《2026年启东中学作业本九年级数学下册苏科版》

三、解答题
8. (2024·宜兴月考)如图，在平面直角坐标系中，已知$△ ABC$三个顶点的坐标分别是$A(2,2)$，$B(4,0)$，$C(4,-4)$.
(1)$△ ABC$的面积是4；
(2)以点$O$为位似中心，将$△ ABC$缩小为原来的$\frac{1}{2}$得到$△ A_{1}B_{1}C_{1}$，请在$y$轴右侧画出$△ A_{1}B_{1}C_{1}$；
(3)请用无刻度的直尺在边$AC$上找一点$P$，使得$∠ PBC=∠ BAC$，并保留作图痕迹.

答案：
8.
(1)4
(2)解:如答图①,$△ A_{1}B_{1}C_{1}$即为所求.
(3)解:如答图②,点P即为所求.

理由:由作图可知$∠ APB=∠ RAC=45^{\circ}$,
$\therefore ∠ CPB=135^{\circ}$.
$\because ∠ ABC=135^{\circ}$,$\therefore ∠ ACB+∠ BAC=45^{\circ}$,$∠ ACB+∠ PBC=45^{\circ}$,$\therefore ∠ PBC=∠ BAC$.

9. (2023·内蒙古)如图，$AB$是$\odot O$的直径，$E$为$\odot O$上的一点，$C$是$\overset{\frown}{AE}$的中点，连接$BC$，过点$C$的直线垂直于$BE$的延长线于点$D$，交$BA$的延长线于点$P$.
(1)求证：$PC$为$\odot O$的切线；
(2)若$PC = 2\sqrt{2}BO$，$PB = 10$，求$BE$的长.

答案：
9.
(1)证明:如答图,连接OC,
$\because C$是$\overset{\frown}{AE}$的中点,$\therefore ∠ ABC=∠ DBC$.
$\because OC=OB$,$\therefore ∠ ABC=∠ OCB$,
$\therefore ∠ DBC=∠ OCB$,$\therefore OC// DB$.
$\because PD⊥ BD$,$\therefore PD⊥ CO$,$\therefore PC$为$\odot O$的切线.

(2)解:如答图,连接AE,设$OB=OC=r$,
$\because PC=2\sqrt{2}BO=2\sqrt{2}r$,$\therefore OP=\sqrt{r^{2}+(2\sqrt{2}r)^{2}}=3r$.
$\because PB=10$,$\therefore 3r+r=10$,即$r=\frac{5}{2}$.
$\because OC// DB$,$\therefore △ PCO∽ △ PDB$,$\therefore \frac{OC}{BD}=\frac{PO}{PB}$,
$\therefore \frac{\frac{5}{2}}{BD}=\frac{\frac{15}{2}}{10}$,$\therefore BD=\frac{10}{3}$.
$\because AB$是$\odot O$的直径,$\therefore AE⊥ BD$,$\therefore AE// PD$,
$\therefore \frac{BE}{BD}=\frac{BA}{BP}$,$\therefore \frac{BE}{\frac{10}{3}}=\frac{5}{10}$,$\therefore BE=\frac{5}{3}$.

10. (2024·上海)如图，在矩形$ABCD$中，$E$为边$CD$上一点，且$AE⊥ BD$.
(1)求证：$AD^{2}=DE· DC$；
(2)$F$为线段$AE$延长线上一点，且满足$EF = CF=\frac{1}{2}BD$，求证：$CE = AD$.

答案：
10.证明:
(1)$\because$在矩形ABCD中,$∠ BAD=90^{\circ}$,$∠ ADE=90^{\circ}$,
$AB=DC$,$\therefore ∠ ABD+∠ ADB=90^{\circ}$.
$\because AE⊥ BD$,$\therefore ∠ DAE+∠ ADB=90^{\circ}$,
$\therefore ∠ ABD=∠ DAE$.
$\because ∠ BAD=∠ ADE=90^{\circ}$,$\therefore △ ADE∽ △ BAD$,
$\therefore \frac{AD}{BA}=\frac{DE}{AD}$,$\therefore AD^{2}=DE· BA$.
$\because AB=DC$,$\therefore AD^{2}=DE· DC$.
(2)如答图,连接AC交BD于点O.$\because$在矩形ABCD中,
$∠ ADE=90^{\circ}$,$\therefore ∠ DAE+∠ AED=90^{\circ}$.
$\because AE⊥ BD$,$\therefore ∠ DAE+∠ ADB=90^{\circ}$,
$\therefore ∠ ADB=∠ AED$.
$\because ∠ FEC=∠ AED$,$\therefore ∠ ADO=∠ FEC$.
$\because$在矩形ABCD中,$OA=OD=\frac{1}{2}BD$,
又$EF=CF=\frac{1}{2}BD$,$\therefore OA=OD=EF=CF$,
$\therefore ∠ ADO=∠ OAD$,$∠ FEC=∠ FCE$.
$\because ∠ ADO=∠ FEC$,
$\therefore ∠ ADO=∠ OAD=∠ FEC=∠ FCE$.
在$△ ODA$和$△ FEC$中,$\begin{cases} ∠ ODA=∠ FEC, \\ ∠ OAD=∠ FCE, \\ OD=FE, \end{cases}$
$\therefore △ ODA≌ △ FEC(\mathrm{AAS})$,$\therefore CE=AD$.