1. 如图，AB是$\odot O$的直径，$AB⊥ AC$，BC交$\odot O$于点P，Q是AC的中点. 求证：QP是$\odot O$的切线.

证明:如答图,连接OP,PA.

$\because$AB是$\odot O$的直径,$\therefore∠ APB=∠ APC=90°$.
在$\mathrm{Rt}△ APC$中,Q是AC的中点,
$\therefore PQ=AQ,\therefore∠ QAP=∠ QPA$.
又$\because OP=OA,\therefore∠ OAP=∠ OPA,\therefore∠ OAQ=∠ QPO$.
$\because AB⊥ AC,\therefore∠ OAQ=90°,\therefore∠ QPO=90°$,
$\therefore QP$是$\odot O$的切线.

2. (1)如图①，$△ ABC$内接于$\odot O$，$AB$为直径，$∠ CAE=∠ B$，求证：$AE$与$\odot O$相切于点$A$；
(2)如图②，若$AB$为非直径的弦，$∠ CAE=∠ B$，$AE$还与$\odot O$相切于点$A$吗？请说明理由。

(1)证明:$\because AB$为直径,$\therefore∠ ACB=90°$,
$\therefore∠ B+∠ BAC=90°$.
$\because∠ CAE=∠ B,\therefore∠ CAE+∠ BAC=90°$,
即$∠ BAE=90°,\therefore OA⊥ AE$,
$\therefore AE$与$\odot O$相切于点A.
(2)解:$AE$还与$\odot O$相切于点A.理由如下:
如答图,作直径AD,连接CD,

$\therefore∠ ACD=90°$,
$\therefore∠ D+∠ DAC=90°$.
$\because∠ B=∠ D,∠ CAE=∠ B$,
$\therefore∠ CAE+∠ DAC=90°$,即$∠ DAE=90°$,
$\therefore OA⊥ AE$.
又$\because OA$为$\odot O$的半径,$\therefore AE$与$\odot O$相切于点A.

3.(2025·盐都区模拟)如图，$△ ABC$内接于$\odot O$，且$AB$为$\odot O$的直径，$∠ BAC$的平分线交$BC$于点$E$，交$\odot O$于点$D$，交过点$B$的一条直线于点$F$，$DF=DE$.
(1)求证：$BF$是$\odot O$的切线；
(2)若$\odot O$的半径为$5$，$AC=6$，求$CE$的长.

(1)证明:如答图,连接DB.
$\because AB$是$\odot O$的直径,$\therefore∠ BDA=∠ C=90°$,
$\therefore BD⊥ FE,∠ AEC+∠ CAE=90°$.
$\because DF=DE,\therefore BD$是$FE$的垂直平分线,$\therefore BF=BE$,
$\therefore∠ F=∠ BED$.
$\because∠ BED=∠ AEC,\therefore∠ F=∠ BED=∠ AEC$,
$\therefore∠ F+∠ CAE=90°$.
$\because AD$平分$∠ CAB$,
$\therefore∠ CAE=∠ BAE$,
$\therefore∠ F+∠ BAE=90°,\therefore∠ FBA=90°$.
$\because AB$是$\odot O$的直径,
$\therefore BF$是$\odot O$的切线.
(2)解:如答图,过点E作$EM⊥ AB$于点M.

$\because AD$平分$∠ CAB,EM⊥ AB,∠ C=90°$,
$\therefore EC=EM,∠ EMA=∠ EMB=∠ C=90°$.
$\because AE=AE,EC=EM$,
$\therefore\mathrm{Rt}△ ACE≌\mathrm{Rt}△ AME$,
$\therefore AM=AC=6.\because\odot O$的半径为5,$\therefore AB=10$,
$\therefore BM=AB-AM=10-6=4$,
$\therefore BC=\sqrt{AB^2-AC^2}=\sqrt{10^2-6^2}=8$,
设$CE=x=EM$,则$BE=8-x$,
在$\mathrm{Rt}△ BEM$中,$\because BE^2=EM^2+BM^2$,
$\therefore(8-x)^2=x^2+4^2$,解得$x=3,\therefore CE=3$.