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23.(13分)已知点$A(0,6)$，$B(-6,0)$，将线段$OA$绕点$O$顺时针旋转至$OC$，连接$AC$，$BC$.
(1)如图1，求$\angle ACB$的度数.
(2)如图2，若$\angle AOC = 60^{\circ}$，$\angle AOB$的平分线$OD$交$BC$于点$D$，求证：$OD + BD = CD$.
(3)如图3，若$\angle AOC = 30^{\circ}$，过点$A$作$AE\bot AC$交$BC$于点$E$，求$BE$的长.
(1)如图1，求$\angle ACB$的度数.
(2)如图2，若$\angle AOC = 60^{\circ}$，$\angle AOB$的平分线$OD$交$BC$于点$D$，求证：$OD + BD = CD$.
(3)如图3，若$\angle AOC = 30^{\circ}$，过点$A$作$AE\bot AC$交$BC$于点$E$，求$BE$的长.
答案：
23.
(1)解：$\because$点$A(0,6)$，$B(-6,0)$，$\therefore OA=OB=6$. 由旋转得，$OC=OA$，$\therefore OC=OB=OA$. 设$\angle AOC=2\alpha$，$\therefore \angle BOC=\angle AOB+\angle AOC=90^{\circ}+2\alpha$. $\therefore \angle OBC=\angle OCB=\frac{1}{2}(180^{\circ}-\angle BOC)=45^{\circ}-\alpha$. $\because \angle AOC=2\alpha$，$\therefore \angle OAC=\angle OCA=90^{\circ}-\alpha$. $\therefore \angle ACB=\angle OCA-\angle OCB=90^{\circ}-\alpha-(45^{\circ}-\alpha)=45^{\circ}$.
(2)证明：如图1，在$BC$上取点$H$，使$CH=BD$. 在$\triangle BOD$和$\triangle COH$中，$\begin{cases}BO=CO,\\\angle OBD=\angle OCH,\\BD=CH,\end{cases}$ $\therefore \triangle BOD\cong\triangle COH(SAS)$. $\therefore \angle BOD=\angle COH$，$OD=OH$. $\because OD$平分$\angle AOB$，$\therefore \angle BOD=\angle AOD=45^{\circ}$. $\therefore \angle COH=\angle BOD=45^{\circ}$. $\because \angle BOC=\angle AOC+\angle AOB=60^{\circ}+90^{\circ}=150^{\circ}$， $\therefore \angle DOH=\angle BOC-\angle BOD-\angle COH=60^{\circ}$. $\therefore \triangle DOH$为等边三角形，$\therefore OD=DH$. $\therefore OD+BD=DH+BD=DH+HC=CD$.
(3)解：如图2，过点$C$作$CN\perp AO$于点$N$，过点$E$作$EM\perp AO$于点$M$，连接$OE$. $\therefore \angle CNA=\angle AME=90^{\circ}$. 由
(1)得，$\angle ACE=45^{\circ}$. $\because AE\perp AC$，$\therefore \angle CAE=90^{\circ}$. $\therefore \angle AEC=90^{\circ}-\angle ACE=45^{\circ}$. $\therefore \angle ACE=\angle AEC$，$\therefore AC=AE$. $\because \angle ACN+\angle NAC=\angle EAM+\angle NAC=90^{\circ}$， $\therefore \angle ACN=\angle EAM$. 在$\triangle AEM$和$\triangle CAN$中，$\begin{cases}\angle AME=\angle CNA,\\\angle EAM=\angle ACN,\\AE=CA,\end{cases}$ $\therefore \triangle AEM\cong\triangle CAN(AAS)$，$\therefore AM=CN$. 在$Rt\triangle CNO$中，$\angle CNO=90^{\circ}$，$\angle AOC=30^{\circ}$，$OB=OA=OC=6$，$\therefore CN=\frac{1}{2}CO=3$. $\therefore AM=3$，$\therefore OM=OA - AM=3$，$\therefore OM=AM$. $\because EM\perp AO$，$\therefore AE=EO$. $\because \angle AOB=90^{\circ}$，$\angle AOC=30^{\circ}$， $\therefore \angle BOC=\angle AOB+\angle AOC=120^{\circ}$. $\therefore \angle CBO=\angle OCB=30^{\circ}$，$\therefore \angle OAC=\angle OCA=75^{\circ}$. $\therefore \angle EAO=\angle EOA=\angle CAE-\angle OAC=90^{\circ}-75^{\circ}=15^{\circ}$. $\therefore \angle BOE=\angle AOB-\angle EOA=75^{\circ}$. $\therefore \angle BEO=180^{\circ}-\angle CBO-\angle BOE=75^{\circ}$. $\therefore \angle BOE=\angle BEO$，$\therefore BE=BO=6$.
