《2025年5年中考3年模拟九年级数学上册北师大版》

8.「2025上海虹口期中,」在正方形网格中,每个小正方形的顶点称为格点,以格点为顶点的三角形叫做格点三角形.如图,$\triangle ABC$是格点三角形,在图中作出格点$\triangle ADE$(不含$\triangle ABC$),使得$\triangle ADE\backsim \triangle ABC$(同一位置的格点$\triangle ADE$只算一个),这样的格点三角形一共有6个.

答案： 6

9.多解法「2025福建泉州七中月考改编,」如图所示的是由4个边长为1的正方形组成的图形.
(1)求证:$\triangle ABD\backsim \triangle BCD$.
(2)求$∠ABC$的度数.

答案：
(1)证明:【证法一】由题意得AB=√10,BC=√5,BD=√2,
∴AB/BC = √2,AD/BD = √2,BD/DC = √2.
∴AB/BC = AD/BD = BD/DC,
∴△ABD∽△BCD.【证法二】根据题意得AD=2,BD=√2,CD=1,
∴AD/BD = BD/DC.由正方形的性质得∠1=45°,
∴∠ADB=∠BDC=135°.
∴△ABD∽△BCD.
(2)
∵△ABD∽△BCD,
∴∠BAD=∠CBD.
∴∠1=∠BAD+∠ABD=∠CBD+∠ABD=45°,即∠ABC=45°.

10.分类讨论「2023河北保定一中月考,」一个钢筋三角形支架边长分别是20 cm,50 cm,60 cm,现在要做一个与其相似的钢筋三角形支架,而只有长为30 cm和50 cm的两根钢筋,要求以其中一根为一边,从另一根上截下两段(允许有余料)作为两边,有几种不同的截法?

答案： 当取30cm为一边长时，设另两边长分别为xcm、ycm(x＜y).若30cm与20cm对应,则x/50 = y/60 = 30/20,解得x=75,y=90.75+90＞50,故此种情况不存在.若30cm与50cm对应,则x/20 = 30/50 = y/60,解得x=12,y=36.12+36=48＜50,故此种情况存在.若30cm与60cm对应，则x/20 = y/50 = 30/60,解得x=10,y=25.10+25=35＜50,故此种情况存在.当取50cm作为一边长时，无法得到符合题意的三角形.综上所述，有两种不同的截法.

11.推理能力如图①,在$\triangle ABC和\triangle A'B'C'$中,D、$D'$分别在AB、$A'B'$上,$\frac {AD}{AB}= \frac {A'D'}{A'B'}$.

(1)当$\frac {CD}{C'D'}= \frac {AC}{A'C'}= \frac {AB}{A'B'}$时,证明$\triangle ABC\backsim \triangle A'B'C'$的途径可以用如下框图表示,请填写其中的空格.

$CD/C'D' = AC/A'C' = AD/A'D'$$\angle A=\angle A'$
(2)当$\frac {CD}{C'D'}= \frac {AC}{A'C'}= \frac {BC}{B'C'}$时,求证:$\triangle ABC\backsim \triangle A'B'C'$.
如图,过点D、D'分别作DE//BC,D'E'//B'C',DE交AC于点E,D'E'交A'C'于点E'.
∵DE//BC,
∴△ADE∽△ABC,
∴AD/AB = DE/BC = AE/AC,同理,A'D'/A'B' = D'E'/B'C' = A'E'/A'C'.又AD/AB = A'D'/A'B',
∴DE/BC = D'E'/B'C',
∴DE/D'E' = BC/B'C'.同理,AE/AC = A'E'/A'C'.
∴(AC - AE)/AC = (A'C' - A'E')/A'C',即EC/AC = E'C'/A'C'.
∴EC/E'C' = AC/A'C'.又CD/C'D' = AC/A'C' = BC/B'C',
∴CD/C'D' = EC/E'C' = DE/D'E',
∴△DCE∽△D'C'E'.
∴∠CED=∠C'E'D'.
∵DE//BC,
∴∠CED+∠ACB=180°.同理,∠C'E'D'+∠A'C'B'=180°.
∴∠ACB=∠A'C'B'.又AC/A'C' = BC/B'C',
∴△ABC∽△A'B'C'.
(3)如图②,M是AC的中点,P,Q是BC的三等分点,AP、AQ分别交BM于点D、点E,则$BD:DE:EM=$5:3:2.

答案：
(1)
∵AD/AB = A'D'/A'B',
∴AD/A'D' = AB/A'B',
∵CD/C'D' = AC/A'C' = AB/A'B',
∴CD/C'D' = AC/A'C' = AD/A'D',
∴△ADC∽△A'D'C',
∴∠A=∠A',
∵AC/A'C' = AB/A'B',
∴△ABC∽△A'B'C'.故答案为CD/C'D' = AC/A'C' = AD/A'D';∠A=∠A'.
(2)如图,过点D、D'分别作DE//BC,D'E'//B'C',DE交AC于点E,D'E'交A'C'于点E'.
∵DE//BC,
∴△ADE∽△ABC,
∴AD/AB = DE/BC = AE/AC,同理,A'D'/A'B' = D'E'/B'C' = A'E'/A'C'.又AD/AB = A'D'/A'B',
∴DE/BC = D'E'/B'C',
∴DE/D'E' = BC/B'C'.同理,AE/AC = A'E'/A'C'.
∴(AC - AE)/AC = (A'C' - A'E')/A'C',即EC/AC = E'C'/A'C'.
∴EC/E'C' = AC/A'C'.又CD/C'D' = AC/A'C' = BC/B'C',
∴CD/C'D' = EC/E'C' = DE/D'E',
∴△DCE∽△D'C'E'.
∴∠CED=∠C'E'D'.
∵DE//BC,
∴∠CED+∠ACB=180°.同理,∠C'E'D'+∠A'C'B'=180°.
∴∠ACB=∠A'C'B'.又AC/A'C' = BC/B'C',
∴△ABC∽△A'B'C'.
(3)5:3:2