《2025年学霸提优大试卷八年级数学上册苏科版》

24. (10分)新题型新定义(2025·宿迁模拟)定义:如图①,点$M,N把线段EF分割成EM,MN和FN$,若$MN^2= EM^2+NF^2$,则称点$M,N是线段EF$的“弦割点”,其中点$M$称为“左弦点”,点$N$称“右弦点”.
(1)如图①,已知点$M,N是线段EF$的弦割点”,若$EM= 1,MN= 2$,求$FN$的长;
(2)如图②,在$Rt\triangle AOB$中,$OA= OB,C,D为线段AB$的“弦割点”,以$CD为斜边在AB上方作等腰直角三角形CDG$,连接$OC,OD$,则$\angle GCO+\angle GDO= $______;
(3)如图③,已知点$P是线段HK$的“左弦点”,请作出线段$HK$的“右弦点”$Q$.(尺规作图,保留作图痕迹)

答案：

(1)
∵已知点M,N是线段EF的“弦割点”,
∴MN² = EM² + FN²,
∴2² = 1² + FN²,
∴FN = $\sqrt{3}$.
(2) 225°　[解析] 如图①,
∵∠AOB = 90°,OA = OB,
∴∠OAB = ∠OBA = 45°.将△BOD绕点O逆时针旋转90°至△AOE,连接CE,
∴∠OAE = ∠OBD = 45°,AE = BD,OE = OD,∠BOD = ∠AOE,
∴∠BAE = ∠OAE + ∠OAB = 90°,
∴CE² = AE² + AC² = BD² + AC².
∵C,D为线段AB的“弦割点”,
∴CD² = BD² + AC²,
∴CD = CE;
∵OC = OC,
∴△COE≌△COD(SSS),
∴∠COE = ∠COD,
∴∠COD = ∠AOE + ∠AOC = ∠BOD + ∠AOC;
∵∠COD + ∠BOD + ∠AOC = 90°,
∴∠COD = 45°.
∵△CDG是等腰直角三角形,
∴∠G = 90°,
∴∠GCO + ∠GDO = 360° - 90° - 45° = 225°.
(3) 如图②,(Ⅰ)作HK的垂直平分线AB,交HK于点B,在AB上截取BA = BH;(Ⅱ)连接AP,作CA⊥AP;(Ⅲ)作∠PAC的平分线AQ,交HK于点Q,则Q是求作的点.

25. (10分)等边三角形$ABC的两边AB,AC所在直线上分别有点M,N,D为\triangle ABC$外一点,且$\angle MDN= 60^{\circ},\angle BDC= 120^{\circ},BD= CD$.当点$M,N分别在直线AB,AC$上移动时,探究$BM,CN,MN之间的数量关系以及\triangle AMN的周长Q与等边三角形ABC的周长L$的关系.
(1)如图①,当点$M,N分别在边AB,AC$上,且$DM= DN$时,$BM,CN,MN$之间的数量关系式为______,此时$\frac{Q}{L}$的值是______.
(2)如图②,当点$M,N分别在边AB,AC$上,且$DM\neq DN$时,猜想(1)问的两个结论还成立吗?写出你的猜想并加以证明.
(3)如图③,当点$M,N分别在边AB,CA$的延长线上时,若$AN= x$,试用含$x,L的代数式表示Q$.

答案：

(1) BM + CN = MN,$\frac{2}{3}$　[解析]
∵DM = DN,∠MDN = 60°,
∴△MDN是等边三角形.
∵△ABC是等边三角形,
∴∠A = 60°.
∵BD = CD,∠BDC = 120°,
∴∠DBC = ∠DCB = 30°,
∴∠MBD = ∠NCD = 90°.
∵DM = DN,BD = CD,
∴Rt△BDM≌Rt△CDN(HL),
∴∠BDM = ∠CDN = 30°,BM = CN,
∴DM = 2BM,DN = 2CN,
∴MN = 2BM = 2CN = BM + CN,
∴AM = AN,
∴△AMN是等边三角形.
∵AB = AM + BM,
∴AM:AB = 2:3,
∴$\frac{Q}{L}$ = $\frac{2}{3}$.
(2) 猜想:结论仍然成立.证明:如图①,延长AC至点$M_1$,使$CM_1$ = BM,连接$DM_1$.由
(1)知∠MBD = ∠$M_1$CD = 90°,
∵BD = CD,
∴△DBM≌△D$M_1$C(SAS),
∴DM = D$M_1$.
∵∠MDN = 60°,∠BDC = 120°,
∴∠$M_1$DN = ∠MDN = 60°,
∴△MDN≌△$M_1$DN(SAS),
∴MN = $M_1$N = $M_1$C + CN = BM + CN,
∴△AMN的周长为AM + MN + AN = AM + BM + CN + AN = AB + AC,
∴$\frac{Q}{L}$ = $\frac{2}{3}$.
(3) 如图②,在CN上截取$CM_2$ = BM,连接$DM_2$.由
(1)知∠MBD = ∠$M_2$CD = 90°,
∵BD = CD,
∴△DBM≌△D$M_2$C(SAS),
∴DM = D$M_2$.
∵∠MDN = 60°,∠BDC = 120°,
∴∠$M_2$DN = ∠MDN = 60°,
∴△MDN≌△$M_2$DN(SAS),
∴MN = $M_2$N,
∴NC - BM = MN.
∵等边三角形ABC的周长为L,
∴AB = $\frac{L}{3}$,△AMN的周长Q = MN + AN + AM = NC - BM + AN + AB + BM = AN + AC + AN + AB = 2AN + 2AB = 2x + $\frac{2}{3}$L,即Q = 2x + $\frac{2}{3}$L.