《2025年多维导学案八年级数学上册人教版》

一、预习导学

答案：
解:如图所示,$△DEF$与$△ABC$全等.
　
三边 $AB = DE$ $BC = EF$ $AC = DF$
SSS

【例1】(人教教材P37例3)在如图所示的三角形钢架中，$AB = AC$，$AD$是连接点$A$与$BC$中点$D$的支架。求证：$AD\perp BC$。

答案： 证明:
∵AD是连接点A与BC中点D的支架,
$\therefore BD = CD$.
在$△ABD$和$△ACD$中,
$\left\{\begin{array}{l} AB = AC,\\ BD = CD,\\ AD = AD,\end{array}\right.$
$\therefore △ABD\cong △ACD(SSS)$.
$\therefore ∠ADB = ∠ADC$.
$\because ∠ADB + ∠ADC = 180^{\circ}$,
$\therefore ∠ADB = ∠ADC = 90^{\circ}$,
即$AD⊥BC$.

【变式1】(人教教材P38T1改编)如图，$AC = BD$，$BC = AD$。求证：$\angle DAC=\angle CBD$。

答案： 解:在$△ABC$和$△BAD$中,
$\left\{\begin{array}{l} AC = BD,\\ AB = BA,\\ BC = AD,\end{array}\right.$
$\therefore △ABC\cong △BAD(SSS)$.
$\therefore ∠ABC = ∠BAD,∠BAC = ∠ABD$.
$\therefore ∠DAC = ∠CBD$.

【例2】如图，点$B$，$F$，$C$，$E$在同一条直线上，点$A$，$D$在直线$BC$的异侧，$AB = DE$，$AC = DF$，$BF = EC$。求证：(1)$\triangle ABC\cong \triangle DEF$；(2)$AC// DF$。

答案： 证明:
(1)$\because BF = EC$,
$\therefore BF + FC = EC + FC$.
$\therefore BC = EF$.
在$△ABC$和$△DEF$中,
$\left\{\begin{array}{l} AB = DE,\\ AC = DF,\\ BC = EF,\end{array}\right.$
$\therefore △ABC\cong △DEF(SSS)$.
(2)由
(1),得$△ABC\cong △DEF$,
$\therefore ∠ACB = ∠DFE$.
$\therefore AC// DF$.

【变式2】如图，点$A$，$C$，$F$，$D$在同一直线上，$AF = DC$，$AB = DE$，$BC = EF$。求证：$BC// EF$。

答案： 证明:$\because AF = DC$,
$\therefore AF - FC = DC - FC$,
即$AC = DF$.
在$△ABC$和$△DEF$中,
$\left\{\begin{array}{l} AB = DE,\\ AC = DF,\\ BC = EF,\end{array}\right.$
$\therefore △ABC\cong △DEF(SSS)$.
$\therefore ∠ACB = ∠DFE$.
又$\because ∠ACB + ∠BCF = 180^{\circ},∠DFE + ∠CFE = 180^{\circ}$,
$\therefore ∠BCF = ∠CFE$.
$\therefore BC// EF$.