《2025年名师帮同步学案九年级数学全一册人教版》

8.(例4)如图,$\odot O$的直径AB为10 cm,弦AC为6 cm,$∠ACB$的平分线交$\odot O$于点D.
(1)求BC,AD,BD的长;
(2)连接OD,求证:$\widehat {AD}=\widehat {BD},OD⊥AB,△ABD$是等腰直角三角形;
(3)过点A作$AF⊥CD$交于点F,求证:$CF=AF=\frac {\sqrt {2}}{2}AC;$
(4)过点D作$DM⊥AC$于点M,作$DN⊥BC$于点N,求证:$CA+CB=\sqrt {2}CD$(提示:证$△DMA\cong △DNB);$
(5)求证:$S_{四边形ADBC}=\frac {1}{2}CD^{2}.$

答案： （例 4）（1）解：$\because AB$是直径，
$\therefore \angle ACB = \angle ADB = 90 ^ { \circ }，$在$\mathrm { Rt } \triangle ABC$中，$AB ^ { 2 } = AC ^ { 2 } + BC ^ { 2 }，$$AB = 10 \mathrm { cm }，$$AC = 6 \mathrm { cm }，$
$\therefore BC ^ { 2 } = AB ^ { 2 } - AC ^ { 2 } = 10 ^ { 2 } - 6 ^ { 2 } = 64，$$\therefore BC = 8 \mathrm { cm }，$
又$\because CD$平分$\angle ACB，$
$\therefore \angle ACD = \angle BCD，$$\therefore \overparen { A D } = \overparen { D B }，$
$\therefore AD = BD，$又$\because $在$\mathrm { Rt } \triangle ABD$中，$AD ^ { 2 } + BD ^ { 2 } = AB ^ { 2 }，$$\therefore AD ^ { 2 } + BD ^ { 2 } = 10 ^ { 2 }，$
$\therefore AD = BD = 5 \sqrt { 2 } \mathrm { cm }；$（2）证明：$\because AB$是直径，$\therefore \angle ACB = \angle ADB = 90 ^ { \circ }，$
又$\because CD$平分$\angle ACB，$
$\therefore \angle ACD = \angle BCD = 45 ^ { \circ }，$$\therefore \overparen { A D } = \overparen { D B }，$
$\therefore DA = DB，$$\therefore \triangle ABD$是等腰直角三角形．
由圆周角定理得$\angle AOD = 2 \angle ACD = 90 ^ { \circ }，$$\angle BOD = 2 \angle BCD = 90 ^ { \circ }，$
$\therefore OD \perp AB；$（3）证明：$\because \angle ACD = \angle BCD = \frac { 1 } { 2 } \angle ACB = 45 ^ { \circ }，$$\therefore $在$\mathrm { Rt } \triangle AFC$中，$CF = AF = \frac { \sqrt { 2 } } { 2 } AC；$
（4）证明：$\because DM \perp CA，$$DN \perp BC，$
$\therefore \angle DMA = \angle DNB = 90 ^ { \circ }，$$\because \angle ACB = 90 ^ { \circ }，$
$\therefore $四边形CMDN是矩形，$\therefore \angle MDN = 90 ^ { \circ }，$
即$\angle ADN + \angle ADM = 90 ^ { \circ }，$
又$\because AB$为直径，
$\therefore \angle ADB = \angle ACB = 90 ^ { \circ }，$$\therefore \angle ADN + \angle BDN = 90 ^ { \circ }，$
$\therefore AM = BN，$CA + CB = CM + CN + BN - AM = CM + CN，在$\mathrm { Rt } \triangle CMD$中，CD平分$\angle ACB，$$\angle ACB = 90 ^ { \circ }，$
$\therefore \angle DCA = 45 ^ { \circ }，$$\angle DCN = 45 ^ { \circ }，$$\therefore CM = CN = \frac { \sqrt { 2 } } { 2 } CD，$
$\therefore CM + CN = \frac { \sqrt { 2 } } { 2 } CD + \frac { \sqrt { 2 } } { 2 } CD = \sqrt { 2 } CD，$即$CA + CB = \sqrt { 2 } CD；$（5）证明：由（4）证得$CM = CN = \frac { \sqrt { 2 } } { 2 } CD，$$CA + CB = \sqrt { 2 } CD，$$\therefore S _ { \text { 四边形 } A D B C } = S _ { \triangle A C D } + S _ { \triangle C D B }，$
又$\because DM = DN，$
$\therefore S _ { \text { 四边形 } A D B C } = \frac { 1 } { 2 } A C \cdot M D + \frac { 1 } { 2 } B C \cdot D N = \frac { 1 } { 2 } ( A C + B C ) \cdot D N = \frac { 1 } { 2 } × \sqrt { 2 } C D \cdot \frac { \sqrt { 2 } } { 2 } C D = \frac { 1 } { 2 } C D ^ { 2 }．$