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8.(秦淮区月考)小丽与爸妈在公园里荡秋千.如图,小丽坐在秋千的起始位置A处,OA与地面垂直,两脚在地面上用力一蹬,妈妈在距地面1 m高的B处接住她后用力一推,爸爸在C处接住她.若妈妈与爸爸到OA的水平距离BD,CE分别为1.4 m和1.8 m,∠BOC=90°.爸爸在C处接住小丽时,小丽距离地面的高度是
1.4
m.答案
8.1.4
9.如图，点E,D,B,F在同一条直线上，AD//CB，∠BAD=∠BCD，DE=BF.
求证:(1)AD=BC；
(2)AE//CF.
求证:(1)AD=BC；
(2)AE//CF.
答案
9. 证明:(1) $\because AD// CB, \therefore ∠ ADB = ∠ CBD$.
在$△ ADB$和$△ CBD$中, $\begin{cases} ∠ ADB = ∠ CBD, \\ ∠ BAD = ∠ DCB, \\ DB = BD, \end{cases}$
$\therefore △ ADB ≌ △ CBD(\mathrm{AAS}), \therefore AD = BC$.
(2) $\because ∠ ADB = ∠ CBD, ∠ ADB + ∠ EDA = 180°, ∠ CBD + ∠ FBC = 180°, \therefore ∠ EDA = ∠ FBC$.
在$△ EDA$和$△ FBC$中, $\begin{cases} DE = BF, \\ ∠ EDA = ∠ FBC, \\ DA = BC, \end{cases}$
$\therefore △ EDA ≌ △ FBC(\mathrm{SAS}), \therefore ∠ E = ∠ F, \therefore AE// CF$.
在$△ ADB$和$△ CBD$中, $\begin{cases} ∠ ADB = ∠ CBD, \\ ∠ BAD = ∠ DCB, \\ DB = BD, \end{cases}$
$\therefore △ ADB ≌ △ CBD(\mathrm{AAS}), \therefore AD = BC$.
(2) $\because ∠ ADB = ∠ CBD, ∠ ADB + ∠ EDA = 180°, ∠ CBD + ∠ FBC = 180°, \therefore ∠ EDA = ∠ FBC$.
在$△ EDA$和$△ FBC$中, $\begin{cases} DE = BF, \\ ∠ EDA = ∠ FBC, \\ DA = BC, \end{cases}$
$\therefore △ EDA ≌ △ FBC(\mathrm{SAS}), \therefore ∠ E = ∠ F, \therefore AE// CF$.
10.(2024春·江宁区月考)如图，AD，BF相交于点O，AB//DF，AC//DE，点E，C在BF上，且BE=CF.
求证:(1)$△ ABC ≌ △ DFE$；
(2)O为BF的中点.
求证:(1)$△ ABC ≌ △ DFE$；
(2)O为BF的中点.
答案
10. 证明:(1) $\because AB// DF, \therefore ∠ B = ∠ F$.
$\because AC// DE, \therefore ∠ ACB = ∠ DEF$.
$\because BE = CF, \therefore BE + EC = CF + EC$, 即 $BC = EF$.
在$△ ABC$和$△ DFE$中, $\begin{cases} ∠ ACB = ∠ DEF, \\ BC = FE, \\ ∠ B = ∠ F, \end{cases}$
$\therefore △ ABC ≌ △ DFE(\mathrm{ASA})$.
(2) $\because △ ABC ≌ △ DFE, \therefore AC = DE$.
在$△ ACO$和$△ DEO$中, $\begin{cases} ∠ AOC = ∠ DOE, \\ ∠ ACO = ∠ DEO, \\ AC = DE, \end{cases}$
$\therefore △ ACO ≌ △ DEO(\mathrm{AAS}), \therefore CO = EO$.
$\because BE = CF, \therefore BE + OE = CF + CO$, 即 $BO = FO$.
$\therefore O$ 为 $BF$ 的中点.
$\because AC// DE, \therefore ∠ ACB = ∠ DEF$.
$\because BE = CF, \therefore BE + EC = CF + EC$, 即 $BC = EF$.
在$△ ABC$和$△ DFE$中, $\begin{cases} ∠ ACB = ∠ DEF, \\ BC = FE, \\ ∠ B = ∠ F, \end{cases}$
$\therefore △ ABC ≌ △ DFE(\mathrm{ASA})$.
