《2025年亮点给力大试卷八年级数学上册苏科版》

25. (8分)
(1) 如图①,E,F分别是正方形ABCD的边AB及边DC延长线上的点,且$BE = CF$,连接EF,交BC于点G,则BG与BC之间的数量关系是BG = $\frac{1}{2}$BC；
(2) 如图②,D,E分别是等腰三角形ABC的边AB及边AC延长线上的点,且$AB = AC$,$BD = CE$,连接DE交BC于点F,过点D作$DG⊥BC$于点G,试判断FG与BC之间的数量关系,并说明理由.
FG = $\frac{1}{2}$BC.理由如下:过点E作EH⊥BC,交BC的延长线于点H,则∠CHE = 90°.因为DG⊥BC,所以∠BGD = ∠DGF = 90°,所以∠DGF = ∠EHF,∠BGD = ∠CHE.因为AB = AC,所以∠B = ∠ACB.因为∠ACB = ∠ECH,所以∠B = ∠ECH.在△DBG和△ECH中,$\left\{\begin{array}{l} ∠BGD = ∠CHE\\ ∠B = ∠ECH\\ BD = CE\end{array}\right.$，所以△DBG≌△ECH(AAS),所以DG = EH,BG = CH,所以BC = BG + GC = CH + GC = GH.在△DGF和△EHF中,$\left\{\begin{array}{l} ∠DFG = ∠EFH\\ ∠DGF = ∠EHF\\ DG = EH\end{array}\right.$，所以△DGF≌△EHF(AAS),所以FG = FH,所以FG = $\frac{1}{2}$GH,所以FG = $\frac{1}{2}$BC.

答案：
(1)BG = $\frac{1}{2}$BC
(2)FG = $\frac{1}{2}$BC.理由如下:过点E作EH⊥BC,交BC的延长线于点H,则∠CHE = 90°.因为DG⊥BC,所以∠BGD = ∠DGF = 90°,所以∠DGF = ∠EHF,∠BGD = ∠CHE.因为AB = AC,所以∠B = ∠ACB.因为∠ACB = ∠ECH,所以∠B = ∠ECH.在△DBG和△ECH中,$\left\{\begin{array}{l} ∠BGD = ∠CHE\\ ∠B = ∠ECH\\ BD = CE\end{array}\right.$，所以△DBG≌△ECH(AAS),所以DG = EH,BG = CH,所以BC = BG + GC = CH + GC = GH.在△DGF和△EHF中,$\left\{\begin{array}{l} ∠DFG = ∠EFH\\ ∠DGF = ∠EHF\\ DG = EH\end{array}\right.$，所以△DGF≌△EHF(AAS),所以FG = FH,所以FG = $\frac{1}{2}$GH,所以FG = $\frac{1}{2}$BC.

26. (8分)新趋势 综合实践
(1) 如图①,在四边形ABCD中,$AB// CD$,E是BC的中点,连接AE.若AE是$∠BAD$的平分线,试判断AB,AD,CD之间的数量关系.
解决此问题可以用如下方法：分别延长AE,DC交于点F,易证$△AEB≌△FEC$,得到$AB = FC$,从而把AB,AD,CD转化到一个三角形中即可判断.
AB,AD,CD之间的数量关系为AD = AB + CD；
(2) 如图②,在四边形ABCD中,$AB// CD$,AF与DC的延长线交于点F,E是BC的中点,连接AE.若AE是$∠BAF$的平分线,试探究AB,AF,CF之间的数量关系,并证明你的结论.

AB = AF + CF.证明如下:分别延长AE,DF交于点H.因为AB//CD,所以∠B = ∠ECH,∠BAE = ∠H.因为E是BC的中点,所以BE = CE.在△AEB和△HEC中,$\left\{\begin{array}{l} ∠BAE = ∠H\\ ∠B = ∠ECH\\ BE = CE\end{array}\right.$，所以△AEB≌△HEC(AAS),所以AB = HC.因为AE平分∠BAF,所以∠BAE = ∠FAE,所以∠H = ∠FAE,所以AF = HF,所以AB = HC = HF + CF = AF + CF.

答案：
(1)AD = AB + CD
(2)AB = AF + CF.证明如下:分别延长AE,DF交于点H.因为AB//CD,所以∠B = ∠ECH,∠BAE = ∠H.因为E是BC的中点,所以BE = CE.在△AEB和△HEC中,$\left\{\begin{array}{l} ∠BAE = ∠H\\ ∠B = ∠ECH\\ BE = CE\end{array}\right.$，所以△AEB≌△HEC(AAS),所以AB = HC.因为AE平分∠BAF,所以∠BAE = ∠FAE,所以∠H = ∠FAE,所以AF = HF,所以AB = HC = HF + CF = AF + CF.