Question

探究：如图①，在矩形ABCD中，过点A作∠EAF＝∠BAD,AE交线段BC于点E，AF交线段CD的延长线于点F.求证：ΔABE∽ΔADF.

拓展：如图②，在四边形ABCD中，∠ABC＋∠ADF＝180º,过点A作∠EAF＝∠BAD,AE交线段BC于点E，AF交线段CD延长线于点F.若AB∶AD＝2∶3，求ΔABE的面积与ΔADF的面积之比.

Answer 1

探究：证明：∵在矩形ABCD中 ∴∠ADC＝∠BAD＝∠B＝90º

                          ∴∠ADF＝∠B＝90º, ∵∠EAF＝∠BAD,

                          ∴∠EAB＋∠EAD＝∠FAD＋∠EAD, ∴∠EAB＝∠FAD

                          ∴ΔABE∽ΔADF

  拓展：解：∵∠ABC＋∠ADC＝180º, ∠ADC＋ADF＝180º.

                      ∴∠ABE＝∠ADF, ∵∠EAF＝∠BAD,

                      ∴∠EAB＋∠EAD＝∠FAD＋∠EAD, ∴∠EAB＝∠FAD

                      ∴ΔABE∽ΔADF. ∴ＳΔABE∶ＳΔADF＝AB²∶AD²＝4∶9