《2025年名校题库九年级数学全一册北师大版》

5. (成华区二诊)在$△ABC$中,$∠ACB= 90^{\circ },BC= 4,AC= 3$,CD 是 AB 边上的中线,E 是 BC 边上的一个动点,连接 DE,将$△BDE$沿直线 DE 翻折得到$△FDE$.
(1)如图 1,线段 DF 与线段 BC 相交于点 G,当 G 是 BC 边的中点时,求 BE 的长.
(2)如图 2,当点 E 与点 C 重合时,线段 EF 与线段 AB 相交于点 P,求 DP 的长.
(3)如图 3,线段 EF 与线段 CD 相交于点 M,是否存在点 E,使得$△DFM$为直角三角形? 若存在,请直接写出 BE 的长;若不存在,请说明理由.

答案：
解:
(1) $\because \angle ACB = 90^{\circ},BC = 4,AC = 3,\therefore AB = \sqrt{AC^{2}+BC^{2}} = \sqrt{3^{2}+4^{2}} = 5$. $\because CD$ 是 $AB$ 边上的中线, $\therefore CD = BD = \frac{1}{2}AB = \frac{5}{2}$. $\because G$ 是 $BC$ 的中点, $D$ 是 $AB$ 的中点, $\therefore DG$ 是 $\triangle ABC$ 的中位线, $\therefore DG = \frac{1}{2}AC = \frac{3}{2},BG = \frac{1}{2}BC = 2$. $\because$ 将 $\triangle BDE$ 沿直线 $DE$ 翻折得到 $\triangle FDE$, $\therefore DF = BD = \frac{5}{2},FE = BE$, $\therefore GF = DF - DG = 1$. $\because DG$ 是 $\triangle ABC$ 的中位线, $\therefore DF// AC$, $\therefore \angle EGF = \angle CGD = 180^{\circ}-\angle ACB = 90^{\circ}$. 设 $BE = EF = x$, 则 $GE = 2 - x$. 在 $Rt\triangle GEF$ 中, $EF^{2} = GF^{2}+GE^{2}$, 即 $x^{2} = 1^{2}+(2 - x)^{2}$, 解得 $x = \frac{5}{4},\therefore BE = \frac{5}{4}$.
(2) 由
(1) 知 $CD = BD,\therefore \angle BCD = \angle B$. $\because \angle PCD = \angle BCD,\therefore \angle PCD = \angle B$. $\because \angle CPD = \angle BPC,\therefore\triangle CPD\backsim\triangle BPC$, $\therefore \frac{DP}{CP} = \frac{PC}{PB} = \frac{CD}{BC} = \frac{\frac{5}{2}}{4} = \frac{5}{8}$. 设 $DP = 5k$, 则 $CP = 8k,PB = 5k + \frac{5}{2},\therefore \frac{PC}{PB} = \frac{8k}{5k + \frac{5}{2}} = \frac{5}{8}$, 解得 $k = \frac{25}{78}$. 经检验, $k = \frac{25}{78}$ 是所列方程的根, $\therefore DP = 5\times\frac{25}{78} = \frac{125}{78}$.
(3) 存在. 如图 1, 当 $\angle FMD = 90^{\circ}$ 时, $\because \angle F = \angle B,\angle FMD = \angle ACB = 90^{\circ}$, $\therefore\triangle FDM\backsim\triangle BAC$, $\therefore \frac{DF}{AB} = \frac{DM}{AC},\therefore \frac{\frac{5}{2}}{5} = \frac{DM}{3},\therefore DM = \frac{3}{2}$, $\therefore CM = CD - DM = 1$. $\because \angle ECM = \angle B,\angle CME = \angle BCA = 90^{\circ}$, $\therefore\triangle CEM\backsim\triangle BAC$, $\therefore \frac{CE}{AB} = \frac{CM}{CB},\therefore \frac{CE}{5} = \frac{1}{4},\therefore CE = \frac{5}{4},\therefore BE = BC - CE = 4 - \frac{5}{4} = \frac{11}{4}$.

如图 2, 当 $\angle FDM = 90^{\circ}$ 时, $\because \angle F = \angle B = \angle BCD,\angle FMD = \angle CME$, $\therefore \angle CEM = \angle FDM = 90^{\circ},\therefore \angle FED = \angle BED = 45^{\circ}$. 过点 $D$ 作 $DH\perp BC$ 于点 $H$, 则 $\triangle BDH\backsim\triangle BAC,\therefore \frac{DH}{AC} = \frac{BH}{BC} = \frac{DB}{AB} = \frac{1}{2}$, $\therefore DH = \frac{3}{2},BH = 2,\therefore EH = DH = \frac{3}{2}$, $\therefore BE = EH + BH = \frac{3}{2} + 2 = \frac{7}{2}$. 综上所述, $BE$ 的长为 $\frac{11}{4}$ 或 $\frac{7}{2}$.