《2025年名校题库九年级数学全一册北师大版》

9. (高新区一诊)在矩形 $ABCD$ 中,$P$ 为 $AB$ 边上一点,将 $\triangle BCP$ 沿直线 $CP$ 翻折,使点 $B$ 落在矩形内的点 $E$ 处,直线 $AE$ 与 $CD$ 边交于点 $F$。
(1)如图 1,当 $P$ 为 $AB$ 的中点时,求证：$AF// CP$；
(2)如图 2,若 $AB= 6$,$BC= 5$,$CF= 1$,求线段 $BP$ 的长；
(3)若直线 $AE$ 与 $CP$ 的延长线交于点 $Q$,$BC= 1$,$AB= n$,当 $S_{\triangle CEP}= S_{\triangle APQ}$ 时,求 $\frac{AF}{CF}$ 的值(用含 $n$ 的代数式表示)。

答案：

(1)证明：如图1，连接$BE$交$CP$于点$G$。$\because$ 将$\triangle BCP$沿直线$CP$翻折，使点$B$落在矩形内的点$E$处，$\therefore PE=PB$，$CE=CB$，$BG=EG$，$BE\perp CP$，$\therefore G$为$BE$的中点。$\because P$为$AB$的中点，$\therefore PG$是$\triangle ABE$的中位线，$\therefore PG// AE$，即$AF// CP$。
(2)解：如图2，过点$E$作$EK\perp AB$于点$K$，连接$BE$交$CP$于点$L$。设$BP=x$，则$AP=6 - x$。$\because$ 四边形$ABCD$是矩形，$\therefore CD=AB=6$，$AD=BC=5$，$\angle BAD=\angle ABC=\angle D=90^{\circ}$。$\because CF=1$，$\therefore DF=CD - CF=6 - 1=5$，$\therefore AD=DF$，$\therefore \triangle ADF$是等腰直角三角形，$\therefore \angle FAD=\angle AFD=45^{\circ}$，$\therefore \angle EAK=90^{\circ}-45^{\circ}=45^{\circ}$，$\therefore \triangle AEK$是等腰直角三角形，$\therefore AK=EK$。由翻折得$BE\perp CP$，$BE=2BL$，$\therefore \angle BCP+\angle CBE=90^{\circ}$。在$Rt\triangle BCP$中，$CP=\sqrt{BC^{2}+BP^{2}}=\sqrt{5^{2}+x^{2}}$。$\because S_{\triangle BCP}=\frac{1}{2}BC\cdot BP=\frac{1}{2}CP\cdot BL$，$\therefore BL=\frac{BC\cdot BP}{CP}=\frac{5x}{\sqrt{x^{2}+25}}$，$\therefore BE=2BL=\frac{10x}{\sqrt{x^{2}+25}}$。$\because \angle CBP=\angle BKE=90^{\circ}$，$\therefore \angle KBE+\angle CBE=90^{\circ}$，$\therefore \angle BCP=\angle KBE$，$\therefore \triangle CPB\backsim\triangle BEK$，$\therefore \frac{EK}{BP}=\frac{BE}{CP}=\frac{BK}{BC}$，即$\frac{EK}{x}=\frac{\frac{10x}{\sqrt{x^{2}+25}}}{\sqrt{x^{2}+25}}=\frac{BK}{5}$，$\therefore EK=\frac{10x^{2}}{x^{2}+25}$，$BK=\frac{50x}{x^{2}+25}$，$\therefore AK=6-\frac{50x}{x^{2}+25}$。$\because AK=EK$，$\therefore 6-\frac{50x}{x^{2}+25}=\frac{10x^{2}}{x^{2}+25}$。整理，得$2x^{2}+25x - 75=0$，解得$x_{1}=-15$，$x_{2}=\frac{5}{2}$。经检验，$x_{1}=-15$，$x_{2}=\frac{5}{2}$均为所列方程的根。$\because x＞0$，$\therefore x=\frac{5}{2}$，$\therefore$ 线段$BP$的长为$\frac{5}{2}$。
(3)设$BP=x$，则$AP=n - x$。如图3，过点$Q$作$QG\perp AB$，交$BA$的延长线于点$G$，连接$BE$交$CP$于点$H$，过点$E$作$EK\perp AB$于点$K$，连接$AC$，则$\angle AGQ=90^{\circ}$。
$\therefore \angle AGQ=\angle ABC$，$\therefore QG// BC$，$\therefore \triangle PQG\backsim\triangle PCB$，$\therefore \frac{QG}{BC}=\frac{PG}{BP}$，即$\frac{QG}{1}=\frac{PG}{x}$，$\therefore PG=x\cdot QG$。$\because S_{\triangle CEP}=S_{\triangle CBP}$，$S_{\triangle CEP}=S_{\triangle APQ}$，$\therefore S_{\triangle CBP}=S_{\triangle APQ}$，$\therefore BC\cdot BP=AP\cdot QG$，$\therefore QG=\frac{BC\cdot BP}{AP}=\frac{1\cdot x}{n - x}=\frac{x}{n - x}$，$\therefore PG=x\cdot QG=\frac{x^{2}}{n - x}$，$\therefore AG=PG - AP=\frac{x^{2}}{n - x}-(n - x)=\frac{2nx - n^{2}}{n - x}$。在$Rt\triangle CPB$中，$CP=\sqrt{BP^{2}+BC^{2}}=\sqrt{x^{2}+1}$。$\because QG// BC$，$BC// AD$，$\therefore QG// AD$，$\therefore \angle FAD=\angle AQG$。又$\because \angle D=\angle G=90^{\circ}$，$\therefore \triangle AFD\backsim\triangle QAG$，$\therefore \frac{QG}{AD}=\frac{AG}{DF}=\frac{AQ}{AF}$，$\therefore \frac{\frac{x}{n - x}}{1}=\frac{\frac{2nx - n^{2}}{n - x}}{DF}$，$\therefore DF=\frac{2nx - n^{2}}{x}$，$\therefore CF=CD - DF=n-\frac{2nx - n^{2}}{x}=\frac{n^{2}-nx}{x}$。$\because S_{\triangle CBP}=S_{\triangle APQ}$，$\therefore S_{\triangle CBP}+S_{\triangle PBQ}=S_{\triangle APQ}+S_{\triangle PBQ}$，即$S_{\triangle AQB}=S_{\triangle CQB}$，$\therefore AC// QB$，$\therefore \angle ACQ=\angle CQB$。又$QC$垂直平分$EB$，$\therefore QE=QB$，$\therefore \angle AQC=\angle CQB$，$\therefore \angle AQC=\angle ACQ$，$\therefore AQ=AC=\sqrt{AB^{2}+BC^{2}}=\sqrt{n^{2}+1}$。$\because \frac{QG}{AD}=\frac{AG}{DF}=\frac{AQ}{AF}$，$\therefore \frac{\frac{x}{n - x}}{1}=\frac{\sqrt{n^{2}+1}}{AF}$，$\therefore AF=\frac{(n - x)\sqrt{n^{2}+1}}{x}$，$\therefore \frac{AF}{CF}=\frac{\frac{(n - x)\sqrt{n^{2}+1}}{x}}{\frac{n^{2}-nx}{x}}=\frac{\sqrt{n^{2}+1}}{n}$。