摘要:证明:如图10.∵ FE⊥轴.FG⊥轴.∠BAD = 90°,∴ 四边形AEFG是矩形 .∴ AE = GF.EF = AG . ∴ S△AEF = S△AFG ,同理S△ABC = S△ACD .∴ S△ABC-S△AEF = S△ACD-S△AFG . 即S1 = S2 . (2)∵FG∥CD , ∴ △AFG ∽ △ACD .

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