8. 完成下面的证明.
如图6,已知CD平分$∠ ACB$,$∠ 4 = ∠ B$. 求证:$∠ 1 = ∠ 2$.
证明:$\because ∠ 4 = ∠ B$(已知),
$\therefore DE //$
$\therefore ∠ 3 =$
$\because CD$平分$∠ ACB$(已知),
$\therefore ∠ 3 =$
$\therefore ∠ 1 = ∠ 2$ (

如图6,已知CD平分$∠ ACB$,$∠ 4 = ∠ B$. 求证:$∠ 1 = ∠ 2$.
证明:$\because ∠ 4 = ∠ B$(已知),
$\therefore DE //$
$BC$
(同位角相等,两直线平行
).$\therefore ∠ 3 =$
$∠1$
(两直线平行,内错角相等
).$\because CD$平分$∠ ACB$(已知),
$\therefore ∠ 3 =$
$∠2$
(角平分线的定义).$\therefore ∠ 1 = ∠ 2$ (
等式的基本事实
).答案
8. $BC$ 同位角相等,两直线平行 $∠1$ 两直线平行,内错角相等 $∠2$ 等式的基本事实
9. 如图7,在方格纸中,每个小正方形的边长均为1个单位长度. 有一个三角形ABC,它的三个顶点均与小正方形的顶点重合.
(1) 将三角形ABC先向右平移3个单位长度,再向下平移1个单位长度得到三角形$A_1B_1C_1$. 请在方格纸中画出三角形$A_1B_1C_1$.
(2) 求三角形$A_1B_1C_1$的面积.

(1) 将三角形ABC先向右平移3个单位长度,再向下平移1个单位长度得到三角形$A_1B_1C_1$. 请在方格纸中画出三角形$A_1B_1C_1$.
(2) 求三角形$A_1B_1C_1$的面积.
答案
9. 解:(1) 如答图1,三角形$A_1B_1C_1$即为所求.
(2) 三角形$A_1B_1C_1$的面积为$2 × 4 - \frac{1}{2} × 1 × 2 - \frac{1}{2} × 1 × 4 - \frac{1}{2} × 2 × 2 = 8 - 1 - 2 - 2 = 3$.
10.(综合探究)如图8,点N在线段CD上,ED与FN交于点M,∠C=∠1,∠2=∠3.
(1)判断AB与CD是否平行,并说明理由;
(2)若∠D=40°,∠EMF=80°,求∠AEP的大小.

(1)判断AB与CD是否平行,并说明理由;
(2)若∠D=40°,∠EMF=80°,求∠AEP的大小.
答案
10. 解:(1) $AB// CD$.
理由:$\because ∠2 = ∠3$,$\therefore PC// FN$(同位角相等,两直线平行).
$\therefore ∠C = ∠FND$(两直线平行,同位角相等).
又$\because ∠C = ∠1$,$\therefore ∠1 = ∠FND$.
$\therefore AB// CD$(内错角相等,两直线平行).
(2) $\because PC// FN$,$\therefore ∠2 = ∠EMF = 80°$(两直线平行,内错角相等).
$\because AB// CD$,$\therefore ∠FED = ∠D = 40°$(两直线平行,内错角相等).
$\therefore ∠BEC = ∠2 + ∠FED = 80° + 40° = 120°$. $\therefore ∠AEP = ∠BEC = 120°$.
理由:$\because ∠2 = ∠3$,$\therefore PC// FN$(同位角相等,两直线平行).
$\therefore ∠C = ∠FND$(两直线平行,同位角相等).
又$\because ∠C = ∠1$,$\therefore ∠1 = ∠FND$.
$\therefore AB// CD$(内错角相等,两直线平行).
(2) $\because PC// FN$,$\therefore ∠2 = ∠EMF = 80°$(两直线平行,内错角相等).
$\because AB// CD$,$\therefore ∠FED = ∠D = 40°$(两直线平行,内错角相等).
$\therefore ∠BEC = ∠2 + ∠FED = 80° + 40° = 120°$. $\therefore ∠AEP = ∠BEC = 120°$.
11.(综合探究)如图9,在三角形ABC中,D是AB上一点,E是BC上一点,点F,G在AC上,$∠ AFD = ∠ DEB$,$∠ DFC + ∠ C = 180°$。
(1)求证:$DE // AC$;
(2)若$∠ C = 38°$,EG平分$∠ DEC$,求$∠ EGC$的度数。

(1)求证:$DE // AC$;
(2)若$∠ C = 38°$,EG平分$∠ DEC$,求$∠ EGC$的度数。
答案
11. (1) 证明:$\because ∠DFC + ∠C = 180°$,$\therefore DF// BC$(同旁内角互补,两直线平行).
$\therefore ∠DEB = ∠EDF$(两直线平行,内错角相等).
$\because ∠AFD = ∠DEB$,$\therefore ∠EDF = ∠AFD$.
$\therefore DE// AC$(内错角相等,两直线平行).
(2) 解:$\because DE// AC$,$\therefore ∠C + ∠DEC = 180°$(两直线平行,同旁内角互补).
$\because ∠C = 38°$,$\therefore ∠DEC = 180° - 38° = 142°$.
$\because EG$平分$∠DEC$,$\therefore ∠DEG = \frac{1}{2}∠DEC = 71°$.
$\because DE// AC$,$\therefore ∠EGC = ∠DEG = 71°$(两直线平行,内错角相等).
$\therefore ∠DEB = ∠EDF$(两直线平行,内错角相等).
$\because ∠AFD = ∠DEB$,$\therefore ∠EDF = ∠AFD$.
$\therefore DE// AC$(内错角相等,两直线平行).
(2) 解:$\because DE// AC$,$\therefore ∠C + ∠DEC = 180°$(两直线平行,同旁内角互补).
$\because ∠C = 38°$,$\therefore ∠DEC = 180° - 38° = 142°$.
$\because EG$平分$∠DEC$,$\therefore ∠DEG = \frac{1}{2}∠DEC = 71°$.
$\because DE// AC$,$\therefore ∠EGC = ∠DEG = 71°$(两直线平行,内错角相等).
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