《2025年5年中考3年模拟八年级数学下册北京课改版北京专版》

8.垂直十字模型(2024四川泸州中考,12,)如图,在边长为6的正方形ABCD中,点E,F分别是边AB,BC上的动点,且满足AE = BF,AF与DE交于点O,点M是DF的中点,G是边AB上的点,AG = 2GB,则OM + $\frac{1}{2}$FG的最小值是 (　　)

A.4　　　　B.5　　　　C.8　　　　D.10

答案：
B
∵四边形ABCD是正方形,
∴AD = AB,∠DAB = ∠ABC = 90°,又
∵AE = BF,
∴△ADE≌△BAF(SAS),
∴∠ADE = ∠BAF,
∴∠DOF = ∠ADO + ∠DAO = ∠BAF + ∠DAO = ∠DAB = 90°,
∵点M是DF的中点,
∴OM = $\frac{1}{2}$DF,如图,在AB延长线上截取BH = BG,连接FH,DH,

∵∠FBG = ∠FBH = 90°,FB = FB,BG = BH,
∴△FBG≌△FBH(SAS),
∴FH = FG,
∴OM + $\frac{1}{2}$FG = $\frac{1}{2}$DF + $\frac{1}{2}$HF = $\frac{1}{2}$(DF + HF),
∴当H、D、F三点共线时,DF + HF有最小值,即OM + $\frac{1}{2}$FG有最小值,最小值为DH长的一半,
∵AG = 2GB,AB = 6,
∴BH = BG = 2,
∴AH = 8,在Rt△ADH中,由勾股定理得DH = $\sqrt{AD^{2}+AH^{2}}$ = 10.
∴OM + $\frac{1}{2}$FG的最小值为5.故选B.

9.(2024北京西城期末,15,)如图,在平面直角坐标系xOy中,点A(3,2$\sqrt{2}$),AB⊥y轴于点B.以AB为边作菱形ABCD,若点C在x轴上,则点D的坐标为 ____________________.

答案：
答案 (2,0)或(4,0)
解析
∵点A(3,2$\sqrt{2}$),AB⊥y轴于点B,
∴AB = 3,OB = 2$\sqrt{2}$,
∵四边形ABCD是菱形,
∴BC = CD = AB = 3,分两种情况:
①当点C在x轴负半轴上时,如图,

OC = $\sqrt{BC^{2}-OB^{2}}$ = $\sqrt{3^{2}-(2\sqrt{2})^{2}}$ = 1,
∴OD = CD - OC = 3 - 1 = 2,
∴D(2,0);
②当点C在x轴正半轴上时,如图,

OC = $\sqrt{BC^{2}-OB^{2}}$ = $\sqrt{3^{2}-(2\sqrt{2})^{2}}$ = 1,
∴OD = CD + OC = 3 + 1 = 4,
∴D(4,0).
综上,点D的坐标为(2,0)或(4,0).

10.(2024北京十一中三模,20,)如图,矩形ABCD中,过点B作BE//AC交DC的延长线于点E.过点D作DF⊥BE于点F,点G为AC的中点,连接FG,BD.
(1)求证:BE = AC.
(2)若AB = 2,BC = 4,求FG的长.

答案： 解析
(1)证明:
∵四边形ABCD是矩形,
∴AB//CD,
∵AC//BE,
∴四边形ABEC是平行四边形,
∴BE = AC.
(2)
∵四边形ABCD是矩形,G为AC的中点,
∴BD = AC,G为BD的中点,
∵DF⊥BE,
∴∠DFB = 90°,
∴GF = $\frac{1}{2}$BD,
∵BD = AC = $\sqrt{2^{2}+4^{2}}$ = 2$\sqrt{5}$,
∴FG = $\sqrt{5}$.

11.推理能力(2024北京门头沟期末)如图,在正方形ABCD中,点E是边AD上的一点(不与点A,D重合),连接CE,点B关于直线CE的对称点是点F,连接CF,DF,直线CE与直线DF交于点P,连接BF与直线CE交于点Q.
(1)依题意补全图形.
(2)求∠CPF的度数.
(3)用等式表示线段PC,PD,PF之间的数量关系,并证明.

答案：
解析
(1)如图.

(2)
∵四边形ABCD是正方形,
∴BC = CD,∠BCD = 90°.
∵点B,F关于直线CP对称,
∴∠CBF = ∠CFB,CP⊥BF,BC = CF,
∴∠BCQ + ∠QBC = ∠BCQ + ∠PCD = 90°,
∴∠QBC = ∠PCD.
∵BC = CF = CD,
∴∠CFD = ∠CDF.
∵∠CFD = ∠CFQ + ∠QFD = ∠CDF = ∠PCD + ∠CPD,
∴∠QFD = ∠CPD,
∵∠PQF = 90°,
∴∠CPD = 45°,即∠CPF = 45°.
(3)PF + PD = $\sqrt{2}$PC.证明:如图,过点C作CH⊥PC交PF的延长线于点H,

∴∠PCH = 90°.
∵∠CPF = 45°,
∴∠H = ∠CPF = 45°,
∴PC = CH,
∵CD = CF,
∴∠CDF = ∠CFD,
∴∠CDP = ∠CFH,
∴△CPD≌△CHF(AAS),
∴PD = HF.在Rt△PCH中,PH = $\sqrt{2}$PC,
∴PF + PD = $\sqrt{2}$PC.