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1. 如图，直线AB与直线CD相交于点O，OE⊥AB，垂足为O. 若∠EOD=$\frac{1}{3}$∠AOC，则∠BOC的度数为（ ）
A. 112.5° B. 135° C. 140° D. 157.5°
A. 112.5° B. 135° C. 140° D. 157.5°
答案
A
2. 如图，OA⊥OB，OC⊥OD，∠BOA：∠AOD=2：3，则∠BOD的度数为__________.
答案
135°
3. 如图，∠B=∠BCD，∠BAC=90°，∠B+∠D=180°，若∠ACB：∠ACD=1：2，则∠BAD=__________.
答案
112.5° 解析：∵$\angle B = \angle BCD$，$\angle B+\angle D = 180^{\circ}$，∴$\angle D+\angle BCD = 180^{\circ}$. ∴$AD// BC$. ∴$\angle B+\angle BAD = 180^{\circ}$，$\angle ACB = \angle CAD$. ∵$\angle ACB:\angle ACD = 1:2$，∴设$\angle ACB = \angle CAD = x$，则$\angle ACD = 2x$. ∴$\angle B = \angle BCD = 3x$. ∴$\angle B+\angle BAD = 3x + 90^{\circ}+x = 180^{\circ}$. ∴$x = 22.5^{\circ}$. ∴$\angle BAD = 90^{\circ}+22.5^{\circ}=112.5^{\circ}$.
4. 如图，直线CD，EF交于点O，OA，OB分别平分∠COE和∠DOE，且∠1+∠2=90°.
(1)若∠2：∠3=2：5，求∠BOF的度数；
(2)试判断AB与CD的位置关系，并说明理由.
(1)若∠2：∠3=2：5，求∠BOF的度数；
(2)试判断AB与CD的位置关系，并说明理由.
答案
(1) ∵$OB$平分$\angle DOE$，∴$\angle BOE=\angle 2$. ∵$\angle 2:\angle 3 = 2:5$，∴设$\angle 2 = 2\alpha$，则$\angle BOE = 2\alpha$，$\angle 3 = 5\alpha$. ∴$\angle BOF=\angle 2+\angle 3 = 7\alpha$. ∵$\angle BOE+\angle BOF = 2\alpha+7\alpha = 9\alpha = 180^{\circ}$，∴$\alpha = 20^{\circ}$.
∴$\angle BOF = 7\alpha = 140^{\circ}$ (2) $AB// CD$ 理由：∵$OA$，$OB$分别平分$\angle COE$和$\angle DOE$，∴$\angle COE = 2\angle AOC$，$\angle DOE = 2\angle 2$.
∵$\angle COE+\angle DOE = 2(\angle AOC+\angle 2)=180^{\circ}$，∴$\angle 2+\angle AOC = 90^{\circ}$. ∵$\angle 1+\angle 2 = 90^{\circ}$，∴$\angle 1=\angle AOC$. ∴$AB// CD$.
∴$\angle BOF = 7\alpha = 140^{\circ}$ (2) $AB// CD$ 理由：∵$OA$，$OB$分别平分$\angle COE$和$\angle DOE$，∴$\angle COE = 2\angle AOC$，$\angle DOE = 2\angle 2$.
∵$\angle COE+\angle DOE = 2(\angle AOC+\angle 2)=180^{\circ}$，∴$\angle 2+\angle AOC = 90^{\circ}$. ∵$\angle 1+\angle 2 = 90^{\circ}$，∴$\angle 1=\angle AOC$. ∴$AB// CD$.
5. 如图，AD//BC，BE平分∠ABC交AD于点E，BD平分∠EBC.
(1)若∠DBC=30°，求∠A的度数.
(2)若点F在线段AE上，且7∠DBC−2∠ABF=180°，则图中是否存在与∠DFB相等的角？
若存在，请写出这个角并证明；若不存在，请说明理由.
(1)若∠DBC=30°，求∠A的度数.
(2)若点F在线段AE上，且7∠DBC−2∠ABF=180°，则图中是否存在与∠DFB相等的角？
若存在，请写出这个角并证明；若不存在，请说明理由.
答案
(1) ∵$BD$平分$\angle EBC$，$\angle DBC = 30^{\circ}$，∴$\angle EBC = 2\angle DBC = 60^{\circ}$. ∵$BE$平分$\angle ABC$，∴$\angle ABC = 2\angle EBC = 120^{\circ}$. ∵$AD// BC$，∴$\angle A+\angle ABC = 180^{\circ}$. ∴$\angle A = 60^{\circ}$
(2) 存在 $\angle DBF=\angle DFB$ 设$\angle DBC = x$. ∵$BD$平分$\angle EBC$，∴$\angle EBC = 2\angle DBC = 2x$. ∵$BE$平分$\angle ABC$，
∴$\angle ABC = 2\angle EBC = 4x$. ∵$7\angle DBC - 2\angle ABF = 180^{\circ}$，
∴$7x - 2\angle ABF = 180^{\circ}$. ∴$\angle ABF=\frac{7}{2}x - 90^{\circ}$. ∴$\angle CBF = \angle ABC-\angle ABF=\frac{1}{2}x + 90^{\circ}$，$\angle DBF=\angle ABC-\angle ABF-\angle DBC = 90^{\circ}-\frac{1}{2}x$. ∵$AD// BC$，∴$\angle DFB+\angle CBF = 180^{\circ}$.
∴$\angle DFB = 90^{\circ}-\frac{1}{2}x$. ∴$\angle DBF=\angle DFB$
(2) 存在 $\angle DBF=\angle DFB$ 设$\angle DBC = x$. ∵$BD$平分$\angle EBC$，∴$\angle EBC = 2\angle DBC = 2x$. ∵$BE$平分$\angle ABC$，
∴$\angle ABC = 2\angle EBC = 4x$. ∵$7\angle DBC - 2\angle ABF = 180^{\circ}$，
∴$7x - 2\angle ABF = 180^{\circ}$. ∴$\angle ABF=\frac{7}{2}x - 90^{\circ}$. ∴$\angle CBF = \angle ABC-\angle ABF=\frac{1}{2}x + 90^{\circ}$，$\angle DBF=\angle ABC-\angle ABF-\angle DBC = 90^{\circ}-\frac{1}{2}x$. ∵$AD// BC$，∴$\angle DFB+\angle CBF = 180^{\circ}$.
∴$\angle DFB = 90^{\circ}-\frac{1}{2}x$. ∴$\angle DBF=\angle DFB$
6. 如图，AB//CD，∠E=90°，∠1=55°，则∠B的度数为__________.
答案
35°
7. 如图，AB//CD，∠ABE=40°. 若CF平分∠ECD，且满足CF//BE，则∠ECD的度数为__________.
答案
80°
8. 如图，在一块长20m、宽14m的长方形草地上有一条宽为2m的曲折小路，则这块草地的绿地面积（空白部分）为__________m².
答案
216
