《2025年课时训练江苏人民出版社八年级数学上册苏科版》

1 如图,△ABC≌△ADE,且点E在BC上,若∠DAB= 30°,则∠CED= 150°.

答案： 150°

2 (2024·上海长宁区延安初级中学期末)如图,在Rt△ABC中,∠BAC= 90°,点D在边AC上,点E在BC的延长线上,BD= DE,EF⊥AC,交AC延长线于点F,AD= EF.
(1)求证：∠ABD= ∠FDE；
(2)求证：AB= CD+EF.

答案：
(1)
∵EF⊥AC,
∴∠F = 90° = ∠BAC,
在Rt△BAD和Rt△DFE中，{BD = DE, AD = EF}
∴Rt△BAD≌Rt△DFE(HL),
∴∠ABD = ∠FDE.
(2)
∵Rt△BAD≌Rt△DFE,
∴AB = DF,∠ADB = ∠DEF.
∵BD = DE,
∴∠DBE = ∠DEB,
∴∠ADB - ∠DBE = ∠DEF - ∠DEB,
∴∠BCD = ∠CEF,
∴∠ECF = ∠CEF,
∴CF = EF,
∴AB = DF = DC + CF = DC + EF.