题目（八年级 · 度量几何学）：如图, $A B / / D E, \angle A=\angle D$, 测得 $A B=D E$.



(1) 求证: $\triangle A B C \cong \triangle D E F$;

(2) 若 $B E=10 \mathrm{~m}, \quad B F=3 \mathrm{~m}$, 求 $F C$ 的长.
选项：
答案：(1) 证明: $\because A B / / D E$,

$\therefore \angle A B C=\angle D E F$,

在 $\triangle A B C$ 与 $\triangle D E F$ 中 $\left\{\begin{array}{c}\angle A B C=\angle D E F \\ A B=D E \\ \angle A=\angle D\end{array}\right.$

$\therefore \triangle A B C \cong \triangle D E F(\mathrm{ASA}) ;$

(2) 解: $\because \triangle A B C \cong \triangle D E F$,

$\therefore B C=E F$,

$\therefore B F+F C=E C+F C$,

$\therefore B F=E C$

$\because B E=10 \mathrm{~cm}, B F=3 \mathrm{~cm}$
$\therefore F C=10-3-3=4 \mathrm{~cm}$

## 25.（1）【旧题重现】

如图(1), $A D 、 A^{\prime} D^{\prime}$ 分别是 $\triangle A B C$ 和 $\triangle A^{\prime} B^{\prime} C^{\prime}$ 的 $B C 、 B^{\prime} C^{\prime}$ 边上的中线,

$A D=A^{\prime} D^{\prime}, A B=A^{\prime} B^{\prime}, B C=B^{\prime} C^{\prime}$. 求证: $\triangle A B C \cong \triangle A^{\prime} B^{\prime} C^{\prime}$.
<ImageHere>

图(1)

证明的途径可以用下面的框图表示，请填写其中的空格．．

<ImageHere>

(2)【深入研究】

如图(2), $A D 、 A^{\prime} D^{\prime}$ 分别是 $\triangle A B C$ 和 $\triangle A^{\prime} B^{\prime} C^{\prime}$ 的 $B C 、 B^{\prime} C^{\prime}$ 边上的中线, $A D=A^{\prime} D^{\prime}, A B=A^{\prime} B^{\prime}, A C=A^{\prime} C^{\prime}$. 判断 $\triangle A B C$ 与 $\triangle A^{\prime} B^{\prime} C^{\prime}$ 是否仍然全等.
<ImageHere>

图(2)

 【解】(1) 证明: $\because A D$ 是 $\triangle A B C$ 的中线,
$\therefore B D=\frac{1}{2} B C$,

$\because A^{\prime} D^{\prime}$ 分别是 $\triangle A^{\prime} B^{\prime} C^{\prime}$ 的中线,

$\therefore B^{\prime} D^{\prime}=\frac{1}{2} B^{\prime} C^{\prime}$,

$\because B C=B^{\prime} C^{\prime}$,

$\therefore B D=B^{\prime} D^{\prime}$,

在 $\triangle A B D$ 和 $\triangle A^{\prime} B^{\prime} D^{\prime}$ 中,

$\left\{\begin{array}{l}B D=B^{\prime} D^{\prime} \\ A D=A^{\prime} D^{\prime}, \\ A B=A^{\prime} B^{\prime}\end{array}\right.$

$\therefore \triangle A B D \cong \triangle A^{\prime} B^{\prime} D^{\prime}(\mathrm{SSS})$

$\therefore \angle B=\angle B^{\prime}$,

在 $\triangle A B C$ 和 $\triangle A^{\prime} B^{\prime} C^{\prime}$ 中,

$\left\{\begin{array}{l}A B=A^{\prime} B^{\prime} \\ \angle B=\angle B^{\prime}, \\ B C=B^{\prime} C^{\prime}\end{array}\right.$

$\therefore \triangle A B C \cong A^{\prime} B^{\prime} C^{\prime}(\mathrm{SAS})$

故答案为: (1) $B D=\frac{1}{2} B C$; (2) $B^{\prime} D^{\prime}=\frac{1}{2} B^{\prime} C^{\prime}$; (3) $A D=A^{\prime} D^{\prime}$; (4) $\angle B=\angle B^{\prime}$;

(2) 解： $\triangle A B C$ 与 $\triangle A^{\prime} B^{\prime} C^{\prime}$ 仍然全等, 理由如下:

