第67页

信息发布者:
B
$(-\sqrt{3},1)$
①②
(1) 证明:
$\because$ 四边形$ABCD$内接于$\odot O,$
$\therefore ∠ ABC+∠ ADC=180°。$
$\because ∠ ABC=60°,$
$\therefore ∠ ADC=120°。$
$\because DB$平分$∠ ADC,$
$\therefore ∠ ADB=∠ CDB=60°。$
$\because \overset{\frown}{AB}=\overset{\frown}{AB},$$\overset{\frown}{BC}=\overset{\frown}{BC},$
$\therefore ∠ ACB=∠ ADB=60°,$$∠ BAC=∠ CDB=60°,$
$\therefore ∠ ABC=∠ ACB=∠ BAC,$
$\therefore △ ABC$是等边三角形。
(2) 过点$A$作$AM⊥ CD,$交$CD$的延长线于点$M。$
$\therefore ∠ AMD=90°。$
$\because ∠ ADC=120°,$
$\therefore ∠ ADM=180°-∠ ADC=60°。$
在$\mathrm{Rt}△ AMD$中,$∠ DAM=30°,$
$\therefore DM=\frac{1}{2}AD=1,$
$\therefore AM=\sqrt{AD^2-DM^2}=\sqrt{3}。$
$\because CD=3,$
$\therefore CM=CD+DM=4。$
在$\mathrm{Rt}△ AMC$中,$AC=\sqrt{AM^2+CM^2}=\sqrt{19}。$
$\because △ ABC$是等边三角形,
$\therefore AB=BC=AC=\sqrt{19},$
$\therefore △ ABC$的周长为$3\sqrt{19}。$

(1) 证明:
$\because \overset{\frown}{BC}=\overset{\frown}{BC},$
$\therefore ∠ BAC=∠ CDB。$
$\because ∠ BAC=∠ ADB,$
$\therefore ∠ CDB=∠ ADB,$即$DB$平分$∠ ADC。$
$\because BD$平分$∠ ABC,$
$\therefore ∠ ABD=∠ CBD。$
$\because △ ABD$与$△ CBD$的内角和均为$180°,$
$\therefore ∠ BAD=∠ BCD。$
$\because$ 四边形$ABCD$是圆内接四边形,
$\therefore ∠ BAD+∠ BCD=180°,$
$\therefore ∠ BAD=∠ BCD=90°。$
(2) $\because ∠ BAD=90°,$
$\therefore BD$是圆的直径。
$\because ∠ ABD=∠ CBD,$
$\therefore \overset{\frown}{AD}=\overset{\frown}{CD},$
$\therefore AD=CD。$
$\because AC=AD,$
$\therefore AC=AD=CD,$
$\therefore △ ADC$是等边三角形,
$\therefore ∠ ADC=60°,$
$\therefore ∠ CDB=\frac{1}{2}∠ ADC=30°。$
在$\mathrm{Rt}△ BCD$中,$BD=2BC。$
$\because CF// AD,$
$\therefore ∠ F+∠ BAD=180°,$
$\therefore ∠ F=90°。$
$\because$ 四边形$ABCD$是圆内接四边形,
$\therefore ∠ ADC+∠ ABC=180°,$
$\therefore ∠ ABC=120°,$
$\therefore ∠ FBC=180°-∠ ABC=60°,$
$\therefore ∠ FCB=90°-60°=30°。$
在$\mathrm{Rt}△ BFC$中,$BF=\frac{1}{2}BC。$
$\because BF=2,$
$\therefore BC=4,$
$\therefore BD=8。$
$\because BD$是圆的直径,
$\therefore$ 该圆的半径为$\frac{1}{2}BD=4。$