《2026年龙江王中王中考总复习数学》

20. (2025广东)如图，点$O$是$Rt\triangle ABC$斜边$AC$上的一点，以$OA$为半径的$\odot O$与边$BC$相切于点$D$. 求证：$AD$平分$\angle BAC$.

答案：
20. 证明: 连接$OD$.
$\because BC$是$\odot O$的切线，$\therefore OD\perp BC$.$\therefore \angle ODC=90^{\circ}$.
$\because \angle ABC=90^{\circ}$，$\therefore \angle ODC=\angle ABC$.
$\therefore OD// AB$.$\therefore \angle ODA=\angle BAD$.
$\because OD=OA$，$\therefore \angle ODA=\angle OAD$.$\therefore \angle DAB=\angle OAD$.
$\therefore AD$平分$\angle BAC$.

21. 如图，$\triangle ABC$内接于$\odot O$，$AB$为$\odot O$的直径，点$D$在$AB$的延长线上，连接$CD$，$\angle BCD = \angle A$，过点$B$作$BE\perp AD$，交$CD$于点$E$.
(1) 求证：$CD$是$\odot O$的切线；
(2) 若点$B$是$AD$的中点，且$BE = 3$，求$\odot O$的半径的长.

答案：
21.
(1)证明:连接$OC$.

$\because AB$为$\odot O$的直径，$\therefore \angle ACB=90^{\circ}$.
$\therefore \angle CAB+\angle ABC=90^{\circ}$.
$\because \angle CAB=\angle BCD$，$\therefore \angle BCD+\angle ABC=90^{\circ}$.
$\because OC=OB$，$\therefore \angle OCB=\angle OBC$.
$\therefore \angle OCD=\angle BCD+\angle OCB=90^{\circ}$.$\therefore OC\perp CD$.
$\because OC$为$\odot O$的半径，$\therefore CD$是$\odot O$的切线.
(2)解:连接$OC$，$\because$点$B$是$AD$的中点，
$\therefore BD=AB=2OC$.$\because OB=OC$，$\therefore OD=3OC$.
由
(1)知$\angle OCD=90^{\circ}$，$\therefore \sin D=\frac{OC}{OD}=\frac{1}{3}$.
$\because BE\perp AD$，$\therefore \angle EBD=90^{\circ}$.
在$Rt\triangle EBD$中，$\sin D=\frac{BE}{ED}=\frac{1}{3}$，
$\therefore ED=9$.$\therefore BD=\sqrt{ED^{2}-BE^{2}}=6\sqrt{2}$.
$\therefore OC=3\sqrt{2}$.即$\odot O$的半径的长为$3\sqrt{2}$.