《2025年名校题库九年级数学全一册北师大版》

8. (金牛区一诊)如图 1,菱形 ABCD 的边长为 5,$DE⊥AB,DF⊥BC$,垂足分别为 E,F,连接 EF,已知$DE= 4.$
(1)求证:$△ADE\cong △CDF;$
(2)求 EF 的长;
(3)连接 AC,与 BD 相交于点 O,将图 1 中的$△DEF$绕点 D 旋转,当点 E 落在线段 OC 上时,如图 2,点 G 在线段 AC 上,连接 FG,与 DE 相交于点 H,$∠EGF= ∠EDF$,求$\frac {GH}{EH}$的值.

答案：

(1) 证明: $\because DE\perp AB,DF\perp BC,\therefore \angle AED = \angle DFC = 90^{\circ}$. $\because$ 四边形 $ABCD$ 是菱形, $\therefore AD = CD,\angle A = \angle C,\therefore\triangle ADE\cong\triangle CDF(AAS)$.
(2) 解: 如图 1, 连接 $AC,BD$, 交于点 $O$.
$\because$ 四边形 $ABCD$ 是菱形, $\therefore AC\perp BD,AC = 2OA,AB = AD = 5,OD = \frac{1}{2}BD$. $\because DE\perp AB,DE = 4,\therefore AE = \sqrt{AD^{2}-DE^{2}} = 3,\therefore BE = AB - AE = 5 - 3 = 2$. $\because DE\perp AB,\therefore \angle DEB = 90^{\circ},\therefore BD = \sqrt{DE^{2}+BE^{2}} = \sqrt{4^{2}+2^{2}} = 2\sqrt{5},\therefore OD = \sqrt{5}$, $\therefore OA = \sqrt{AD^{2}-OD^{2}} = \sqrt{5^{2}-(\sqrt{5})^{2}} = 2\sqrt{5},\therefore AC = 4\sqrt{5}$. 由
(1) 知, $\triangle ADE\cong\triangle CDF,\therefore CF = AE = 3$, $\therefore BE = BF = 2,\therefore \frac{BE}{AB} = \frac{BF}{BC} = \frac{2}{5}$. $\because \angle EBF = \angle ABC,\therefore\triangle BEF\backsim\triangle BAC$, $\therefore \frac{EF}{AC} = \frac{BE}{AB},\therefore \frac{EF}{4\sqrt{5}} = \frac{2}{5},\therefore EF = \frac{8\sqrt{5}}{5}$.
(3) 解: 如图 2, 过点 $D$ 作 $DV// GH$, 交 $AC$ 于点 $V$, 作 $DW\perp EF$ 于点 $W$, 过点 $E$ 作 $EQ\perp DF$ 于点 $Q$, $\therefore\triangle EGH\backsim\triangle EVD,\therefore \frac{GH}{EH} = \frac{DV}{DE},\angle DVO = \angle EGF = \angle EDF$.
$\because \angle DOV = \angle DQE = 90^{\circ},\therefore\triangle DOV\backsim\triangle EQD$, $\therefore \frac{OD}{DV} = \frac{EQ}{DE}$. 由
(2) 知, $DE = DF = 4,EF = \frac{8\sqrt{5}}{5},\therefore EW = FW = \frac{1}{2}EF = \frac{4\sqrt{5}}{5},\therefore DW = \sqrt{DE^{2}-EW^{2}} = \sqrt{4^{2}-(\frac{4\sqrt{5}}{5})^{2}} = \frac{8\sqrt{5}}{5}$. 由 $S_{\triangle DEF} = \frac{1}{2}DF\cdot EQ = \frac{1}{2}EF\cdot DW$, 得 $EQ = \frac{EF\cdot DW}{DF} = \frac{\frac{8\sqrt{5}}{5}\times\frac{8\sqrt{5}}{5}}{4} = \frac{16}{5}$, $\therefore \frac{OD}{DV} = \frac{EQ}{DE} = \frac{\frac{16}{5}}{4} = \frac{4}{5},\therefore \frac{\sqrt{5}}{DV} = \frac{4}{5},\therefore DV = \frac{5\sqrt{5}}{4},\therefore \frac{GH}{EH} = \frac{DV}{DE} = \frac{\frac{5\sqrt{5}}{4}}{4} = \frac{5\sqrt{5}}{16}$.