《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》

3.如图，在矩形ABCD中，E，F两点在对角线AC上，AE=CF.
(1)求证：四边形BEDF是平行四边形.
(2)若AD=15，AB=20，DE⊥AC，求四边形BEDF的面积.

答案：
解析
(1)证明:如图,连接DB,交AC于O点,
∵四边形ABCD为矩形,

∴OA = OC,OB = OD,
∵AE = CF,
∴OA - AE = OC - CF,
∴OE = OF,
∵OB = OD,
∴四边形BEDF是平行四边形.
(2)
∵四边形ABCD是矩形,
∴∠ADC = 90°,CD = AB = 20,
∴AC = √(AD² + CD²) = 25,
∵DE⊥AC,
∴S△ADC = 1/2AD·CD = 1/2AC·DE,
∴DE = AD·CD/AC = 15×20/25 = 12,
∴AE = √(AD² - DE²) = 9,
∴EF = 25 - 9 - 9 = 7,
∴四边形BEDF的面积 = 2×1/2DE·EF = 12×7 = 84.

4.如图，在□ABCD中，O为线段AD的中点，延长BO交CD的延长线于点E，连接AE，BD，∠BDC=90°.
(1)求证：四边形ABDE是矩形.
(2)连接OC，若AB=2，BD=2√2，求OC的长.
　　　　　　　　　　　　

答案：
解析
(1)证明:
∵O为AD的中点,
∴AO = DO,
∵四边形ABCD是平行四边形,
∴AB//CD,
∴∠BAO = ∠EDO,
又
∵∠AOB = ∠DOE,
∴△AOB≌△DOE(ASA),
∴AB = DE,
∴四边形ABDE是平行四边形,
∵∠BDC = 90°,
∴∠BDE = 90°,
∴平行四边形ABDE是矩形.
(2)如图,过点O作OF⊥DE于点F,
∵四边形ABDE是矩形,

∴DE = AB = 2,OD = 1/2AD,OB = OE = 1/2BE,AD = BE,
∴OD = OE,
∵OF⊥DE,
∴DF = EF = 1/2DE = 1,
∴OF为△BDE的中位线,
∴OF = 1/2BD = √2,
∵四边形ABCD是平行四边形,
∴CD = AB = 2,
∴CF = CD + DF = 3,
在Rt△OCF中,由勾股定理得OC = √(OF² + CF²) = √((√2)² + 3²) = √11,即OC的长为√11.

5.(2024北京中关村中学期中)正方形ABCD中，点P是射线BD上一动点，连接AP，过P作PE⊥AP，交射线CD于E，连接AE.
(1)如图①，请补全图形.
(2)如图②，当点E在CD的延长线上时，试确定线段BP与CE之间的数量关系，并说明理由.
(3)如图③，点P在BD的延长线上，若AB=3，DP=$\sqrt{2}$，则四边形ADPE的面积=________.

答案：
解析
(1)补全图形如图.

(2)√2BP = CE,理由如下:
过点P作PH⊥AB于H,延长HP交CD于F,过点P作PG⊥BC于G,如图,

∵四边形ABCD为正方形,
∴AD = CD,∠ADB = ∠CDB = 45°,∠ABC = ∠BCD = 90°,
易得四边形PHBG为正方形,四边形PGCF为矩形,
∴PG = BG = CF,PH//AD,
在Rt△PBG中,由勾股定理得BP = √(PG² + BG²) = √2BG,
∴√2/2BP = PG,
在△ADP和△CDP中,{AD = CD,∠ADP = ∠CDP,DP = DP},
∴△ADP≌△CDP(SAS),
∴∠DAP = ∠DCP,
设∠DAP = ∠DCP = α,则∠CPF = 90° - α,
∵PH//AD,
∴∠APH = ∠DAP = α,
∵PE⊥AP,
∴∠EPF = 90° - α,
∴∠CPF = ∠EPF = 90° - α,
在△CPF和△EPF中,{∠CPF = ∠EPF,PF = PF,∠CFP = ∠EFP},
∴△CPF≌△EPF(ASA),
∴CP = EP,CF = EF,
∴PG = CF = EF = 1/2CE,
∴√2/2BP = PG = 1/2CE,
∴√2BP = CE.
(3)10.详解:过点P作PT⊥CE于T,连接PC,如图,

同
(2)可得CP = EP,则CT = ET,
∵∠PDT = ∠CDB = 45°,
∴△PDT为等腰直角三角形,
∴PT = DT,
在Rt△PTD中,由勾股定理得DP = √(PT² + DT²) = √2DT,
∵DP = √2,
∴DT = PT = 1,
∵AB = CD = AD = 3,
∴CT = ET = CD + DT = 4,
∴ED = ET + DT = 4 + 1 = 5,
∵S△ADE = 1/2DE·AD = 1/2×5×3 = 7.5,S△PDE = 1/2DE·PT = 1/2×5×1 = 2.5,
∴S四边形ADPE = S△ADE + S△PDE = 7.5 + 2.5 = 10.