(1)解：$\because$点$A(0,6)$，$B(-6,0)$，$\therefore OA=OB=6$. 由旋转得，$OC=OA$，$\therefore OC=OB=OA$. 设$\angle AOC=2\alpha$，$\therefore \angle BOC=\angle AOB+\angle AOC=90^{\circ}+2\alpha$. $\therefore \angle OBC=\angle OCB=\frac{1}{2}(180^{\circ}-\angle BOC)=45^{\circ}-\alpha$. $\because \angle AOC=2\alpha$，$\therefore \angle OAC=\angle OCA=90^{\circ}-\alpha$. $\therefore \angle ACB=\angle OCA-\angle OCB=90^{\circ}-\alpha-(45^{\circ}-\alpha)=45^{\circ}$.
(2)证明：如图1，在$BC$上取点$H$，使$CH=BD$. 在$\triangle BOD$和$\triangle COH$中，$\begin{cases}BO=CO,\\\angle OBD=\angle OCH,\\BD=CH,\end{cases}$ $\therefore \triangle BOD\cong\triangle COH(SAS)$. $\therefore \angle BOD=\angle COH$，$OD=OH$. $\because OD$平分$\angle AOB$，$\therefore \angle BOD=\angle AOD=45^{\circ}$. $\therefore \angle COH=\angle BOD=45^{\circ}$. $\because \angle BOC=\angle AOC+\angle AOB=60^{\circ}+90^{\circ}=150^{\circ}$， $\therefore \angle DOH=\angle BOC-\angle BOD-\angle COH=60^{\circ}$. $\therefore \triangle DOH$为等边三角形，$\therefore OD=DH$. $\therefore OD+BD=DH+BD=DH+HC=CD$.
(3)解：如图2，过点$C$作$CN\perp AO$于点$N$，过点$E$作$EM\perp AO$于点$M$，连接$OE$. $\therefore \angle CNA=\angle AME=90^{\circ}$. 由
(1)得，$\angle ACE=45^{\circ}$. $\because AE\perp AC$，$\therefore \angle CAE=90^{\circ}$. $\therefore \angle AEC=90^{\circ}-\angle ACE=45^{\circ}$. $\therefore \angle ACE=\angle AEC$，$\therefore AC=AE$. $\because \angle ACN+\angle NAC=\angle EAM+\angle NAC=90^{\circ}$， $\therefore \angle ACN=\angle EAM$. 在$\triangle AEM$和$\triangle CAN$中，$\begin{cases}\angle AME=\angle CNA,\\\angle EAM=\angle ACN,\\AE=CA,\end{cases}$ $\therefore \triangle AEM\cong\triangle CAN(AAS)$，$\therefore AM=CN$. 在$Rt\triangle CNO$中，$\angle CNO=90^{\circ}$，$\angle AOC=30^{\circ}$，$OB=OA=OC=6$，$\therefore CN=\frac{1}{2}CO=3$. $\therefore AM=3$，$\therefore OM=OA - AM=3$，$\therefore OM=AM$. $\because EM\perp AO$，$\therefore AE=EO$. $\because \angle AOB=90^{\circ}$，$\angle AOC=30^{\circ}$， $\therefore \angle BOC=\angle AOB+\angle AOC=120^{\circ}$. $\therefore \angle CBO=\angle OCB=30^{\circ}$，$\therefore \angle OAC=\angle OCA=75^{\circ}$. $\therefore \angle EAO=\angle EOA=\angle CAE-\angle OAC=90^{\circ}-75^{\circ}=15^{\circ}$. $\therefore \angle BOE=\angle AOB-\angle EOA=75^{\circ}$. $\therefore \angle BEO=180^{\circ}-\angle CBO-\angle BOE=75^{\circ}$. $\therefore \angle BOE=\angle BEO$，$\therefore BE=BO=6$.
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