《2026年1号卷中考试题精编九年级数学安徽专版》

22. 如图1,$\triangle ABC$的两条角平分线$AD$,$BE$相交于点$I$,$\angle ABC = 2\angle C$.
(1) 求证:$BI = BD$.
(2) 若$AB = 6$,$BD = 5$,求$BE$的长.
(3) 如图2,$AF\perp AC$交$BC$于点$F$,求证:$CF = 2AB$.

答案： 22.解:
(1)$\because AD$平分$\angle BAC$,$BE$平分$\angle ABC$，
$\therefore\angle BAD=\angle CAD,\angle ABE=\angle CBE$，
$\because\angle BDI=\angle C+\angle CAD=\angle C+\frac{1}{2}\angle BAC,\angle BID=\angle ABI+\angle BAI=\angle ABI+\frac{1}{2}\angle BAC$，$\because\angle ABC=2\angle C$，$\therefore\angle ABI=\angle C$，
$\therefore\angle BDI=\angle BID$，$\therefore BI=BD$. (4分)
(2)如图1,在$AC$上截取$AG=AB$,连接$DG$,易证$\triangle ABD\cong\triangle AGD$，
$\therefore DG=BD,\angle ABD=\angle AGD$，
$\because\angle ABC=2\angle C$，$\therefore\angle AGD=2\angle C$，
$\therefore CG=DG=BD$，$\therefore AC=AG+CG=AB+BD$.
$\because AB=6,BD=5$，$\therefore AC=11$.
$\because\angle ABE=\frac{1}{2}\angle ABC$，$\angle C=\frac{1}{2}\angle ABC$，$\therefore\angle ABE=\angle C$.
又$\because\angle BAE=\angle CAB$，$\therefore\triangle ABE\backsim\triangle ACB$，
$\therefore\frac{AB}{AC}=\frac{AE}{AB}$，$\therefore AB^{2}=AE· AC$，
$\because\angle EBC=\angle C$，$\therefore BE=CE$，
$\therefore AB^{2}=AC(AC-BE)$，$\therefore BE=\frac{85}{11}$. (9分)
(3)如图2,取$CF$中点$M$,连接$AM$.
在$Rt\triangle ACF$中,$AM=CM=\frac{1}{2}CF$，$\therefore\angle AMF=2\angle C$，
又$\because\angle ABC=2\angle C$，$\therefore\angle ABC=\angle AMF$，
$\therefore AM=AB$，$\therefore CF=2AB$. (12分)

23. 在平面直角坐标系中,$O$为坐标原点,抛物线$y = ax^{2} + bx + c$($a$,$b$,$c$为常数,且$a\neq 0$)与$x$轴交于$A(-1,0)$,$B(3,0)$两点,与$y$轴交于点$C(0,3)$.
(1) 求抛物线的函数表达式.
(2) $P$,$Q$两点均在抛物线上,$PD\perp x$轴于点$D$,$QE\perp x$轴于点$E$,$AP$与$y$轴交于点$F$.已知$P$,$Q$两点的横坐标分别为$t$和$t + 1$,且$0 \lt t \lt 2$.
(ⅰ) 分别记$\triangle AOF$和$\triangle CPF$的面积为$S_{\triangle AOF}$,$S_{\triangle CPF}$,求$S_{\triangle AOF} + S_{\triangle CPF}$的最小值.
(ⅱ) 分别记$\triangle OPD$和$\triangle OQE$的面积为$S_{\triangle OPD}$,$S_{\triangle OQE}$,若$S_{\triangle OPD} = S_{\triangle OQE}$,求$t$的值.

答案： 23.解:
(1)由题意得，$\begin{cases}a-b+c=0,\\9a+3b+c=0,\\c=3,\end{cases}$解得$\begin{cases}a=-1,\\b=2,\\c=3,\end{cases}$
$\therefore$抛物线的函数表达式为$y=-x^{2}+2x+3$. (4分)
(2)(i)如图1,设$P(t,-t^{2}+2t+3)$,直线$PA$的函数表达式为$y=mx+n$，
$\because$直线$PA$经过$A(-1,0),P(t,-t^{2}+2t+3)$，$\therefore\begin{cases}-m+n=0,\\mt+n=-t^{2}+2t+3,\end{cases}$
$\therefore mt+m=-t^{2}+2t+3$，
即$m(t+1)=-(t+1)(t-3)$.
$\because0＜t＜2$，$\therefore t+1＞0$，
$\therefore m=-t+3$，$\therefore n=-t+3$，
$\therefore$直线$PA$的函数表达式为$y=(-t+3)x+(-t+3)$，
$\therefore OF=-t+3,CF=3-(-t+3)=t$，
$\therefore S_{\triangle AOF}=\frac{1}{2}(-t+3),S_{\triangle CPF}=\frac{1}{2}t^{2}$，
$\therefore S_{\triangle AOF}+S_{\triangle CPF}=\frac{1}{2}t^{2}+\frac{1}{2}(-t+3)=\frac{1}{2}(t-\frac{1}{2})^{2}+\frac{11}{8}$，
$\because0＜t＜2$，
$\therefore S_{\triangle AOF}+S_{\triangle CPF}$的最小值为$\frac{11}{8}$. (9分)
(ii)如图2,$\because P(t,-t^{2}+2t+3),Q(t+1,-(t+1)^{2}+2(t+1)+3)$，即$Q(t+1,-t^{2}+4)$，
$\therefore S_{\triangle OPD}=\frac{1}{2}t(-t^{2}+2t+3),S_{\triangle OQE}=\frac{1}{2}(t+1)·(-t^{2}+4)$，
$\because S_{\triangle OPD}=S_{\triangle OQE}$，
$\therefore\frac{1}{2}t(-t^{2}+2t+3)=\frac{1}{2}(t+1)(-t^{2}+4)$，
即$t(t+1)(t-3)=(t+1)(t^{2}-4)$，
$\because0＜t＜2$，$\therefore t+1＞0$，
$\therefore t^{2}-3t=t^{2}-4$，解得$t=\frac{4}{3}$. (14分)