《2025年学霸题中题八年级数学上册人教版》

9. 在等边$\triangle ABC$的两边$AB$,$AC$所在直线上分别有两点$M$,$N$,$D$为$\triangle ABC$外一点,且$\angle MDN = 60^{\circ}$,$\angle BDC = 120^{\circ}$,$BD = DC$. 探究:当$M$,$N$分别在直线$AB$,$AC$上移动时,$BM$,$NC$,$MN$之间的数量关系及$\triangle AMN$的周长$Q$与等边$\triangle ABC$的周长$L$的关系.
(1) 如图①所示,当点$M$,$N$在边$AB$,$AC$上,且$DM = DN$时,$BM$,$NC$,$MN$之间的数量关系是____;此时$\frac{Q}{L} = $____.(不必证明)
(2) 如图②所示,点$M$,$N$在边$AB$,$AC$上,且当$DM \neq DN$时,猜想(1)问的两个结论还成立吗? 写出你的猜想并加以证明.
(3) 如图③所示,当$M$,$N$分别在边$AB$,$CA$的延长线上时,猜想$BM$,$NC$,$MN$之间的数量关系并加以证明.
(4) 在(3)问的条件下,若此时$AN = x$,则$Q = $____(用含有$x$,$L$的式子表示).

答案：

(1) $BM + NC = MN$ $\frac{2}{3}$
(2)
(1) 问的两个结论仍然成立。证明：如图①，在 $AC$ 的延长线上截取 $CP = BM$，连接 $DP$，在等边 $\triangle ABC$ 中，$\angle ABC = \angle ACB = 60^{\circ}$。$\because \angle BDC = 120^{\circ}$，$BD = DC$，$\therefore \angle DBC = \angle DCB = 30^{\circ}$，
$\therefore \angle DBM = \angle DCP = 90^{\circ}$。在 $\triangle DBM$ 与 $\triangle DCP$ 中，
$\begin{cases}BM = CP,\\\angle DBM = \angle DCP,\\BD = CD,\end{cases}$ $\therefore \triangle DBM \cong \triangle DCP(SAS)$，
$\therefore \angle BDM = \angle CDP$，$DM = DP$。$\because \angle BDC = 120^{\circ}$，$\angle MDN = 60^{\circ}$，
$\therefore \angle PDN = \angle CDP + \angle CDN = \angle BDM + \angle CDN = 120^{\circ} - 60^{\circ} = 60^{\circ}$。在 $\triangle DMN$ 与 $\triangle DPN$ 中，$\begin{cases}DM = DP,\\\angle MDN = \angle PDN,\\DN = DN,\end{cases}$ $\therefore \triangle DMN \cong \triangle DPN(SAS)$，
$\therefore MN = PN = NC + PC = NC + BM$，$\therefore Q = AM + MN + AN = (AM + BM) + (CN + AN) = AB + AC = 2AB$。$\because L = AB + AC + BC = 3AB$，
$\therefore \frac{Q}{L} = \frac{2}{3}$。

(3) $CN = BM + MN$。证明：如图②，在 $NC$ 上截取 $CH = BM$，连接 $DH$，由
(2) 知，$\angle ABD = \angle ACD = 90^{\circ}$，
$\therefore \angle MBD = 180^{\circ} - 90^{\circ} = 90^{\circ}$。在 $\triangle DBM$ 和 $\triangle DCH$ 中，$\begin{cases}BD = CD,\\\angle DBM = \angle DCH,\\BM = CH,\end{cases}$ $\therefore \triangle DBM \cong \triangle DCH(SAS)$，
$\therefore \angle BDM = \angle CDH$，$DM = DH$。
$\because \angle MDN = \angle BDM + \angle BDN = \angle CDH + \angle BDN = 60^{\circ}$，$\angle BDC = 120^{\circ}$，
$\therefore \angle NDH = 60^{\circ}$。在 $\triangle MDN$ 和 $\triangle HDN$ 中，$\begin{cases}MD = HD,\\\angle MDN = \angle HDN,\\DN = DN,\end{cases}$
$\therefore \triangle MDN \cong \triangle HDN(SAS)$，$\therefore MN = HN$。$\because CN = NH + CH$，$CH = BM$，
$\therefore CN = BM + MN$。
(4) $2x + \frac{2}{3}L$ 解析：$\triangle AMN$ 的周长 $Q = AN + AM + MN = AN + AB + BM + AN + AC - BM = 2AN + 2AB$。$\because AN = x$，$AB = \frac{1}{3}L$，$\therefore Q = 2x + \frac{2}{3}L$。

10. 如图,在$Rt\triangle ABC$中,$\angle BAC = 90^{\circ}$,点$D$,$E$在边$BC$上,$\angle CAE = \angle B$,$E$是$CD$的中点,且$AD$平分$\angle BAE$,试问:$BD$与$AC$相等吗? 请说说你的理由.

