《2025年课堂专题导学精练七年级数学下册人教版》

1. 求角
(1)如图①，直线AB、CD相交于点O，$\angle AOC:\angle BOC = 2:3$，则$\angle BOD = $______$^{\circ}$；
(2)如图②，$DE // BC$，$\angle D:\angle DBC = = 2:1$，$\angle 1 = \angle 2$，求$\angle E$的度数.
$\begin{array}{cc}图① & 图② \\\begin{array}{c}\begin{tikzpicture}\draw (0,0) -- (2,0) node[right] {B};\draw (0,0) -- (-2,0) node[left] {A};\draw (0,0) -- (0,-2) node[below] {D};\draw (0,0) -- (1,1) node[above] {C};\node at (-0.5,0.5) {O};\end{tikzpicture}\end{array} &\begin{array}{c}\begin{tikzpicture}\draw (0,0) -- (3,0) node[right] {C};\draw (0,0) -- (1,1) node[above] {B};\draw (1,1) -- (3,1) node[above] {E};\draw (3,1) -- (2,0) node[right] {D};\node at (1.5,0.5) {1};\node at (2.5,0.5) {2};\end{tikzpicture}\end{array}\end{array}$

答案：
(1)72
(2)$30^{\circ}$
解析：
(1)设$\angle AOC = 2k$，$\angle BOC = 3k$，$2k + 3k = 180^{\circ}$，$k = 36^{\circ}$，$\angle BOD = \angle AOC = 72^{\circ}$。
(2)设$\angle DBC = x$，$\angle D = 2x$，$DE // BC$得$\angle D + \angle DBC = 180^{\circ}$，$3x = 180^{\circ}$，$x = 60^{\circ}$，$\angle 1 = \angle 2 = 30^{\circ}$，$\angle E = \angle 1 = 30^{\circ}$。

3. (1)如图①，$AB // CD$，$\angle 2:\angle 1 = 3:2$，$\angle B = 130^{\circ}$，则$\angle A = $______$^{\circ}$；
(2)如图②，直线AC、EF相交于点O，OD是$\angle AOB$的平分线，OE在$\angle BOC$内，$\angle BOE = \dfrac{1}{2}\angle EOC$，$\angle DOE = 72^{\circ}$，求$\angle AOF$的度数.
$\begin{array}{cc}图① & 图② \\\begin{array}{c}\begin{tikzpicture}\draw (0,1) -- (3,1) node[right] {B};\draw (0,0) -- (3,0) node[right] {A};\draw (1,1) -- (1,0) node[left] {1};\draw (2,1) -- (2,0) node[right] {2};\end{tikzpicture}\end{array} &\begin{array}{c}\begin{tikzpicture}\draw (-2,0) -- (2,0) node[right] {C};\draw (0,-1) -- (0,2) node[above] {E};\draw (-1,-1) -- (1,2) node[above] {F};\draw (-1,2) -- (1,-1) node[below] {D};\node at (0,0) {O};\end{tikzpicture}\end{array}\end{array}$

答案：
(1)72
(2)$36^{\circ}$
解析：
(1)设$\angle 1 = 2k$，$\angle 2 = 3k$，$AB // CD$得$\angle 1 + \angle 2 = 180^{\circ}$，$k = 36^{\circ}$，$\angle 1 = 72^{\circ}$，$\angle A = \angle 1 = 72^{\circ}$。
(2)设$\angle BOE = x$，$\angle EOC = 2x$，$\angle BOC = 3x$，$\angle AOB = 180^{\circ} - 3x$，OD平分$\angle AOB$得$\angle BOD = \dfrac{180^{\circ} - 3x}{2}$，$\angle DOE = \angle BOD + x = 72^{\circ}$，解得$x = 36^{\circ}$，$\angle AOF = \angle EOC = 72^{\circ} - 36^{\circ} = 36^{\circ}$。