《2025年千里马单元测试卷八年级数学下册人教版》

5. 如图,将长方形ABCD沿直线EF折叠,使点C与点A重合,折痕交AD于点E,交BC于点F,连接CE.
(1)求证:AE = AF = CE = CF;
(2)设AE = a,ED = b,DC = c,请写出a,b,c三者之间的数量关系式.

答案：
(1)证明:由题意,知$AF = CF$,$AE = CE,\angle AFE=\angle CFE$.
$\because$四边形$ABCD$是长方形,
$\therefore AD// BC,\therefore \angle AEF=\angle CFE$,
$\therefore \angle AFE=\angle AEF$,
$\therefore AE = AF$,
$\therefore AE = AF = CE = CF$.
(2)解:由题意,知$AE = CE = a$.
由$\angle D = 90^{\circ}$知$ED^{2}+DC^{2}=CE^{2}$,
即$b^{2}+c^{2}=a^{2}$.