如图13,在平行四边形ABCD中,AC是一条对角线,且$AB=AC=5,BC=6$,$E,F$是$AD$边上两点, 点 $F$ 在点 $E$ 的右侧, $A E=D F$, 连接 $C E, C E$ 的延长线与 $B A$ 的延长线相交于点 $G$.
(1) 如图 13-1, $M$ 是 $B C$ 边上一点, 连接 $A M, M F, M F$ 与 $C E$ 相交于点 $N$.
(1) 若 $A E=\frac{3}{2}$, 求 $A G$ 的长;
(2)在满足(1)的条件下, 若 $E N=N C$, 求证: $A M \perp B C$;
(2) 如图 13-2, 连接 $G F, H$ 是 $G F$ 上一点, 连接 $E H$. 若 $\angle E H G=\angle E F G+\angle C E F$, 且 $H F=2 G H$, 求 $E F$ 的长.
【答案】 (1) $\because$ 四边形 $A B C D$ 是平行四边形, $\therefore A B / / C D, A D / / B C, D C=A B=5$,
$$
A D=B C=6, \quad \therefore \angle G A E=\angle C D E, \angle A G E=\angle D C E \text {, }
$$

$$
\begin{aligned}
& \therefore \triangle A G E \sim \triangle D C E, \therefore \frac{A G}{D C}=\frac{A E}{D E}, \\
& \therefore A G \cdot D E=D C \cdot A E . \\
& \because A E=\frac{3}{2}, \therefore D E=A D-A E=6-\frac{3}{2}=\frac{9}{2}, \\
& \therefore \frac{9}{2} A G=5 \times \frac{3}{2}, \therefore A G=\frac{5}{3}
\end{aligned}
$$



(2)证明: $\because A D / / B C, \therefore \angle E F N=\angle C M N$,
$$
\begin{aligned}
& \because \angle E N F=\angle C N M, \quad E N=N C, \therefore \triangle E N F \cong \triangle C N M, \quad \therefore E F=C M . \\
& \because A E=\frac{3}{2}, A E=D F, \quad \therefore D F=\frac{3}{2}, \quad \therefore E F=A D-A E-D F=3 . \\
& \therefore C M=3 . \because B C=6, \quad \therefore B M=B C-C M=3, \\
& \therefore B M=M C . \because A B=A C, \quad \therefore A M \perp B C, \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \text { (8 分) }
\end{aligned}
$$


(2) 如答案图 5 .
连接 $C F . \because A B=A C, A B=D C, \therefore A C=D C$,
$$
\begin{aligned}
& \therefore \angle C A D=\angle C D A . \because A E=D F, \\
& \therefore \triangle A E C \cong \triangle D F C, \therefore C E=C F, \\
& \therefore \angle C F E=\angle C E F \therefore \angle E H G=\angle E F G+\angle C E F, \\
& \therefore \angle E H G=\angle E F G+\angle C F E=\angle C F G . \\
& \therefore E H / / C F, \therefore \frac{G H}{H F}=\frac{G E}{E C} . \because H F=2 G H,
\end{aligned}
$$


$$
\begin{aligned}
& \therefore \frac{G E}{E C}=\frac{1}{2} \therefore A B / / C D, \therefore \angle G A E=\angle C D E, \angle A G E=\angle D C E \\
& \therefore \triangle A G E \sim \triangle D C E, \therefore \frac{A E}{D E}=\frac{G E}{C E}, \therefore \frac{A E}{D E}=\frac{1}{2}
\end{aligned}
$$

$\therefore D E=2 A E$. 设 $A E=x$, 则 $D E=2 x . \because A D=6, \therefore x+2 x=6, \therefore x=2$,
即 $A E=2, \therefore D F=2 , \therefore E F=A D-A E-D F=2 $



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