8. (教材变式)已知图中的两个三角形全等,则$\angle 1$的度数是()

A. $50^{\circ}$
B. $54^{\circ}$
C. $60^{\circ}$
D. $76^{\circ}$

A

9. (教材变式)如图,在$\triangle ABC$中,$AB = 7\ cm$,$BC = 5\ cm$,$AC = 4\ cm$,沿过点$B$的直线折叠这个三角形,使点$C落在AB边上的点E$处,折痕为$BD$,则$\triangle AED$的周长为______$cm$。

6

10. (教材变式)如图,$\triangle ACB \cong \triangle DCE$,$AB与ED交于点F$,若$\angle DCB = 30^{\circ}$,$\angle ACE = 120^{\circ}$,则$\angle BFE$的度数是______。

$45^{\circ}$

11. (教材变式)如图,$\triangle ABE \cong \triangle DCF$,点$B$,$F$,$E$,$C$在同一条直线上。
(1)求证：$AB // CD$；
(2)若$BC = 10$,$EF = 7$,求$BE$的长度。

解：(1)$\because \triangle ABE \cong \triangle DCF$，$\therefore \angle B = \angle C$，$\therefore AB // CD$；(2)$\because \triangle ABE \cong \triangle DCF$，$\therefore BE = CF$，$\therefore BE - EF = CF - EF$，$\therefore CE = BF$. $\because BC = 10$，$EF = 7$，$\therefore CE = BF = \frac{1}{2} \times (10 - 7) = 1.5$，$\therefore BE = BC - CE = 10 - 1.5 = 8.5$.

12. (教材变式)如图,$\triangle ABC \cong \triangle ADE$,$BC的延长线交DE于点F$,$\angle B = 30^{\circ}$,$\angle AED = 110^{\circ}$,$\angle DAC = 10^{\circ}$,求$\angle DFB$的度数。

解：$\because \triangle ABC \cong \triangle ADE$，$\therefore \angle ACB = \angle AED = 110^{\circ}$，$\angle D = \angle B = 30^{\circ}$，$\therefore \angle CAB = 180^{\circ} - \angle B - \angle ACB = 40^{\circ}$，$\therefore \angle DAB = \angle DAC + \angle CAB = 50^{\circ}$. 设$BF$与$AD$交于点$G$，则$\angle DGB = \angle DFB + \angle D = \angle DAB + \angle B$，$\therefore \angle DFB = \angle DAB = 50^{\circ}$.

13. 如图,已知长方形$ABCD的边长AB = 20\ cm$,$BC = 16\ cm$,点$E在边AB$上,$AE = 6\ cm$,如果点$P从点B出发在线段BC上以2\ cm/s的速度向点C$运动,同时,点$Q在线段CD上从点C到点D$运动,它们运动的时间为$t\ s$。当$\triangle BPE与\triangle CQP$全等时,求$t$的值。

解：$\because AB = 20\mathrm{cm}$，$AE = 6\mathrm{cm}$，$BC = 16\mathrm{cm}$，点$P$的运动速度为$2\mathrm{cm/s}$，运动时间为$t\mathrm{s}$，$\therefore BE = 14\mathrm{cm}$，$BP = 2t\mathrm{cm}$，$PC = (16 - 2t)\mathrm{cm}$. 当$\triangle BPE \cong \triangle CQP$时，则有$BE = PC$，即$14 = 16 - 2t$，解得$t = 1$；当$\triangle BPE \cong \triangle CPQ$时，则有$BP = PC$，即$2t = 16 - 2t$，解得$t = 4$，故$t$的值为$1$或$4$.