《2025年诚成教育学业评价八年级数学下册人教版》

25. (8分)在正方形ABCD中，E，F分别是AD，CD上的点，连接AF，BE交于点G，DE = CF.
(1) 如图①，求证∠AGB = 90°;
(2) 如图②，E是AD的中点，连接CG，求证CG = CB;
(3) 若正方形ABCD的边长为2，直接写出DG长度的最小值.

答案：
解：
(1)证明：$\because$四边形$ABCD$是正方形，
$\therefore\angle BAD = \angle D = 90^{\circ}$，$AB = AD = CD$.
$\because DE = CF$，
$\therefore AD - DE = CD - CF$.
$\therefore AE = DF$.
$\therefore\triangle BAE\cong\triangle ADF(SAS)$.
$\therefore\angle ABE = \angle DAF$.
$\therefore\angle ABE + \angle BAF = \angle DAF + \angle BAF = \angle BAD = 90^{\circ}$.
$\therefore\angle AGB = 90^{\circ}$.
(2)证明：如图，分别延长$AF$，$BC$交于点$H$.
$\because$四边形$ABCD$是正方形，
$\therefore CB = AD = CD$，$\angle D = \angle BCD = \angle HCF = 90^{\circ}$.
$\because E$是$AD$的中点，
$\therefore DE=\frac{1}{2}AD=\frac{1}{2}CD$.
$\because CF = DE$，
$\therefore CF=\frac{1}{2}CD$.
$\therefore CF = DF$.
$\because\angle AFD = \angle HFC$，$\angle D = \angle HCF = 90^{\circ}$，
$\therefore\triangle ADF\cong\triangle HCF(ASA)$.
$\therefore CH = AD = CB$.
由
(1)知$\angle AGB = 90^{\circ}$，
$\therefore\angle BGH = 90^{\circ}$.
$\therefore CG=\frac{1}{2}BH$.
$\therefore CG = CB$.

(3)$DG$长度的最小值为$\sqrt{5}-1$.