题目内容
已知:如图①,在矩形ABCD中,AB=5,AD=
,AE⊥BD,垂足是E.点F是点E关于AB的对称点,连接AF、BF.
(1)求AE和BE的长;
(2)若将△ABF沿着射线BD方向平移,设平移的距离为m(平移距离指点B沿BD方向所经过的线段长度).当点F分别平移到线段AB、AD上时,直接写出相应的m的值.
(3)如图②,将△ABF绕点B顺时针旋转一个角
(0°<
<180°),记旋转中的△ABF为△A′BF′,在旋转过程中,设A′F′所在的直线与直线AD交于点P.与直线BD交于点Q.是否存在这样的P、Q两点,使△DPQ为等腰三角形?若存在,求出此时DQ的长;若不存在,请说明理由.
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(1)求AE和BE的长;
(2)若将△ABF沿着射线BD方向平移,设平移的距离为m(平移距离指点B沿BD方向所经过的线段长度).当点F分别平移到线段AB、AD上时,直接写出相应的m的值.
(3)如图②,将△ABF绕点B顺时针旋转一个角
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(1)4,3;(2)当点F在线段AB上时,
;当点F在线段AD上时,
;
(3)存在,
.
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(3)存在,
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试题分析:(1)由勾股定理求得BD的长,根据三角形面积公式求出AE的长,再应用勾股定理即可求得BE的长.
(2)根据平移的性质求解即可.
(3)分DP=DQ(考虑点Q在线段BD的延长线和点Q在线段BD上两种情况),QP=QD,PD=PQ三种情况求解即可.
试题解析:(1)∵AB=5,AD=
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∵
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∴
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(2)当点F在线段AB上时,
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(3)存在,理由如下:
①当DP=DQ时,若点Q在线段BD的延长线上时,如答图1,有∠Q=∠1,则∠2=∠1+∠Q=2∠Q.
∵∠3=∠4+∠Q,∠3=∠2,∴∠4+∠Q=2∠Q.∴∠4=∠Q.
∴A′Q=A′B=5.∴F′Q=4+5=9.
在Rt△BF′Q中,
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若点Q在线段BD上时,如答图2,有∠1=∠2=∠4,
∵∠1=∠3,∴∠3=∠4.
∵∠3=∠5+∠A′,∠A′=∠CBD,∴∠3=∠5+∠CBD=∠A′BQ.∴∠4=∠∠A′BQ.∴A′Q= A′B=5.
∴F′Q=5-4=1.∴
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②当QP=QD时,如答图3,有∠P=∠1,
∵∠A′=∠1,∠2=∠3,∴∠4=∠P.∴∠4=∠A′.∴QB="Q" A′.
设QB="Q" A′=x,
在Rt△BF′Q中,
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③当PD=PQ时,如答图4,有∠1=∠2=∠3,
∵∠1=∠A′,∴∠3=∠A′.∴BQ=A′B=5.
∴
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综上所述,当△DPQ为等腰三角形时,DQ的长为
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