《2025年思维新观察九年级数学上册人教版》

【典例1】(2024·武汉)如图,$△ABC内接于\odot O$,$AB= AC$,点$P在\overset{\frown}{AB}$上,连接$PA$,过$A作AN⊥PC于N$点.
(1)求证：$AP平分△BPC的外角∠CPE$；
(2)若$PB= 1$,$PC= 3$,求$PN$的长.

(1)证明:略;
(2)过 A 作 $ AM \perp PB $ 于 M,
易证 $ \triangle APM \cong \triangle APN $, $ \triangle ABM \cong \triangle ACN $,
$ PM = PN $, $ BM = CN $,
$\therefore PN = $1.

答案： 解:
(1)证明:略;
(2)过 A 作 $ AM \perp PB $ 于 M,
易证 $ \triangle APM \cong \triangle APN $, $ \triangle ABM \cong \triangle ACN $,
$ PM = PN $, $ BM = CN $,
$\therefore PN = 1$.

变式.如图,在$\odot O$中,$\overset{\frown}{AC}= \overset{\frown}{BC}$,点$P在\odot O$上,$CD⊥PA于E$,交$\odot O于点D$.
(1)求证：$PC平分∠APB$；
(2)若$AE= 1$,$PE= 3$,求$PB$的长.

(1)证明：$\because \overset{\frown}{AC} = \overset{\frown}{BC}$，$\therefore ∠CPA=∠CPB$，$\therefore PC$平分$∠APB$；
(2)解：连接$CA$，$CB$，
方法一：在$PA$上取点$M$满足$PM = PB$，连接$CM$，
$\because ∠CPA = ∠CPB$，$PC=PC$，$\therefore \triangle CPM \cong \triangle CPB(SAS)$，$\therefore CM = CB$，$\because \overset{\frown}{AC} = \overset{\frown}{BC}$，$\therefore CA=CB$，$\therefore CM=CA$，又$CD \perp PA$，$\therefore AE = ME = 1$，$\because PE=3$，$\therefore PM=PE-ME=3-1=2$，$\therefore PB=PM=$2；
方法二：作$CH \perp PB$交$PB$的延长线于$H$，$\because PC$平分$∠APB$，$CD⊥PA$，$CH⊥PB$，$\therefore CE=CH$，$\because \overset{\frown}{AC} = \overset{\frown}{BC}$，$\therefore CA=CB$，$\therefore \text{Rt}\triangle CAE \cong \text{Rt}\triangle CBH(HL)$，$\therefore AE = BH=1$，$\because PC=PC$，$CE=CH$，$\therefore \text{Rt}\triangle PCE \cong \text{Rt}\triangle PCH(HL)$，$\therefore PE=PH=3$，$\therefore PB=PH-BH=3-1=$2.

答案：
(1)解:$\because \overset{\frown}{AC} = \overset{\frown}{BC}$,
$\therefore PC$ 平分 $ \angle APB $;
(2)解:连接 $ CA $, $ CB $,
方法一:在 $ PA $ 上取点 $ M $ 满足 $ PM = PB $,连接 $ CM $,
$\because \angle CPA = \angle CPB$,
易证 $ \triangle CPM \cong \triangle CPB $,
$\therefore CM = CB = CA $,又 $ CD \perp PA $,
$\therefore AE = ME = 1$,
$\therefore PB = PM = 2$;
方法二:作 $ CH \perp PB $ 交 $ PB $ 的延长线于 $ H $,证 $ \triangle CAE \cong \triangle CBH $,
$\therefore AE = BH $,易证 $ \triangle PCE \cong \triangle PCH $,
$\therefore PB = 3 - 1 = 2$.

【典例2】(教材P90T14变式)已知$\odot O为等腰△ABC$的外接圆,$AB= AC$,点$P为\overset{\frown}{BC}$上任一点,连接$PB$,$PC$.
(1)问题背景：如图1,若$∠BAC= 60^{\circ}$,求证：$PB+PC= PA$；
(2)迁移运用：如图2,若$∠BAC= 90^{\circ}$,求$\frac {PB+PC}{PA}$的值为$\sqrt{2}$；
(3)拓展运用：如图3,若$∠BAC= 120^{\circ}$,$PA= 2$,直接写出$PB+PC$的值为$2\sqrt{3}$.

答案：
(1)证明:在 $ PA $ 上取 $ PM = PC $,连接 $ CM $,则 $ \triangle PMC $ 为等边三角形,证 $ \triangle ACM \cong \triangle BCP $,
$\therefore AM = PB $,
$\therefore PA = PB + PC $.
(2)解:过 $ A $ 作 $ AM \perp AP $ 交直线 $ PB $ 于 $ M $,
$ \triangle ABM \cong \triangle ACP $,
$\therefore BM = PC $,$\therefore \frac{PB + PC}{PA} = \sqrt{2}$.
(3)$ 2\sqrt{3} $.