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九年级数学解答题一般
题目
如图,在▱ABCDABCD中,对角线ACAC的垂直平分线分别交ADADBCBC于点FFEE,点OO为垂足,连接AEAE,FCFC.求证:四边形AECFAECF是菱形.
知识点:线段垂直平分线的性质、矩形的性质、菱形的判定与性质章节:第1章 特殊的平行四边形 / 1.1 菱形的性质与判定

答案与解析

答案

证明:\because四边形ABCDABCD是平行四边形,
AD\therefore ADBCBC
EAO=FCO\therefore \angle EAO=\angle FCO
EF\because EF垂直平分ACAC
OA=OC.EA=EC\therefore OA=OC.EA=EC
AOE\triangle AOECOF\triangle COF中,
{EAO=FCOOA=OCAOE=COF\left\{\begin{array}{l}{∠EAO=∠FCO}\\{OA=OC}\\{∠AOE=∠COF}\end{array}\right.
AOE\therefore \triangle AOECOF(ASA)\triangle COF\left(ASA\right)
OE=OF\therefore OE=OF
\therefore四边形AECFAECF是平行四边形,
EA=EC\because EA=EC
\therefore平行四边形AECFAECF是菱形.

解析

证明:\because四边形ABCDABCD是平行四边形,
AD\therefore ADBCBC
EAO=FCO\therefore \angle EAO=\angle FCO
EF\because EF垂直平分ACAC
OA=OC.EA=EC\therefore OA=OC.EA=EC
AOE\triangle AOECOF\triangle COF中,
{EAO=FCOOA=OCAOE=COF\left\{\begin{array}{l}{∠EAO=∠FCO}\\{OA=OC}\\{∠AOE=∠COF}\end{array}\right.
AOE\therefore \triangle AOECOF(ASA)\triangle COF\left(ASA\right)
OE=OF\therefore OE=OF
\therefore四边形AECFAECF是平行四边形,
EA=EC\because EA=EC
\therefore平行四边形AECFAECF是菱形.

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