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1. (2024·泰州期末)数学活动:折纸与证明.
折纸,常常能为证明一个命题提供思路和方法.
如图①,在$\triangle ABC$中,$AB＞AC$,怎样证明$∠C＞∠B$呢?
如图②,把$AC沿∠A的平分线AD$翻折,因为$AB＞AC$,所以点$C落在AB上的点C'$处.于是,由$∠AC'D= ∠C,∠AC'D＞∠B$,可得$∠C＞∠B$.
感悟与应用:
(1)如图③,$AD是\triangle ABC$的高,$∠C= 2∠B$.若$AC= 10,CD= 4$,求$BD$的长.小龙同学的解法是将$\triangle ADC沿AD$折叠,点$C落在BC边上的点C'$处……画出图形并写出完整的解题过程;
(2)如图④,$AD是\triangle ABC$的角平分线,$∠C= 2∠B$.线段$AB,AC,CD$之间有怎样的数量关系?写出你的猜想并证明.
折纸,常常能为证明一个命题提供思路和方法.
如图①,在$\triangle ABC$中,$AB＞AC$,怎样证明$∠C＞∠B$呢?
如图②,把$AC沿∠A的平分线AD$翻折,因为$AB＞AC$,所以点$C落在AB上的点C'$处.于是,由$∠AC'D= ∠C,∠AC'D＞∠B$,可得$∠C＞∠B$.
感悟与应用:
(1)如图③,$AD是\triangle ABC$的高,$∠C= 2∠B$.若$AC= 10,CD= 4$,求$BD$的长.小龙同学的解法是将$\triangle ADC沿AD$折叠,点$C落在BC边上的点C'$处……画出图形并写出完整的解题过程;
(2)如图④,$AD是\triangle ABC$的角平分线,$∠C= 2∠B$.线段$AB,AC,CD$之间有怎样的数量关系?写出你的猜想并证明.
答案
(1)将$\triangle ADC$沿$AD$折叠，点$C$落在$BC$边上的点$C'$处，如图①，$\because AD\perp BC$，点$C$落在$BD$上的点$C'$处，$\therefore AC = AC' = 10$，$CD = C'D = 4$，$\angle AC'D = \angle C$。$\because \angle AC'D = \angle C'AB + \angle B$，$\angle C = 2\angle B$，$\therefore \angle B = \angle BAC'$，$\therefore C'A = C'B = AC = 10$，$\therefore BD = BC' + C'D = 10 + 4 = 14$。
(2)$AB = AC + CD$，证明：如图②，把$AC$沿$\angle A$的平分线$AD$翻折，使点$C$落在$AB$上的点$C'$处，$\therefore AC = AC'$。$\because \angle AC'D = \angle C = \angle B + \angle C'DB$，又$\angle C = 2\angle B$，$\therefore \angle B = \angle C'DB$，$\therefore C'B = C'D$，$\therefore AB - AC = AB - AC' = BC' = C'D = CD$，即$AB = AC + CD$。
2. (2024·鞍山校级月考)折纸是同学们喜欢的手工活动之一.
按照下面过程折一折,并探究其蕴含的数学知识:
如图①:把边长为4的正方形纸片$ABCD$对折,使边$AB与CD$重合,展开后得到折痕$EF$;
如图②:点$M为CF$上一点,将正方形纸片$ABCD沿直线DM$折叠,使点$C落在折痕EF上的点N$处,展开后连接$DN,MN,AN$.
试探究:
(1)判断$\triangle AND$的形状,并给出证明;
(2)说明线段$NF与CM$的数量关系.
按照下面过程折一折,并探究其蕴含的数学知识:
如图①:把边长为4的正方形纸片$ABCD$对折,使边$AB与CD$重合,展开后得到折痕$EF$;
如图②:点$M为CF$上一点,将正方形纸片$ABCD沿直线DM$折叠,使点$C落在折痕EF上的点N$处,展开后连接$DN,MN,AN$.
试探究:
(1)判断$\triangle AND$的形状,并给出证明;
(2)说明线段$NF与CM$的数量关系.
答案
(1)$\triangle AND$是等边三角形，证明：$\because$把边长为$4$的正方形纸片$ABCD$对折，使边$AB$与$CD$重合，展开后得到折痕$EF$，$\therefore FE$是$AD$的垂直平分线，$\therefore AN = DN$。$\because$将正方形纸片$ABCD$沿直线$DM$折叠，使点$C$落在折痕$EF$上的点$N$处，$\therefore DC = DN = 4$，$\therefore AN = DN = 4 = AD$，$\therefore \triangle AND$是等边三角形。
(2)$\because$把边长为$4$的正方形纸片$ABCD$对折，使边$AB$与$CD$重合，展开后得到折痕$EF$，$\therefore DE = \frac{1}{2}AD = 2$，$\angle DEN = 90^{\circ}$。$\because$将正方形纸片$ABCD$沿直线$DM$折叠，使点$C$落在折痕$EF$上的点$N$处，$\therefore DC = DN = 4$，$\angle C = \angle DNM = 90^{\circ}$，$MN = CM$，$\therefore DE = \frac{1}{2}DN$，$\therefore \angle DNE = 30^{\circ}$，$\therefore \angle MNF = 180^{\circ} - \angle DNE - \angle DNM = 60^{\circ}$。$\because \angle NFC = 90^{\circ}$，$\therefore \angle FMN = 30^{\circ}$，$\therefore NF = \frac{1}{2}MN$，$\therefore NF = \frac{1}{2}CM$。
(2)$\because$把边长为$4$的正方形纸片$ABCD$对折，使边$AB$与$CD$重合，展开后得到折痕$EF$，$\therefore DE = \frac{1}{2}AD = 2$，$\angle DEN = 90^{\circ}$。$\because$将正方形纸片$ABCD$沿直线$DM$折叠，使点$C$落在折痕$EF$上的点$N$处，$\therefore DC = DN = 4$，$\angle C = \angle DNM = 90^{\circ}$，$MN = CM$，$\therefore DE = \frac{1}{2}DN$，$\therefore \angle DNE = 30^{\circ}$，$\therefore \angle MNF = 180^{\circ} - \angle DNE - \angle DNM = 60^{\circ}$。$\because \angle NFC = 90^{\circ}$，$\therefore \angle FMN = 30^{\circ}$，$\therefore NF = \frac{1}{2}MN$，$\therefore NF = \frac{1}{2}CM$。
