《2025年通成学典课时作业本七年级数学上册苏科版宿迁专版》

22.(8分)如图,在长方形ABCD中,AD//BC,E为边BC上一点,将长方形沿AE折叠(AE为折痕),使点B与点F重合,EG平分∠CEF,交CD于点G,过点G作HG⊥EG,交AD于点H. (1)试说明HG//AE; (2)若∠CEG= 20°,求∠DHG的度数.

答案： （1）由折叠,知∠AEB = ∠AEF = ½∠BEF.因为EG平分∠CEF,所以∠FEG = ∠CEG = ½∠CEF.因为∠BEF + ∠CEF = 180°,所以∠AEG = ∠AEF + ∠FEG = ½(∠BEF + ∠CEF) = 90°.所以AE⊥EG.因为HG⊥EG,所以HG//AE （2）因为∠CEG = 20°,∠AEG = 90°,所以∠AEB = 70°.因为四边形ABCD是长方形,所以AD//BC.所以∠AEB = ∠DAE = 70°.因为HG//AE,所以∠DHG = ∠DAE = 70°

23.(8分)如图,线段AB= 6 cm,延长BA到点C,D是BC的中点. (1)若AC= 4 cm,求线段AD的长. (2)若AC的长逐渐增大,关于AD的长的变化趋势,给出下列结论:①变小;②变大;③先变小,后变大;④先变大,后变小.其中,正确的是______③(填序号). (3)若AD= 2 cm,求线段AC的长.
（1）因为AB = 6 cm,AC = 4 cm,所以BC = AB + AC = 6 + 4 = 10(cm).因为D是BC的中点,所以CD = ½BC = ½×10 = 5(cm).所以AD = CD - AC = 5 - 4 = 1(cm).所以线段AD的长为1 cm （3）① 当点D在AB上时,因为AB = 6 cm,AD = 2 cm,所以BD = AB - AD = 6 - 2 = 4(cm).因为D是BC的中点,所以BC = 2BD = 2×4 = 8(cm).所以AC = BC - AB = 8 - 6 = 2(cm).② 当点D在BA的延长线上时,因为AB = 6 cm,AD = 2 cm,所以BD = AB + AD = 6 + 2 = 8(cm).因为D是BC的中点,所以BC = 2BD = 2×8 = 16(cm).所以AC = BC - AB = 16 - 6 = 10(cm).综上所述,线段AC的长为2 cm或10 cm

答案： （1）因为AB = 6 cm,AC = 4 cm,所以BC = AB + AC = 6 + 4 = 10(cm).因为D是BC的中点,所以CD = ½BC = ½×10 = 5(cm).所以AD = CD - AC = 5 - 4 = 1(cm).所以线段AD的长为1 cm （2）③ （3）① 当点D在AB上时,因为AB = 6 cm,AD = 2 cm,所以BD = AB - AD = 6 - 2 = 4(cm).因为D是$BC\n$的中点,所以BC = 2BD = 2×4 = 8(cm).所以AC = BC - AB = 8 - 6 = 2(cm).② 当点D在BA的延长线上时,因为AB = 6 cm,AD = 2 cm,所以BD = AB + AD = 6 + 2 = 8(cm).因为D是BC的中点,所以BC = 2BD = 2×8 = 16(cm).所以AC = BC - AB = 16 - 6 = 10(cm).综上所述,线段AC的长为2 cm或10 cm