《2025年鹏教图书精彩假期暑假篇七年级》

7. 【问题背景】
在$△ABC$中,$AC= BC$,$∠C= 90^{\circ }$,点D为直线AC上一点,连接BD,在BD右侧作$BE⊥BD且BE= BD$,过点E作$EF⊥AB$交直线AB于点F,交直线BC于点H。
【初步探究】
(1)如图(a),当点D在线段AC上时,求证：$AB= EH$。
【推广探究】
(2)如图(b),当点D为CA延长线上一点时,其他条件不变,(1)中的结论是否仍然成立？若成立,请证明；若不成立,请说明理由。
【拓展应用】
(3)若$AB= 8$,$BF= 3$,其他条件不变,直接写出EF的长。
（1）证明：如图1，
$\because AC = BC$，$∠C = 90^{\circ}$
$\therefore ∠A = ∠ABC = 45^{\circ}$
$\therefore ∠FBH = ∠ABC = 45^{\circ}$
$\because BE⊥BD$，$EF⊥AB$，
$\therefore ∠DBE = ∠BFE = ∠BFH = 90^{\circ}$
$\therefore ∠ABD = ∠E = 90^{\circ}-∠EBF$
$∠H = ∠FBH = 45^{\circ}$
$\therefore ∠A = ∠H$
在$\triangle ABD$和$\triangle HEB$中，$\because ∠ABD = ∠E$，$∠A = ∠H$，$BD = EB$，
$\therefore \triangle ABD\cong \triangle HEB(AAS)$
$\therefore AB = EH$
（2）解：成立
理由如下：如图2
$\because ∠DBE = ∠BFE = 90^{\circ}$，
$\therefore ∠ABD = ∠E = 90^{\circ}-∠EBF$
$\because ∠FHB = ∠ABC = ∠BAC = 45^{\circ}$，
$\therefore ∠BAD = ∠EHB = 180^{\circ}-45^{\circ}= 135^{\circ}$
在$\triangle ABD$和$\triangle HEB$中，$\because ∠ABD = ∠E$，$∠BAD = ∠EHB$，$BD = EB$，
$\therefore \triangle ABD\cong \triangle HEB(AAS)$
$\therefore AB = EH$
（3）5或11

答案：
（1）证明：如图1，
$\because AC = BC$，$∠C = 90^{\circ}$
$\therefore ∠A = ∠ABC = 45^{\circ}$

$\therefore ∠FBH = ∠ABC = 45^{\circ}$
$\because BE⊥BD$，$EF⊥AB$，
$\therefore ∠DBE = ∠BFE = ∠BFH = 90^{\circ}$
$\therefore ∠ABD = ∠E = 90^{\circ}-∠EBF$
$∠H = ∠FBH = 45^{\circ}$
$\therefore ∠A = ∠H$
在$\triangle ABD$和$\triangle HEB$中，$\because ∠ABD = ∠E$，$∠A = ∠H$，$BD = EB$，
$\therefore \triangle ABD\cong \triangle HEB(AAS)$
$\therefore AB = EH$
（2）解：成立
理由如下：如图2
$\because ∠DBE = ∠BFE = 90^{\circ}$，

$\therefore ∠ABD = ∠E = 90^{\circ}-∠EBF$
$\because ∠FHB = ∠ABC = ∠BAC = 45^{\circ}$，
$\therefore ∠BAD = ∠EHB = 180^{\circ}-45^{\circ}= 135^{\circ}$
在$\triangle ABD$和$\triangle HEB$中，$\because ∠ABD = ∠E$，$∠BAD = ∠EHB$，$BD = EB$，
$\therefore \triangle ABD\cong \triangle HEB(AAS)$
$\therefore AB = EH$
（3）5或11