Question

初二上册数学 两道关于菱形的题目 请求帮忙解答 要快哦

1. 三角形ABC中，角C=90度，AD平分角BAC，ED垂直BC，DF//AB，求证：AD与EF互相垂直平分

2. 平行四边形ABCD的对角线AC的垂直平分线与边AD、BC分别交于E、F，四边形AFCE是否是菱形？为什么？

（抱歉，没图）

Answer 1

1、因为：DE⊥BC,AC⊥BC
所以：AC‖DE,
而DF‖AB
所以：四边形AFDE是平行四边形
又：∠EAD=∠FAD (由AD平分∠BAC得）
∠EAD=∠FDA
所以：∠FAD=∠FDA
所以：AF=DF
所以：四边形AFDE是棱形
所以：AD,EF互相垂直平分

2、是菱形。
由于四边形ABCD是平行四边形
所以:AE‖CF,
所以：∠EAO=∠FCO (O是AC与EF的交点）
而AO=CO,∠AOE=∠COF
所以：△AOE≌△COF
所以：AE=CF
而由线段的垂直平分线定理得：AE=EC,AF=CF
所以：AE=EC=CF=FA
所以：四边形AFCE是菱形

Answer 2

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