7.(2025·宿迁宿城区期末)如图，AB为$\odot O$的直径，D，T是圆上的两点，且AT平分$∠ BAD$，过点T作AD延长线的垂线PQ，垂足为C.
(1)求证：PQ是$\odot O$的切线；
(2)若$\odot O$的半径为2，TC=$\sqrt{3}$，求弦AD的长.

7.(1)证明:如答图,连接OT.
$\because OT=OA,\therefore ∠ ATO=∠ OAT$.
$\because AT$平分$∠ BAD,\therefore ∠ TAC=∠ BAT$,
$\therefore ∠ ATO=∠ TAC,\therefore OT// AC$.
$\because AC⊥ PQ,\therefore OT⊥ PQ$.
又$\because$点T在$\odot O$上,$\therefore PQ$是$\odot O$的切线.
(2)解:如答图,过点O作$OM⊥ AC$于点M,则$AM=MD$.
$\because ∠ OTC=∠ ACT=∠ OMC=90°$,
$\therefore$四边形OTCM为矩形,$\therefore OM=TC=\sqrt{3}$.
在$\mathrm{Rt}△ AOM$中,$AM=\sqrt{OA^2-OM^2}=1$,
$\therefore$弦AD的长为2.

8. 如图①，AB是$\odot O$的直径，AD与$\odot O$相切于点A，DE与$\odot O$相切于点E，C为DE延长线上一点，且$CE=CB$.
(1)求证：BC为$\odot O$的切线；
(2)连接AE，AE的延长线与BC的延长线交于点G（如图②所示）.若$\odot O$的半径为$\sqrt{5}$，AD=2，求线段CE和GE的长.

8.(1)证明:如答图①,连接OE,OC.
$\because$在$△ OBC$和$△ OEC$中,$\begin{cases} CB=CE,\\ OB=OE,\\ OC=OC, \end{cases}$
$\therefore △ OBC≌△ OEC(\mathrm{SSS}),\therefore ∠ OBC=∠ OEC$.
又$\because DE$与$\odot O$相切于点E,$\therefore ∠ OEC=90°$,
$\therefore ∠ OBC=90°$.
$\because OB$为$\odot O$的半径,$\therefore BC$为$\odot O$的切线.
(2)解:如答图②,过点D作$DF⊥ BC$于点F,连接BE,
易知$DF=AB=2\sqrt{5}$.
$\because AB$为$\odot O$的直径,$\therefore ∠ AEB=90°$.
$\because AD,DC,BG$分别切$\odot O$于点A,E,B,
$\therefore DA=DE,CE=CB$. 设BC的长为x,
则$CF=x-2,DC=x+2$.
在$\mathrm{Rt}△ DFC$中,$(x+2)^2-(x-2)^2=(2\sqrt{5})^2$,
解得$x=\frac{5}{2},\therefore CE=BC=\frac{5}{2}$.
易知$AD// BG,\therefore ∠ DAE=∠ EGC$.
$\because DA=DE,\therefore ∠ DAE=∠ AED$.
$\because ∠ AED=∠ CEG,\therefore ∠ EGC=∠ CEG$,
$\therefore CG=CE=CB=\frac{5}{2},\therefore BG=5$,
$\therefore AG=\sqrt{(2\sqrt{5})^2+5^2}=3\sqrt{5}$,
$\therefore S_{△ ABG}=\frac{1}{2}AB· BG=\frac{1}{2}AG· BE,\therefore BE=\frac{10}{3}$.
在$\mathrm{Rt}△ BEG$中,$EG=\sqrt{BG^2-BE^2}=\frac{5}{3}\sqrt{5}$.

9. 如图，AB是$\odot O$的直径，AC是$\odot O$的弦，P为AB延长线上一点，连接CP，$∠ BCP=∠ A$，$∠ ACB$的平分线与直径AB交于点E，交$\odot O$于点D。
(1)求证：CP是$\odot O$的切线；
(2)求证：$PE=PC$；
(3)探究$AC+BC$与$CD$之间的数量关系，并说明理由。

9.(1)证明:如答图①,连接OC.
$\because OA=OC,\therefore ∠ A=∠ ACO$.
又$\because ∠ BCP=∠ A,\therefore ∠ ACO=∠ BCP$.
$\because AB$为$\odot O$的直径,$\therefore ∠ ACO+∠ BCO=90°$,
$\therefore ∠ PCB+∠ BCO=90°$,即$∠ OCP=90°$.
$\because OC$为$\odot O$的半径,$\therefore CP$是$\odot O$的切线.
(2)证明:$\because CE$平分$∠ ACB,\therefore ∠ ACD=∠ BCD$.
$\because ∠ PCE=∠ PCB+∠ BCE,∠ PEC=∠ ACD+∠ A$,
且$∠ PCB=∠ A,\therefore ∠ PCE=∠ PEC,\therefore PE=PC$.
(3)解:$AC+BC=\sqrt{2}CD$. 理由如下:
如答图②,连接AD,BD,过点D作$DM⊥ AC$于点M,过点D作$DN⊥ CB$交CB的延长线于点N.
$\because CD$平分$∠ ACB,DM⊥ AC,DN⊥ CB$,
$\therefore DM=DN,\overset{\frown}{AD}=\overset{\frown}{BD},\therefore AD=BD$.
$\therefore \mathrm{Rt}△ AMD≌\mathrm{Rt}△ BND,\therefore AM=BN$.
易证四边形CMDN为正方形,$\therefore CD=\sqrt{2}CN$.
而$AC+BC=CM+AM+CB=CM+CB+BN=CM+CN=2CN,\therefore AC+BC=\sqrt{2}CD$.