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如图,在\(▱ABCD\)中,过点\(A\)作\(AE \bot DC\)交\(DC\)的延长线于点\(E\)过点\(D\)作\(DF \bot BA\),交\(BA\)的延长线于点\(F\),求证:四边形\(AEDF\)是矩形。
参考答案:证明:\(\because \)四边形\(ABCD\)是平行四边形,
\(\therefore AB//DC\),\(AF//ED\),
\(\because AE \bot DC\),
\(∴AE⊥BA\),
∵\(DF \bot BA\),
\(\therefore DF//EA\),
\(\therefore \)四边形\(AEDF\)是平行四边形,
\(\because AE \bot DE\),
\(\therefore \angle E = 90^\circ \),
\(\therefore \)四边形\(AEDF\)是矩形。