《2026年启东中学作业本九年级数学下册苏科版》

4. 如图，在□ABCD中，AM⊥BC，AN⊥CD，垂足分别为M，N.
求证：(1)△AMB∽△AND；
(2)$\frac{AM}{AB}=\frac{MN}{AC}$.

答案： 证明:
(1) $ \because $ 四边形 $ ABCD $ 是平行四边形,
$\therefore ∠ B = ∠ D $.
$\because AM ⊥ BC $, $ AN ⊥ CD $,
$\therefore ∠ AMB = ∠ AND = 90^{\circ}$,
$\therefore △ AMB ∼ △ AND $.
(2) $ \because △ AMB ∼ △ AND $, $ \therefore \frac{AM}{AN} = \frac{AB}{AD} $, $ ∠ BAM = ∠ DAN $.
又 $ \because AD = BC $, $ \therefore \frac{AM}{AN} = \frac{AB}{BC} $.
$\because ∠ B + ∠ BCD = ∠ MAN + ∠ BCD = 180^{\circ}$,
$\therefore ∠ B = ∠ MAN $, $ \therefore △ AMN ∼ △ BAC $, $ \therefore \frac{AM}{AB} = \frac{MN}{AC} $.

5. 如图，P为矩形ABCD的对角线BD上一点，PE⊥AP交BC于点E，若AD=10，DC=6，求$\frac{PA}{PE}$的值.

答案：
解: 如答图, 过点 $ P $ 作 $ AD $ 的垂线 $ MN $, 交 $ AD $ 于点 $ M $, 交 $ BC $ 于点 $ N $. $ \because ∠ APM + ∠ NPE = 180^{\circ} - ∠ APE = 90^{\circ} $, $ ∠ APM + ∠ PAM = 180^{\circ} - ∠ AMP = 90^{\circ} $,
$\therefore ∠ NPE = ∠ PAM $.
又 $ \because ∠ AMP = ∠ PNE = 90^{\circ} $.
$\therefore △ APM ∼ △ PEN $, $ \therefore \frac{PA}{PE} = \frac{AM}{PN} $.
$\because ∠ PNB = ∠ DCB = 90^{\circ} $, $ ∠ PBN = ∠ DBC $,
$\therefore △ BPN ∼ △ BDC $, $ \therefore \frac{BN}{PN} = \frac{BC}{CD} $.
$\because $ 四边形 $ AMNB $, 四边形 $ ADCB $ 均为矩形,
$\therefore AM = BN $, $ AD = BC $,
$\therefore \frac{AM}{PN} = \frac{AD}{CD} = \frac{10}{6} = \frac{5}{3} $, $ \therefore \frac{PA}{PE} = \frac{5}{3} $.

6. 已知正方形ABCD，E是对角线AC上一点.
(1)如图①，连接BE，DE. 求证：△ABE≌△ADE；
(2)如图②，F是DE延长线上一点，DF交AB于点G，BF⊥BE. 判断△FBG的形状并说明理由；
(3)在(2)的条件下，若BE=BF=2，求$\frac{AE}{AB}$的值.

答案：
(1) 证明: $ \because $ 四边形 $ ABCD $ 是正方形,
$\therefore AB = AD = CB = CD $, $ ∠ ABC = ∠ ADC = 90^{\circ} $,
$\therefore ∠ BAC = ∠ BCA = ∠ DAC = ∠ DCA = 45^{\circ} $.
在 $ △ ABE $ 和 $ △ ADE $ 中, $ \begin{cases} AB = AD, \\ ∠ BAE = ∠ DAE, \\ AE = AE, \end{cases} $
$\therefore △ ABE ≌ △ ADE (\mathrm{SAS}) $.
(2) 解: $ △ FBG $ 是等腰三角形, 理由如下:
$\because △ ABE ≌ △ ADE $,
$\therefore ∠ ABE = ∠ ADE $, $ \therefore ∠ ABC - ∠ ABE = ∠ ADC - ∠ ADE $, $ \therefore ∠ EBC = ∠ EDC $.
$\because AB // CD $, $ \therefore ∠ FGB = ∠ EDC $, $ \therefore ∠ FGB = ∠ EBC $.
$\because BF ⊥ BE $, $ \therefore ∠ FBE = 90^{\circ} $, $ \therefore ∠ FBG = ∠ EBC = 90^{\circ} - ∠ ABE $, $ \therefore ∠ FGB = ∠ FBG $, $ \therefore BF = GF $, $ \therefore △ FBG $ 是等腰三角形.
(3) 解: $ \because BE = BF = 2 $, $ ∠ FBE = 90^{\circ} $, $ \therefore ∠ F = ∠ BEF = 45^{\circ} $, $ \therefore ∠ BAC = ∠ F $, $ \therefore ∠ AEG = ∠ FBG $.
$\because ∠ AGE = ∠ FGB $ 且 $ ∠ FGB = ∠ FBG $,
$\therefore ∠ AGE = ∠ AEG $, $ \therefore AE = AG $.
$\because EF = \sqrt{BE^2 + BF^2} = \sqrt{2^2 + 2^2} = 2\sqrt{2} $, $ BF = GF = 2 $,
$\therefore GE = EF - GF = 2\sqrt{2} - 2 $.
$\because △ ABE ≌ △ ADE $, $ \therefore BE = DE = 2 $.
$\because AG // CD $, $ \therefore △ AGE ∼ △ CDE $,
$\therefore \frac{AG}{CD} = \frac{GE}{DE} = \frac{2\sqrt{2} - 2}{2} = \sqrt{2} - 1 $, $ \therefore \frac{AE}{AB} = \sqrt{2} - 1 $.