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.

(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,
$\because BE=AB$,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,
$\begin{cases} PB=PB, \\ ∠ PBG=∠ PBF, \\ BG=BF, \end{cases}$
$\therefore △ PBG ≌ △ PBF(\mathrm{SAS})$,$\therefore PG=PF$.

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,

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$(0,-2-\sqrt{7})$.