【2021中考真题19】菏泽压轴——第23题(动点路径)
![](http://pic.ikafan.com/imgp/L3Byb3h5L2h0dHBzL2ltYWdlMTA5LjM2MGRvYy5jbi9Eb3dubG9hZEltZy8yMDIxLzA3LzIwMjMvMjI2NjQ5OTQxXzVfMjAyMTA3MjAxMTEzMDM0ODk=.jpg)
双平等腰模型
![](http://pic.ikafan.com/imgp/L3Byb3h5L2h0dHBzL2ltYWdlMTA5LjM2MGRvYy5jbi9Eb3dubG9hZEltZy8yMDIxLzA3LzIwMjMvMjI2NjQ5OTQxXzZfMjAyMTA3MjAxMTEzMDM3ODY=.jpg)
△EPF为等腰三角形不变
![](http://pic.ikafan.com/imgp/L3Byb3h5L2h0dHBzL2ltYWdlMTA5LjM2MGRvYy5jbi9Eb3dubG9hZEltZy8yMDIxLzA3LzIwMjMvMjI2NjQ5OTQxXzdfMjAyMTA3MjAxMTEzMDM4NjQ=.jpg)
因为AE=CD,故DE=BF,所以DE=EH=BF,又因为PE=PF,所以PH=PB
![](http://pic.ikafan.com/imgp/L3Byb3h5L2h0dHBzL2ltYWdlMTA5LjM2MGRvYy5jbi9Eb3dubG9hZEltZy8yMDIxLzA3LzIwMjMvMjI2NjQ5OTQxXzhfMjAyMTA3MjAxMTEzMDM5MjY=.jpg)
根据PH=PB,可证△PHM≌△PBM,则∠HPM=∠BPM,又因为PE=PF,由“三线合一”可知PO是EF的垂直平分线,故点M在EF的垂直平分线上
![](http://pic.ikafan.com/imgp/L3Byb3h5L2h0dHBzL2ltYWdlMTA5LjM2MGRvYy5jbi9Eb3dubG9hZEltZy8yMDIxLzA3LzIwMjMvMjI2NjQ5OTQxXzlfMjAyMTA3MjAxMTEzMDQ0.jpg)
变中不变,动线过定点,连接BD交EF于点Q,易证BQ=DQ,故EF必过定点Q
![](http://n4.ikafan.com/assetsj/blank.gif)
旋转变换想隐圆,连接QG、QC,则QG=QC=5,则点G在以Q为圆心,5为半径的圆弧上,E在A时点G与点C重合,当E到AD中点P时,点G与B重合,故点G的运动路径是下图中弧BC的长,圆心角∠BQC=120°,半径QC=5,根据弧长公式计算即可求解
![](http://n4.ikafan.com/assetsj/blank.gif)
本题关键是确定动线EF在运动过程中过矩形ABCD的中心Q是解题的关键,体现了解决动态运动问题的根本策略即“变中不变”
![](http://n4.ikafan.com/assetsj/blank.gif)
![](http://n4.ikafan.com/assetsj/blank.gif)
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