2025年通城学典课时作业本七年级数学下册人教版南通专版第17页答案
8. 如图,$\angle ABC+\angle BCD+\angle CDE = 360^{\circ}$,直线$FG$分别交$AB$,$DE$于点$F$,$G$. 若$\angle 1 = 120^{\circ}$,则$\angle 2 =$_______.
            第8题

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60
9. 如图,如果$AB// CD$,$\angle \alpha = 145^{\circ}$,$\angle \beta = 60^{\circ}$,那么$\angle \gamma$的度数是_______.
       第9题

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25° 解析:如图,过点 E 作 EF//AB,∴ ∠BAE + ∠AEF = 180°. ∵ ∠BAE = ∠α = 145°,∴ ∠AEF = 35°. ∵ ∠AED = ∠β = 60°,∴ ∠DEF = 25°. ∵ AB//CD,EF//AB,∴ EF//CD. ∴ ∠γ = ∠DEF = 25°.
第9题
10. 如图,$AE\perp BC$于点$M$,$FG\perp BC$于点$N$,$\angle 1 = \angle 2$.
(1)试说明:$AB// CD$;
(2)若$\angle D = \angle 3 + 50^{\circ}$,$\angle CBD = 60^{\circ}$,求$\angle C$的度数.
    第10题

答案

(1) ∵ AE⊥BC,FG⊥BC,∴ ∠AMB = ∠GNB = 90°. ∴ AE//FG. ∴ ∠A = ∠2. 又∵ ∠2 = ∠1,∴ ∠A = ∠1. ∴ AB//CD (2) ∵ AB//CD,∴ ∠D + ∠CBD + ∠3 = 180°. ∵ ∠D = ∠3 + 50°,∠CBD = 60°,∴ ∠3 = 35°. ∵ AB//CD,∴ ∠C = ∠3 = 35°
11. 如图,$\angle 1 = \angle BDE$,$\angle 2+\angle 3 = 180^{\circ}$.
(1)试说明:$AD// EF$;
(2)若$DA$平分$\angle BDE$,$FE\perp AF$于点$F$,$\angle 1 = 50^{\circ}$,求$\angle BAC$的度数.
    第11题

答案

(1) ∵ ∠1 = ∠BDE,∴ AC//DE. ∴ ∠2 = ∠ADE. ∵ ∠2 + ∠3 = 180°,∴ ∠3 + ∠ADE = 180°. ∴ AD//EF (2) ∵ ∠1 = ∠BDE,∠1 = 50°,∴ ∠BDE = 50°. ∵ DA 平分∠BDE,∴ ∠ADE = $\frac{1}{2}$∠BDE = 25°. ∴ ∠2 = ∠ADE = 25°. ∵ FE⊥AF,∴ ∠F = 90°. 由(1)得,AD//EF,∴ ∠BAD = ∠F = 90°. ∴ ∠BAC = ∠BAD - ∠2 = 90° - 25° = 65°
12. 如图,在三角形$ABC$中,点$D$,$E$分别在$AB$,$AC$上,$EF$交$DC$于点$F$,$\angle 3+\angle 2 = 180^{\circ}$,$\angle 1 = \angle B$.
(1)试说明:$DE// BC$;
(2)若$DE$平分$\angle ADC$,$\angle 3 = 3\angle B$,求$\angle 2$的度数.
    B4第12题

答案

(1) ∵ ∠DFE + ∠2 = 180°,∠3 + ∠2 = 180°,∴ ∠DFE = ∠3. ∴ BD//EF. ∴ ∠1 = ∠ADE. ∵ ∠1 = ∠B,∴ ∠ADE = ∠B. ∴ DE//BC (2) 由(1)知,∠ADE = ∠B,BD//EF,∴ ∠2 = ∠ADC. ∵ DE 平分∠ADC,∴ ∠ADC = 2∠ADE = 2∠B. ∵ ∠3 + ∠ADC = 180°,∠3 = 3∠B,∴ 3∠B + 2∠B = 180°,解得∠B = 36°. ∴ ∠ADC = 72°. ∴ ∠2 = 72°