《2025年黄冈状元成才路状元大课堂八年级数学下册人教版》

例 ★★☆ (1)如图①,在□ABCD中,O是对角线AC的中点,过点O的直线EF分别交AD,BC于点E,F.若□ABCD的面积是8,则四边形CDEF的面积是____.
(2)探究:如图②,在菱形ABCD中,对角线AC,BD相交于点O,过点O的直线EF分别交AD,BC于点E,F.若AB = 5,BD = 8,求四边形ABFE的面积.
(3)应用:如图③,在Rt△ABC中,∠BAC = 90°,延长BC至点D,使DC = BC,连接AD.若AC = 1,AD = √5,求△ABD的面积.

分析:(1)考虑证△AOE≌△COF 转化 S_{四边形CDEF}=S_{△ACD};(2)考虑证△AOE≌△COF 转化 S_{四边形ABFE}=S_{△ABC};(3)遇中点,考虑用倍长中线法构造全等三角形 转化 S_{△ABD}=S_{△ADE}.
解:(1)4
(2)∵四边形ABCD是菱形,
∴AD//BC,AO = CO,BO = 1/2BD = 4,AC⊥BD.∴AO = √(AB² - BO²)=√(5² - 4²)= 3,AC = 2AO = 6,∠OAE = ∠OCF,∠OEA = ∠OFC.
在△AOE和△COF中,{∠OEA = ∠OFC,∠OAE = ∠OCF,AO = CO,
∴△AOE≌△COF(AAS),
∴S_{△AOE}=S_{△COF}.
∴S_{四边形ABFE}=S_{△AOE}+S_{四边形ABFO}=S_{△COF}+S_{四边形ABFO}=S_{△ABC}=1/2AC·BO = 1/2×6×4 = 12.
(3)如图③,延长AC至点E,使CE = AC = 1,连接DE.
在△ABC和△EDC中,{AC = EC,∠ACB = ∠ECD,BC = DC,
∴△ABC≌△EDC(SAS), 倍长中线可以构造全等三角形.
∴S_{△ABC}=S_{△EDC},∠E = ∠BAC = 90°.
∵AE = AC + CE = 1 + 1 = 2,
∴在Rt△ADE中,DE = √(AD² - AE²)=√((√5)² - 2²)= 1.
∴S_{△ABD}=S_{△ABC}+S_{△ACD}=S_{△EDC}+S_{△ACD}=S_{△ADE}=1/2AE·DE = 1/2×2×1 = 1.

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