“A”型结构证相似
一、知识准备
相似三角形的性质、判定方法
二、拓展导学
【问题呈现】
如图所示，在△ABC中，AB=8cm，BC=16cm.点P从点A出发沿AB向点B以2cm/s的速度运动，点Q从点B出发沿BC向点C以4cm/s的速度运动.
如果P，Q分别从A，B出发，问几秒钟后△PBQ与△ABC相似？
【思路点拨】
△PBQ与△ABC已经有∠B这个公共角了，只要使夹∠B的两边对应成比例即可使△PBQ与△ABC相似.点P、Q在运动，需分情况讨论.
【问题解决】
解：设x秒钟后△PBQ与△ABC相似，此时AP=2x，BP=8-2x,BQ=4x
情况(1)：若使△BPQ∽△BAC，则需：![](https://wkrtcs.bdimg.com/rtcs/image?w=69.33&md5sum=4bccbe17e9351f75034aa936f2bec916&sign=265974b2f3&rtcs_flag=2&rtcs_ver=3.1&l=webapp&bucketNum=89&ipr={"t":"img","w":"69.33","h":"41.33","dataType":"gif","c":"word/media/image3.gif"})
即：![](https://wkrtcs.bdimg.com/rtcs/image?w=80.00&md5sum=4bccbe17e9351f75034aa936f2bec916&sign=265974b2f3&rtcs_flag=2&rtcs_ver=3.1&l=webapp&bucketNum=89&ipr={"t":"img","w":"80.00","h":"41.33","dataType":"gif","c":"word/media/image4.gif"})
，解得x=2
情况(2)：若使△BQP∽△BAC，则需：![](https://wkrtcs.bdimg.com/rtcs/image?w=70.67&md5sum=4bccbe17e9351f75034aa936f2bec916&sign=265974b2f3&rtcs_flag=2&rtcs_ver=3.1&l=webapp&bucketNum=89&ipr={"t":"img","w":"70.67","h":"41.33","dataType":"gif","c":"word/media/image6.gif"})
.即：![](https://wkrtcs.bdimg.com/rtcs/image?w=80.00&md5sum=4bccbe17e9351f75034aa936f2bec916&sign=265974b2f3&rtcs_flag=2&rtcs_ver=3.1&l=webapp&bucketNum=89&ipr={"t":"img","w":"80.00","h":"41.33","dataType":"gif","c":"word/media/image7.gif"})
,解得
∴综上，2秒或![](https://wkrtcs.bdimg.com/rtcs/image?w=16.00&md5sum=4bccbe17e9351f75034aa936f2bec916&sign=265974b2f3&rtcs_flag=2&rtcs_ver=3.1&l=webapp&bucketNum=89&ipr={"t":"img","r":"r_5","w":"16.00","h":"41.33","dataType":"gif","c":"word/media/image9.gif"})
秒后△PBQ与△ABC相似.
【经验分享】
“A”型结构是指如图所示的图形，可以构造相似三角形，有平行相似和不平行相似两种情况.
三、配套练习
1. 如图，在△ABC中，AB=6，AC=3，∠BAC=120°，∠BAC的平分线交BC于点D.求AD的长.
2．如图，在直角坐标系中，OB=6，OA=8.D为Rt△AOB的斜边AB的中点，过点D作直线截△AOB，使截得的三角形与△AOB相似.求满足这样条件的直线的函数表达式.
四、学习感悟
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文档为doc格式
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