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在\( \mathrm{R}\mathrm{t}△ABC\)中,\(\mathrm{\angle }C=90°\),\(D\),\(E\),\(F\)分别是\( AC,AB,BC\)的中点,连接\(ED\),\(EF\)。求证四边形\( DEFC\)是矩形。
参考答案:证明:
\(\because D\),\(E\),\(F\)分别是\(AC\),\(AB\),\(BC\)的中点
\(\therefore DE\)、\(EF\)分别是\(\Delta ABC\)的中位线
\(\therefore DE//BC\),\(EF//AC\)
\(\therefore \)四边形\(DEFC\)是平行四边形
\(\because \angle C = 90^\circ \)
\(\therefore \)平行四边形\(DEFC\)是矩形