《2025年时代新课程九年级数学上册苏科版》

11. 如图,已知△ABC中,AC=5,AB=6,BC=7,求△ABC内切圆的半径.

答案： 11.$\frac{2\sqrt{6}}{3}$

12. 如图,⊙O为△ABC的内切圆,∠C=90°,AO的延长线交BC于点D,AC=4,DC=1,求⊙O的半径.

答案： 12.$\frac{4}{5}$

13. (2025春·宁波月考)如图,点A的坐标为(0,4),以O点为圆心,以OA为半径的圆交x轴于点B,点C为第一象限圆上一动点,CD⊥x轴于D点,点I为△OCD的内心,则AI的最小值为$2\sqrt{10}-2\sqrt{2}$.

答案： 13.$2\sqrt{10}-2\sqrt{2}$

14. (2023秋·武昌区期末)如图,点I是△ABC的内心,AI的延长线和△ABC的外接圆⊙O相交于点D.
(1) 求证:DB=DI;
(2) 如果OI⊥AD,IM⊥AB于M.求证:BC=2AM.

答案： 14.
(1)连接$BI$,
$\because$点$I$是$\triangle ABC$的内心,
$\therefore\angle BAD = \angle CAD,\angle ABI = \angle CBI$,
$\therefore\angle BAD + \angle ABI = \angle CAD + \angle CBI$,
$\because\angle CAD = \angle CBD$,
$\therefore\angle BAD + \angle ABI = \angle CBD + \angle CBI$,
$\because\angle BID = \angle BAD + \angle ABI,\angle IBD = \angle CBD + \angle CBI$,
$\therefore\angle BID = \angle IBD,\therefore DB = DI$.
(2)连接$OD$交$BC$于点$E$,
$\because\angle BAD = \angle CAD,\therefore\overset{\frown}{BD} = \overset{\frown}{CD}$,
$\therefore OD\perp BC,BE = CE$,
$\because OI\perp AD,IM\perp AB$,
$\therefore\angle BED = \angle AMI = 90^{\circ},IA = DI$,
$\because DB = DI,\therefore DB = IA$,
$\because\angle DBE = \angle DAC,\angle IAM = \angle DAC$,
$\therefore\angle DBE = \angle IAM$,
$\therefore\triangle DBE\cong\triangle IAM(AAS)$,
$\therefore BE = AM,\therefore 2BE = 2AM$,
$\because BC = 2BE,\therefore BC = 2AM$.