延长 $A D$ 至 $E$, 使 $D E=A D$, 连接 $B E$, 延长 $A^{\prime} D^{\prime}$ 至 $E^{\prime}$, 使 $D^{\prime} E^{\prime}=A^{\prime} D^{\prime}$, 连接 $B^{\prime} E^{\prime}$.
<ImageHere>

$\because A D$ 和 $A^{\prime} D^{\prime}$ 分别是 $\triangle A B C$ 和 $\triangle A^{\prime} B^{\prime} C^{\prime}$ 的 $B C$ 和 $B^{\prime} C^{\prime}$ 边上的中线,
$\therefore B D=C D, \quad B^{\prime} D^{\prime}=C^{\prime} D^{\prime}$.

在 $\triangle A D C$ 和 $\triangle E D B$ 中, $\left\{\begin{array}{l}A D=D E \\ \angle A D C=\angle B D E, \\ B D=C D\end{array}\right.$

$\therefore \triangle A D C \cong \triangle E D B(\mathrm{SAS})$.

$\therefore A C=E B, \quad \angle D A C=\angle E$,

同理 $A^{\prime} C^{\prime}=E^{\prime} B^{\prime}, \angle D^{\prime} A^{\prime} C^{\prime}=\angle E^{\prime}$.

$\because A C=A^{\prime} C^{\prime}$,

$\therefore E B=E^{\prime} B^{\prime}$.

$\because A D=A^{\prime} D^{\prime}, \quad A D=D E, \quad A^{\prime} D^{\prime}=D^{\prime} E^{\prime}$,

$\therefore A E=A^{\prime} E^{\prime}$.

$\because A B=A^{\prime} B^{\prime}$,

$\therefore \triangle A B E \cong A^{\prime} B^{\prime} E^{\prime}(\mathrm{SSS})$.

$\therefore \angle B A E=\angle B^{\prime} A^{\prime} E^{\prime}, \quad \angle E=\angle E^{\prime}$.

$\therefore \angle D A C=\angle D^{\prime} A^{\prime} C^{\prime}$.

$\therefore \angle B A C=\angle B^{\prime} A^{\prime} C^{\prime}$,

又 $A B=A^{\prime} B^{\prime}, A C=A^{\prime} C^{\prime}$,

$\therefore \triangle A B C \cong A^{\prime} B^{\prime} C^{\prime}(\mathrm{SAS})$
解析：(1) 证明: $\because A B / / D E$,

$\therefore \angle A B C=\angle D E F$,

在 $\triangle A B C$ 与 $\triangle D E F$ 中 $\left\{\begin{array}{c}\angle A B C=\angle D E F \\ A B=D E \\ \angle A=\angle D\end{array}\right.$

$\therefore \triangle A B C \cong \triangle D E F(\mathrm{ASA}) ;$

(2) 解: $\because \triangle A B C \cong \triangle D E F$,

$\therefore B C=E F$,

$\therefore B F+F C=E C+F C$,

$\therefore B F=E C$

$\because B E=10 \mathrm{~cm}, B F=3 \mathrm{~cm}$
$\therefore F C=10-3-3=4 \mathrm{~cm}$

## 25.（1）【旧题重现】

如图(1), $A D 、 A^{\prime} D^{\prime}$ 分别是 $\triangle A B C$ 和 $\triangle A^{\prime} B^{\prime} C^{\prime}$ 的 $B C 、 B^{\prime} C^{\prime}$ 边上的中线,

$A D=A^{\prime} D^{\prime}, A B=A^{\prime} B^{\prime}, B C=B^{\prime} C^{\prime}$. 求证: $\triangle A B C \cong \triangle A^{\prime} B^{\prime} C^{\prime}$.


图(1)

证明的途径可以用下面的框图表示，请填写其中的空格．．



(2)【深入研究】

如图(2), $A D 、 A^{\prime} D^{\prime}$ 分别是 $\triangle A B C$ 和 $\triangle A^{\prime} B^{\prime} C^{\prime}$ 的 $B C 、 B^{\prime} C^{\prime}$ 边上的中线, $A D=A^{\prime} D^{\prime}, A B=A^{\prime} B^{\prime}, A C=A^{\prime} C^{\prime}$. 判断 $\triangle A B C$ 与 $\triangle A^{\prime} B^{\prime} C^{\prime}$ 是否仍然全等.


