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
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![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](/uploads/image/z/12885228-36-8.jpg?t=%E5%88%9D%E4%BA%8C%E4%B8%8A%E5%86%8C%E6%95%B0%E5%AD%A6+%E4%B8%A4%E9%81%93%E5%85%B3%E4%BA%8E%E8%8F%B1%E5%BD%A2%E7%9A%84%E9%A2%98%E7%9B%AE+%E8%AF%B7%E6%B1%82%E5%B8%AE%E5%BF%99%E8%A7%A3%E7%AD%94+%E8%A6%81%E5%BF%AB%E5%93%A61.+%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%2C%E8%A7%92C%3D90%E5%BA%A6%2CAD%E5%B9%B3%E5%88%86%E8%A7%92BAC%2CED%E5%9E%82%E7%9B%B4BC%2CDF%2F%2FAB%2C%E6%B1%82%E8%AF%81%EF%BC%9AAD%E4%B8%8EEF%E4%BA%92%E7%9B%B8%E5%9E%82%E7%9B%B4%E5%B9%B3%E5%88%862.+%E5%B9%B3%E8%A1%8C%E5%9B%9B%E8%BE%B9%E5%BD%A2ABCD%E7%9A%84%E5%AF%B9%E8%A7%92%E7%BA%BFAC%E7%9A%84%E5%9E%82%E7%9B%B4%E5%B9%B3%E5%88%86%E7%BA%BF%E4%B8%8E%E8%BE%B9AD%E3%80%81BC%E5%88%86%E5%88%AB%E4%BA%A4)
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
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)
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
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
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