4. 如图,在四边形ABCD中,AC,BD相交于点E,且AB=AC=AD,经过A,C,D三点的$\odot O$交BD于点F,连接CF.
(1)求证:$CF=BF$;
(2)若$CD=CB$,求证:CB是$\odot O$的切线.

(1)证明:$\because AB=AC,\therefore∠ ACB=∠ ABC$.
$\because AB=AD,\therefore∠ ADB=∠ ABD$.
又$\because∠ ADB=∠ ACF,\therefore∠ ACF=∠ ABD$,
$\therefore∠ ACB-∠ ACF=∠ ABC-∠ ABD$,
即$∠ BCF=∠ CBF,\therefore CF=BF$.
(2)如答图,连接CO并延长交$\odot O$于点G,连接GF.

$\because CG$为$\odot O$的直径,$\therefore∠ GFC=90°$,
$\therefore∠ G+∠ GCF=90°$.
$\because∠ CDB=∠ G,\therefore∠ CDB+∠ GCF=90°$.
$\because CD=CB,\therefore∠ CDB=∠ CBD$.
$\because∠ BCF=∠ CBF,\therefore∠ BCF=∠ CDB$,
$\therefore∠ BCF+∠ GCF=90°,\therefore∠ BCG=90°$,
$\therefore CG⊥ BC.\because CG$为$\odot O$的直径,
$\therefore CB$是$\odot O$的切线.

5. 如图，O为正方形ABCD的对角线AC上一点，以点O为圆心，OA长为半径的$\odot O$与BC相切于点E.
(1)求证:CD是$\odot O$的切线；
(2)若正方形ABCD的边长为10，求$\odot O$的半径.

(1)证明:如答图,连接OE,过点O作$OF⊥ CD$于点F. $\because BC$切$\odot O$于点E,$\therefore OE⊥ BC,OE=OA$. 又$\because AC$为正方形ABCD的对角线,$\therefore∠ ACB=∠ ACD,\therefore OF=OE=OA,\therefore CD$是$\odot O$的切线.

(2)解:$\because$正方形ABCD的边长为10,
$\therefore AB=BC=10,∠ B=90°,∠ ACB=45°$,
$\therefore AC=\sqrt{AB^2+BC^2}=\sqrt{10^2+10^2}=10\sqrt{2}$.
$\because OE⊥ BC,\therefore OE=EC$.
设$OA=r$,则$OE=EC=r,\therefore OC=\sqrt{OE^2+EC^2}=\sqrt{r^2+r^2}=\sqrt{2}r$.
$\because OA+OC=AC,\therefore r+\sqrt{2}r=10\sqrt{2}$,
解得$r=20-10\sqrt{2}$,即$\odot O$的半径为$20-10\sqrt{2}$.

6. 如图，$AO$是$△ ABC$的中线，$\odot O$与$AB$边相切于点$D$.
(1)要使$\odot O$与$AC$边也相切，应增加的条件是________；(任写一个)
(2)增加条件后，请你证明$\odot O$与$AC$边相切.

(1)$∠ B=∠ C$(答案不唯一)
(2)(答案不唯一)
证明:如答图,连接OD,过点O作$OE⊥ AC$,垂足为E,
则$∠ CEO=90°$.

$\because\odot O$与AB相切于点D,$\therefore∠ BDO=∠ CEO=90°$.
$\because AO$是$△ ABC$的中线,$\therefore OB=OC$.
又$\because∠ B=∠ C,\therefore△ BDO≌△ CEO,\therefore OD=OE$.
$\because OD$是$\odot O$的半径,$\therefore OE$是$\odot O$的半径,
$\therefore\odot O$与AC边相切.