《2025年轻松暑假复习加预习中国海洋大学出版社八年级数学54制》

18.(10分)(2024·湖北模拟)如图所示,为了测量一栋楼的高度$OE$,小明同学先在操场上$A$处放一面镜子,向后退到$B$处,恰好在镜子中看到楼的顶部$E$；再将镜子放到$C$处,然后后退到$D$处,恰好再次在镜子中看到楼的顶部$E$($O$,$A$,$B$,$C$,$D$在同一条直线上),测得$AC= 2m$,$BD= 2.1m$,如果小明眼睛距地面高度$BF$,$DG为1.6m$,试确定楼的高度$OE$.

答案：
解：

令$OE = a$，$AO = b$，$CB = x$，
则由$\triangle GDC \backsim \triangle EOC$得$\frac{GD}{EO} = \frac{CD}{OC}$，
即$\frac{1.6}{a} = \frac{2.1 - x}{2 + b}$，
整理得：$3.2 + 1.6b = 2.1a - ax$， ①
由$\triangle FBA \backsim \triangle EOA$得$\frac{FB}{EO} = \frac{AB}{OA}$，
即$\frac{1.6}{a} = \frac{2 - x}{b}$，
整理得：$1.6b = 2a - ax$， ②
将②代入①，得$3.2 + 2a - ax = 2.1a - ax$，
$\therefore a = 32$，
即$OE = 32$，
答：楼的高度$OE$为 32 米.

19.(12分)(2023·温州)如图所示,已知矩形$ABCD$,点$E在CB$延长线上,点$F在BC$延长线上,过点$F作FH\perp EF交ED的延长线于点H$,连接$AF交EH于点G$,$GE= GH$.
(1)求证：$BE= CF$；
证明：$\because FH \perp EF$，$GE = GH$，
$\therefore$ $GE = GF = GH$.
$\therefore$ $\angle GFE = \angle E$.
$\because$ 四边形$ABCD$是矩形，
$\therefore$ $AB = CD$，$\angle ABC = \angle DCB = 90^{\circ}$.
$\therefore$ $\triangle ABF \cong \triangle DCE(AAS)$.
$\therefore$ $BF = CE$.
$\therefore$ $BF - BC = CE - BC$，即$BE = CF$；
(2)当$\frac{AB}{FH}= \frac{5}{6}$,$AD= 4$时,求$EF$的长.
解：$\because$ $CD // FH$，
$\therefore$ $\triangle DCE \backsim \triangle HFE$.
$\therefore$ $\frac{EC}{EF} = \frac{CD}{FH}$.
$\because$ $CD = AB$，
$\therefore$ $\frac{CD}{FH} = \frac{AB}{FH} = \frac{5}{6}$.
设$BE = CF = x$，
$\because$ $BC = AD = 4$，
$\therefore$ $CE = x + 4$，$EF = 2x + 4$.
$\because$ $\frac{x + 4}{2x + 4} = \frac{5}{6}$.
解得 $x = 1$，
$\therefore$ $EF = 6$.

答案： 解：
(1)$\because FH \perp EF$，$GE = GH$，
$\therefore GE = GF = GH$.
$\therefore \angle GFE = \angle E$.
$\because$ 四边形$ABCD$是矩形，
$\therefore AB = CD$，$\angle ABC = \angle DCB = 90^{\circ}$.
$\therefore \triangle ABF \cong \triangle DCE(AAS)$.
$\therefore BF = CE$.
$\therefore BF - BC = CE - BC$，即$BE = CF$；
(2)$\because CD // FH$，
$\therefore \triangle DCE \backsim \triangle HFE$.
$\therefore \frac{EC}{EF} = \frac{CD}{FH}$.
$\because CD = AB$，
$\therefore \frac{CD}{FH} = \frac{AB}{FH} = \frac{5}{6}$.
设$BE = CF = x$，
$\because BC = AD = 4$，
$\therefore CE = x + 4$，$EF = 2x + 4$.
$\because \frac{x + 4}{2x + 4} = \frac{5}{6}$.
解得$x = 1$，
$\therefore EF = 6$.

20.(14分)已知：如图1所示,$AB\perp BD$,$CD\perp BD$,垂足分别为$B$,$D$,$AD和BC相交于点E$,$EF\perp BD$,垂足为$F$,我们可以证明$\frac{1}{AB}+\frac{1}{CD}= \frac{1}{EF}$成立(不要求考生证明).


若将图中的垂线改为斜交,如图2所示,$AB// CD$,$AD$,$BC相交于点E$,过点$E作EF// AB交BD于点F$,则：
(1)$\frac{1}{AB}+\frac{1}{CD}= \frac{1}{EF}$还成立吗？如果成立,请给出证明；如果不成立,请说明理由；
(2)请找出$S_{\triangle ABD}$,$S_{\triangle BED}和S_{\triangle BDC}$间的关系式,并给出证明.

答案：

(1)成立.
证明：$\because AB // EF$，
$\therefore \frac{EF}{AB} = \frac{DF}{DB}$.
$\because CD // EF$，
$\therefore \frac{EF}{CD} = \frac{BF}{DB}$.
$\therefore \frac{EF}{AB} + \frac{EF}{CD} = \frac{DF}{DB} + \frac{BF}{DB} = \frac{DB}{DB} = 1$.
$\therefore \frac{1}{AB} + \frac{1}{CD} = \frac{1}{EF}$；
(2)关系式为：$\frac{1}{S_{\triangle ABD}} + \frac{1}{S_{\triangle BDC}} = \frac{1}{S_{\triangle BED}}$.
证明如下：分别过$A$作$AM \perp BD$于点$M$，过点$E$作$EN \perp BD$于点$N$，过点$C$作$CK \perp BD$交$BD$的延长线于点$K$，

由题设可得：$\frac{1}{AM} + \frac{1}{CK} = \frac{1}{EN}$，
$\therefore \frac{2}{BD \cdot AM} + \frac{2}{BD \cdot CK} = \frac{2}{BD \cdot EN}$.
即$\frac{1}{\frac{1}{2} \cdot BD \cdot AM} + \frac{1}{\frac{1}{2} \cdot BD \cdot CK} = \frac{1}{\frac{1}{2} \cdot BD \cdot EN}$
又$\because \frac{1}{2} \cdot BD \cdot AM = S_{\triangle ABD}$，$\frac{1}{2} \cdot BD \cdot CK = S_{\triangle BDC}$，$\frac{1}{2} \cdot BD \cdot EN = S_{\triangle BED}$，
$\therefore \frac{1}{S_{\triangle ABD}} + \frac{1}{S_{\triangle BDC}} = \frac{1}{S_{\triangle BED}}$.