图(2)

 【解】(1) 证明: $\because A D$ 是 $\triangle A B C$ 的中线,
$\therefore B D=\frac{1}{2} B C$,

$\because A^{\prime} D^{\prime}$ 分别是 $\triangle A^{\prime} B^{\prime} C^{\prime}$ 的中线,

$\therefore B^{\prime} D^{\prime}=\frac{1}{2} B^{\prime} C^{\prime}$,

$\because B C=B^{\prime} C^{\prime}$,

$\therefore B D=B^{\prime} D^{\prime}$,

在 $\triangle A B D$ 和 $\triangle A^{\prime} B^{\prime} D^{\prime}$ 中,

$\left\{\begin{array}{l}B D=B^{\prime} D^{\prime} \\ A D=A^{\prime} D^{\prime}, \\ A B=A^{\prime} B^{\prime}\end{array}\right.$

$\therefore \triangle A B D \cong \triangle A^{\prime} B^{\prime} D^{\prime}(\mathrm{SSS})$

$\therefore \angle B=\angle B^{\prime}$,

在 $\triangle A B C$ 和 $\triangle A^{\prime} B^{\prime} C^{\prime}$ 中,

$\left\{\begin{array}{l}A B=A^{\prime} B^{\prime} \\ \angle B=\angle B^{\prime}, \\ B C=B^{\prime} C^{\prime}\end{array}\right.$

$\therefore \triangle A B C \cong A^{\prime} B^{\prime} C^{\prime}(\mathrm{SAS})$

故答案为: (1) $B D=\frac{1}{2} B C$; (2) $B^{\prime} D^{\prime}=\frac{1}{2} B^{\prime} C^{\prime}$; (3) $A D=A^{\prime} D^{\prime}$; (4) $\angle B=\angle B^{\prime}$;

(2) 解： $\triangle A B C$ 与 $\triangle A^{\prime} B^{\prime} C^{\prime}$ 仍然全等, 理由如下:

延长 $A D$ 至 $E$, 使 $D E=A D$, 连接 $B E$, 延长 $A^{\prime} D^{\prime}$ 至 $E^{\prime}$, 使 $D^{\prime} E^{\prime}=A^{\prime} D^{\prime}$, 连接 $B^{\prime} E^{\prime}$.


$\because A D$ 和 $A^{\prime} D^{\prime}$ 分别是 $\triangle A B C$ 和 $\triangle A^{\prime} B^{\prime} C^{\prime}$ 的 $B C$ 和 $B^{\prime} C^{\prime}$ 边上的中线,
$\therefore B D=C D, \quad B^{\prime} D^{\prime}=C^{\prime} D^{\prime}$.

在 $\triangle A D C$ 和 $\triangle E D B$ 中, $\left\{\begin{array}{l}A D=D E \\ \angle A D C=\angle B D E, \\ B D=C D\end{array}\right.$

$\therefore \triangle A D C \cong \triangle E D B(\mathrm{SAS})$.

$\therefore A C=E B, \quad \angle D A C=\angle E$,

同理 $A^{\prime} C^{\prime}=E^{\prime} B^{\prime}, \angle D^{\prime} A^{\prime} C^{\prime}=\angle E^{\prime}$.

$\because A C=A^{\prime} C^{\prime}$,

$\therefore E B=E^{\prime} B^{\prime}$.

$\because A D=A^{\prime} D^{\prime}, \quad A D=D E, \quad A^{\prime} D^{\prime}=D^{\prime} E^{\prime}$,

$\therefore A E=A^{\prime} E^{\prime}$.

$\because A B=A^{\prime} B^{\prime}$,

$\therefore \triangle A B E \cong A^{\prime} B^{\prime} E^{\prime}(\mathrm{SSS})$.

$\therefore \angle B A E=\angle B^{\prime} A^{\prime} E^{\prime}, \quad \angle E=\angle E^{\prime}$.

$\therefore \angle D A C=\angle D^{\prime} A^{\prime} C^{\prime}$.

$\therefore \angle B A C=\angle B^{\prime} A^{\prime} C^{\prime}$,

又 $A B=A^{\prime} B^{\prime}, A C=A^{\prime} C^{\prime}$,

$\therefore \triangle A B C \cong A^{\prime} B^{\prime} C^{\prime}(\mathrm{SAS})$