(2) $\because △ ABC ≌ △ DFE, \therefore AC = DE$.
在$△ ACO$和$△ DEO$中, $\begin{cases} ∠ AOC = ∠ DOE, \\ ∠ ACO = ∠ DEO, \\ AC = DE, \end{cases}$
$\therefore △ ACO ≌ △ DEO(\mathrm{AAS}), \therefore CO = EO$.
$\because BE = CF, \therefore BE + OE = CF + CO$, 即 $BO = FO$.
$\therefore O$ 为 $BF$ 的中点.
11.(云龙区月考)如图①，AB=7 cm，AC⊥AB，BD⊥AB，垂足分别为A，B，AC=5 cm.点P在线段AB上以2 cm/s的速度由点A向点B运动，同时点Q在射线BD上运动，它们运动的时间为t(s).(当点P运动结束时，点Q运动随之结束)
(1)若点Q的运动速度与点P的运动速度相等，当t=1时，△ACP与△BPQ是否全等？并判断此时线段PC和线段PQ的位置关系，请分别说明理由.
(2)如图②，若“AC⊥AB，BD⊥AB”改为“∠CAB=∠DBA”，点Q的运动速度为x cm/s，其他条件不变，当点P，Q运动到何处时，有△ACP与△BPQ全等？并求出相应的x的值.
(1)若点Q的运动速度与点P的运动速度相等，当t=1时，△ACP与△BPQ是否全等？并判断此时线段PC和线段PQ的位置关系，请分别说明理由.
(2)如图②，若“AC⊥AB，BD⊥AB”改为“∠CAB=∠DBA”，点Q的运动速度为x cm/s，其他条件不变，当点P，Q运动到何处时，有△ACP与△BPQ全等？并求出相应的x的值.
答案
11. 解:(1) $△ ACP ≌ △ BPQ, PC ⊥ PQ$. 理由如下:
$\because AC ⊥ AB, BD ⊥ AB, \therefore ∠ A = ∠ B = 90°$.
$\because AP = BQ = 2, \therefore BP = 5, \therefore BP = AC$.
在$△ ACP$和$△ BPQ$中, $\begin{cases} AP = BQ, \\ ∠ A = ∠ B, \\ AC = BP, \end{cases}$
$\therefore △ ACP ≌ △ BPQ(\mathrm{SAS}), \therefore ∠ C = ∠ BPQ$.
$\because ∠ C + ∠ APC = 90°, \therefore ∠ APC + ∠ BPQ = 90°$,
$\therefore ∠ CPQ = 90°$, 即 $PC ⊥ PQ$.
(2) 若$△ ACP ≌ △ BPQ$, 则 $AC = BP, AP = BQ$,
可得 $5 = 7 - 2t, 2t = xt$, 解得 $x = 2, t = 1$;
若$△ ACP ≌ △ BQP$, 则 $AC = BQ, AP = BP$,
可得 $5 = xt, 2t = 7 - 2t$,
解得 $x = \frac{20}{7}, t = \frac{7}{4}$.
综上所述, 当$△ ACP$与$△ BPQ$全等时, $x$ 的值为2或$\frac{20}{7}$.
$\because AC ⊥ AB, BD ⊥ AB, \therefore ∠ A = ∠ B = 90°$.
$\because AP = BQ = 2, \therefore BP = 5, \therefore BP = AC$.
在$△ ACP$和$△ BPQ$中, $\begin{cases} AP = BQ, \\ ∠ A = ∠ B, \\ AC = BP, \end{cases}$
$\therefore △ ACP ≌ △ BPQ(\mathrm{SAS}), \therefore ∠ C = ∠ BPQ$.
$\because ∠ C + ∠ APC = 90°, \therefore ∠ APC + ∠ BPQ = 90°$,
$\therefore ∠ CPQ = 90°$, 即 $PC ⊥ PQ$.
(2) 若$△ ACP ≌ △ BPQ$, 则 $AC = BP, AP = BQ$,
可得 $5 = 7 - 2t, 2t = xt$, 解得 $x = 2, t = 1$;
若$△ ACP ≌ △ BQP$, 则 $AC = BQ, AP = BP$,
可得 $5 = xt, 2t = 7 - 2t$,
解得 $x = \frac{20}{7}, t = \frac{7}{4}$.
综上所述, 当$△ ACP$与$△ BPQ$全等时, $x$ 的值为2或$\frac{20}{7}$.
