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如图,\( E\)是\(▱ABCD\)的边\( AD\)延长线上一点,\( BE\)交\( DC\)于\( F\)。求证:\( {S}_{△ADF}={S}_{△EFC}\)。
参考答案:\(\because ▱ABCD\mathrm{},\)
\(\therefore AB//CD,AD//BC,\)
\(\therefore AB//CD,\mathrm{}AE//BC\mathrm{}\)
连接\( BD\)
\( \therefore {S}_{△BDF}={S}_{△ADF},\)
\(\because {S}_{△BCE}={S}_{△BCD},\)
\(\therefore {S}_{△CFE}={S}_{△BFD},\)
\(\therefore {S}_{△FDB}={S}_{△FDA}\)。