答案：
$BD = AC$。理由如下：如图，由于 $AD$ 平分 $\angle BAE$，可将 $\triangle ABD$ 沿 $AD$ 所在直线翻折到 $\triangle AFD$ 的位置，则 $\triangle ABD \cong \triangle AFD$，$\therefore \angle F = \angle B$，$BD = DF$。又 $\because \angle CAE = \angle B$，$\therefore \angle CAE = \angle F$。又 $\because \angle AEC = \angle FED$，且 $E$ 是 $CD$ 的中点，$\therefore \triangle ACE \cong \triangle FDE$，
$\therefore DF = AC$，$\therefore BD = AC$。

11. 如图,正方形$ABCD$中,点$E$是$BC$上一点,点$F$是$DC$上一点,$\angle EAF = 45^{\circ}$.
(1) 如图①,若$BE = DF = 1$,求$EF$的长;
(2) 如图②,求证:$BE + DF = EF$;
(3) 如图③,点$E$为$CB$延长线上一点,点$F$为$DC$延长线上一点,$\angle EAF = 45^{\circ}$.请直接写出线段$BE$,$DF$,$EF$的数量关系.

答案：

(1) 如图①，将 $\triangle ADF$ 绕着点 $A$ 按顺时针方向旋转 $90^{\circ}$，得 $\triangle ABH$。
$\because$ 四边形 $ABCD$ 是正方形，$\therefore AD = AB = BC = CD$，$\angle ABC = \angle ADF = \angle C = 90^{\circ}$。$\because DF = BE = BH = 1$，由旋转可得 $AF = AH$，$\angle DAF = \angle BAH$。
$\because \angle EAF = 45^{\circ}$，$\therefore \angle DAF + \angle BAE = 45^{\circ} = \angle BAH + \angle BAE = \angle HAE = \angle EAF$。又 $\because AE = AE$，$\therefore \triangle AEF \cong \triangle AEH(SAS)$，$\therefore HE = EF = 2$。

(2) 将 $\triangle ADF$ 绕着点 $A$ 按顺时针方向旋转 $90^{\circ}$，得 $\triangle ABF'$，如图②，则 $\angle ABF' = \angle D$，$AF = AF'$，$\angle BAF' = \angle DAF$。$\because$ 四边形 $ABCD$ 是正方形，$\therefore \angle D = \angle ABC = 90^{\circ}$，$\therefore \angle ABF' = 90^{\circ}$，$\therefore \angle F'BC = 180^{\circ}$，$\therefore$ 点 $F'$，$B$，$E$ 在一条直线上。$\because \angle EAF = 45^{\circ}$，$\therefore \angle DAF + \angle BAE = 45^{\circ} = \angle F'AB + \angle BAE = \angle F'AE = \angle EAF$。又 $\because AE = AE$，$\therefore \triangle AF'E \cong \triangle AFE(SAS)$，$\therefore EF = EF'$，$\therefore EF = F'E = BE + DF$。
(3) $EF = DF - BE$。解析：将 $\triangle ABE$ 绕着点 $A$ 按逆时针方向旋转 $90^{\circ}$ 得 $\triangle ADH$，$\therefore \triangle ADH \cong \triangle ABE$，$\therefore \angle DAH = \angle BAE$，$AE = AH$。$\because \angle EAF = 45^{\circ}$，$\therefore \angle BAE + \angle BAF = 45^{\circ} = \angle DAH + \angle BAF$，$\therefore \angle FAH = 90^{\circ} - \angle DAH - \angle BAF = 45^{\circ} = \angle EAF$。又 $\because AF = AF$，$AE = AH$，$\therefore \triangle AEF \cong \triangle AHF(SAS)$，$\therefore EF = FH$，$\therefore EF = FH = DF - DH = DF - BE$。
归纳总结
本题与第 9 题均属于半角模型，又称角含半角模型，即一个大角包含一个它一半大小的小角。角含半角模型中的角度，通常为特殊角，$30^{\circ}$ 和 $60^{\circ}$，$45^{\circ}$ 和 $90^{\circ}$，$60^{\circ}$ 和 $120^{\circ}$。
证明方法：①旋转法，②翻折法，③截长补短法。