4. 如图,$AB$为$\odot O$的直径,点$C$在$\odot O$上,$∠ ACB$的平分线交$\odot O$于点$D$,过点$D$作$DE// AB$,交$CB$的延长线于点$E$.
(1)求证:$ED$是$\odot O$的切线;
(2)若$AC=3\sqrt{2}$,$BC=\sqrt{2}$,求$BD$,$CD$的长.

4.(1)证明:如答图,连接$OD$.
$\because CD$是$∠ACB$的平分线,$\therefore ∠ACD=∠BCD$,
$\therefore ∠AOD=∠BOD$.
$\because AB$为$\odot O$的直径,
$\therefore ∠AOD=∠BOD=\frac{1}{2}×180°=90°$,
$\therefore OD⊥AB$.
$\because DE// AB$,$\therefore OD⊥DE$.
$\because OD$为$\odot O$的半径,$\therefore ED$是$\odot O$的切线.
第4题答图
(2)解:$\because AB$为$\odot O$的直径,
$\therefore ∠ACB=90°$,$∠ADB=90°$.
$\because AC=3\sqrt{2}$,$BC=\sqrt{2}$,$\therefore AB=\sqrt{AC^2+BC^2}=\sqrt{(3\sqrt{2})^2+(\sqrt{2})^2}=2\sqrt{5}$.
$\because ∠ACD=∠BCD$,$\therefore \overset{\frown}{AD}=\overset{\frown}{BD}$,$\therefore AD=BD$.
$\because AD^2+BD^2=AB^2$,$\therefore AD=BD=\sqrt{10}$.
如答图,过点$B$作$BH⊥CD$于点$H$.
$\because ∠BCD=\frac{1}{2}∠ACB=45°$,$\therefore BH=CH$.
由$BH^2+CH^2=CB^2$,得$BH=CH=1$,
$\therefore DH=\sqrt{BD^2-BH^2}=\sqrt{(\sqrt{10})^2-1^2}=3$,
$\therefore CD=CH+DH=1+3=4$.

5. (2025·镇江期中) 如图①,点 A,B,C,D 在$\odot O$上,且$\overset{\frown}{AD}=\overset{\frown}{BC}$,E 是 AB(不是直径)延长线上一点,且$BE=AB$,F 是 EC 的中点.
(1)探索 BF 与 BD 之间的数量关系,并说明理由；
(2)如图②,设 G 是 BD 的中点,过点 B 作$BP ⊥ AE$交$\odot O$于点 P,连接 PG,PF,求证:$PG=PF$.

5.(1)解:$BF=\frac{1}{2}BD$,
理由:如答图①,连接$AC$,
第5题答图①
$\because BE=AB$,$F$是$EC$的中点,$\therefore BF=\frac{1}{2}AC$.
$\because \overset{\frown}{AD}=\overset{\frown}{BC}$,$\therefore \overset{\frown}{BD}=\overset{\frown}{AC}$,$\therefore BD=AC$,$\therefore BF=\frac{1}{2}BD$.
(2)证明:如答图②,连接$AC$.由(1)知,$BF=\frac{1}{2}BD$,
$\because G$是$BD$的中点,$\therefore BG=\frac{1}{2}BD$,$\therefore BG=BF$.
$\because BE=AB$,$F$是$EC$的中点,$\therefore BF// AC$,$\therefore ∠1=∠3$.
$\because \overset{\frown}{AD}=\overset{\frown}{BC}$,$\therefore ∠1=∠2$,$\therefore ∠2=∠3$.
$\because BP⊥AE$,$\therefore ∠PBA=∠PBE=90°$,
$\therefore ∠PBG=∠PBF$,
在$△ PBG$和$△ PBF$中,$\begin{cases} PB=PB, \\ ∠PBG=∠PBF, \\ BG=BF, \end{cases}$
$\therefore △ PBG≌△ PBF(\mathrm{SAS})$,$\therefore PG=PF$.
第5题答图②

6. 在平面直角坐标系中,已知点 $A(3,0),B(-1,0)$,C 是 y 轴上一动点, 当 $∠ BCA=45°$ 时,求点C的坐标.

6.解:如答图,
第6题答图
先作等腰直角$△ PAB$,再以点$P$为圆心,$PA$的长为半径作$\odot P$交$y$轴于点$C$.
作$PD⊥y$轴于点$D$,可得点$P(1,2)$,$PA=2\sqrt{2}$,
$\therefore PC=2\sqrt{2}$,
$\therefore CD=\sqrt{(2\sqrt{2})^2-1^2}=\sqrt{7}$,
$\therefore OC=2+\sqrt{7}$,
$\therefore C(0,2+\sqrt{7})$,同理可得$C'(0,-2-\sqrt{7})$.
综上所述,点$C$的坐标为$(0,2+\sqrt{7})$或$(0,-2-\sqrt{7})$.