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已知:如图,点E在直线DF上,点B在直线AC上,∠1=∠2,∠3=∠4.

求证:∠A=∠F.

证明:∵∠1=∠2(已知)

∠2=∠DGF(                  ▲                        

∴∠1=∠DGF(等量代换)

∴BD                  ▲                                          ▲                        

∴∠3+∠                  ▲                        =180°(                  ▲                        

又∵∠3=∠4(已知)

∴∠4+∠C=180°(等量代换)

                  ▲                        DF(                  ▲                        

∴∠A=∠F(                  ▲                        

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