CD}, \therefore AD=CD, \because AB=CE, \therefore △ ABD≌△ \mathrm{CED}(\mathrm{SAS}), \therefore BD=ED$.
(2)如图,过点A作$AN ⊥ BD$于点N,过点D作$DM ⊥ BE$于点M,
$\because BC=6,CE=AB=4, \therefore BE=BC+CE=10$,
$\because BD=ED,DM ⊥ BE, \therefore BM=ME=\frac{1}{2} BE=5$,
$\because ∠ ABC=60°,∠ 1=∠ 2, \therefore ∠ 1=∠ 2=30°$,
$\therefore DM=\frac{1}{2} BD,AN=\frac{1}{2} AB=2$,
$\because DM^2+BM^2=BD^2, \therefore DM=\frac{5\sqrt{3}}{3}$(舍负), $\therefore BD=\frac{10\sqrt{3}}{3}$,
$\therefore S_{\mathrm{四边形}ABCD}=S_{△ ABD}+S_{△ BCD}=\frac{1}{2} BD · AN + \frac{1}{2} BC · DM = \frac{1}{2} × \frac{10\sqrt{3}}{3} × 2 + \frac{1}{2} × 6 × \frac{5\sqrt{3}}{3} = \frac{25\sqrt{3}}{3}$.

22. 核心素养推理能力「★★☆」(10分)如图所示，$\odot O$的弦$AC=BD$，$AC$，$BD$交于点$E$，$F$为$\overset{\frown}{BC}$上一点，连接$AF$，$BF$，$AB$，$AD$.
(1)求证:$AE=BE$.
(2)若$AC ⊥ BD$，$\overset{\frown}{CF}=\overset{\frown}{CD}$，$AB=\sqrt{2}$，求$BF+CE$的值.

22.解析 (1)证明:$\because AC=BD, \therefore \overset{\frown}{AC}=\overset{\frown}{BD}$,
$\therefore \overset{\frown}{AC}-\overset{\frown}{CD}=\overset{\frown}{BD}-\overset{\frown}{CD}, \therefore \overset{\frown}{AD}=\overset{\frown}{BC}$,
$\therefore ∠ ABD=∠ BAC, \therefore AE=BE$.
(2)设AF交BD于点G,连接CD,如图,
$\because \overset{\frown}{CD}=\overset{\frown}{CF}, \therefore ∠ CAD=∠ CAF$,
$\because AC ⊥ BD, \therefore ∠ AED=∠ AEG=90°, \therefore ∠ ADE=∠ AGE$,
$\therefore AG=AD, \therefore △ ADG$为等腰三角形,$DE=GE$,
$\because ∠ ADE=∠ F,∠ AGE=∠ BGF, \therefore ∠ F=∠ BGF, \therefore BG=BF$,
$\because ∠ BDC=∠ BAC,∠ C=∠ ABD,∠ ABD=∠ BAC$,
$\therefore ∠ BDC=∠ C, \therefore CE=DE$.
$\therefore CE=GE, \therefore BF+CE=BG+GE=BE$.
$\because △ ABE$为等腰直角三角形,$AB=\sqrt{2}$,
$\therefore BE=AE=1, \therefore BF+CE=1$.

23.「2026浙江杭州上城期中，★★☆」(12分)如图，AD是$△ ABC$的外角$∠ EAC$的平分线，与$△ ABC$的外接圆交于点D，$∠ EAC=120°$.
(1)连接OB,OC,求$∠ OCB$的度数.
(2)连接DB,DC,求证:$DB=DC$.
(3)探究线段AD,AB,AC之间的数量关系,并证明你的结论.

23.解析 (1)$\because ∠ EAC=120°, \therefore ∠ BAC=60°$,
$\therefore ∠ BOC=2∠ BAC=120°$,
$\because OB=OC, \therefore ∠ OCB=\frac{180°-∠ BOC}{2}=30°$.
(2)证明:$\because AD$是$∠ EAC$的平分线,
$\therefore ∠ DAC=\frac{1}{2} ∠ EAC=\frac{1}{2} × 120°=60°$,
$\therefore ∠ DBC=∠ DAC=60°$,
$\because ∠ BDC=∠ BAC=60°, \therefore ∠ BDC=∠ DBC=60°$,
$\therefore △ BDC$是等边三角形,$\therefore BD=CD$.
(3)$AC=AD+AB$.证明如下:
如图,延长AD至点F,使DF=AB,连接CF,
$\because$ 四边形ABCD是$\odot O$的内接四边形,
$\therefore ∠ ADC+∠ ABC=180°$,
$\because ∠ ADC+∠ CDF=180°$,
$\therefore ∠ ABC=∠ CDF$,
由(2)知$△ BDC$是等边三角形,
$\therefore BC=CD,∠ BCD=60°, \therefore △ FDC≌△ \mathrm{ABC}(\mathrm{SAS})$,
$\therefore ∠ ACB=∠ DCF,AC=CF, \therefore ∠ ACF=∠ BCD=60°$,
$\therefore △ ACF$是等边三角形,$\therefore AC=AF=AD+DF=AD+AB$.