梯形ABCD中,AD ∥CB,对角线AC,BD相交于点O,设梯形ABCD的面S,△AOD,...
2个回答

∵AD∥BC,∴S(△ABD)＝S(△ACD),∴S(△AOD)＋S(△AOB)＝S(△AOD)＋S(△COD),

∴S(△AOB)＝S(△COD)＝S3.

∴S1＋S2＋2S3＝S.······①

∵AD∥BC,∴AO/CO＝DO/BO,∴（AO/CO）（BO/DO）＝1.

显然有：S3/S1＝BO/DO、S3/S2＝AO/CO,∴（S3/S1）（S3/S2）＝1,∴S3^2＝S1S2,

∴√S1√S2＝S3.······②

由①、②,得：S1＋2√S1√S2＋S2＝S,∴（√S1＋√S2）^2＝S,∴√S1＋√S2＝√S.······③

由②、③,结合韦达定理可知：√S1、√S2是方程x^2－√Sx＋S3＝0的两实数根.

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