旋转于中考的矩形
2022-05-30周贵胜
周贵胜
图形的旋转是各地中考数学试卷中永恒的话题,本文选取一道中考题为例进行分析.
例 (2021·浙江·嘉兴)小王在学习浙教版九上课本第[72]页例[2]后,进一步开展探究活动:将一个矩形[ABCD]绕点[A]顺时针旋转[α](0° < [α] ≤ 90°),得到矩形[AB'C'D'],连接[BD].
(1)探究1:如图[1],当[α=90°]时,点[C']恰好在[DB]延长线上[.]若[AB=1],求[BC]的长.
(2)探究2:如图[2],连接[AC'],过点[D']作[D'M]∥[AC']交[BD]于点[M.] 线段[D'M]与[DM]相等吗?请说明理由.
(3)探究3:在探究[2]的条件下,射线[DB]分别交[AD'],[AC']于点[P],[N(]如图[3)],发现线段[DN],[MN],[PN]存在一定的数量关系,请写出这个关系式,并加以证明.
分析:(1)设BC = [x],由旋转的性质得出矩形ABCD ≌ 矩形AB′C′D′,所以C′D′ = CD = AB,AD′ = AD = BC = [x],由△BAD ∽ △BD′C′,可列比例式求解;
(2)如图4,连接DD′,由旋转可知△AC′D′ ≌ △DBA,可得[∠D'AC'=∠ADB],由MD′∥AC′,得到∠MD′A = ∠D′AC′,再证出∠MDD′ = ∠MD′D,则得到DM = D′M;
(3) 由(2)可证明△NPA ∽ △NAD,[ANNP=NDAN],连接AM,如图5,可推导出∠NMA = ∠NAM = 2∠MDA,得出MN = AN,进而可得出结论:MN2 = NP·ND.
解:(1)设[BC=x],
∵矩形[ABCD]绕点[A]顺时针旋转90°得到矩形[AB'C'D'],
∴点[A],[B],[D']在同一直线上,矩形ABCD ≌ 矩形AB′C′D′,
∴∠BAD = ∠D′ = 90°,[AD'=AD=BC=x],D′C′ = DC = AB = 1,
∴[D'B=AD'-AB=x-1],
又∵点[C']在[DB]的延长线上,∴∠D′BC′ = ∠ABD,
∴[△D'C'B] ∽ [△ADB],
∴[D'C'AD=D'BAB],∴[1x=x-11],
解得[x1=1+52],[x2=1-52](不合题意,舍去),
∴[BC=1+52].
(2)[D'M=DM].
证明:如圖4,连接[DD'],
[∵D'M∥AC'],[∴∠AD'M=∠D'AC'],
[∵AD'=DA],[∠AD'C'=∠DAB=90°],[D'C'=AB],
[∴△AC'D'] ≌ [△DBA](SAS),
[∴∠D'AC'=∠ADB],[∴∠ADB=∠AD'M],
[∵AD'=AD],[∴∠ADD'=∠AD'D],
[∴∠MDD'=∠MD'D],[∴D'M=DM].
(3)关系式为[MN2=PN?DN].
证明:如图5,连接AM,
∵MD = MD′,AM = AM,AD = AD′,
∴△AMD ≌ △AMD′,∴∠MAD = ∠MAD′,
∵∠NMA = ∠ADM + ∠MAD,∠NAM = ∠NAP + ∠D′AM,
∴∠NMA = ∠NAM,∴MN = AN.
在[△NAP]和[△NDA]中,[∠ANP=∠DNA],[∠NAP=∠NDA],
∴△NPA ∽ △NAD,∴[PNAN=ANDN],
[∴AN2=PN?DN],[∴MN2=PN?DN].
点评:本题是矩形结合图形旋转的综合题,解这类问题的关键是利用图形旋转的性质得到角相等、线段相等等条件.
(作者单位:东北育才学校初中部)