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The last sprint king of junior high school mathematics
As shown in the figure, it is known that ∠ ABC = ∠ DCB, and one of the following conditions is added: ① ∠ A = ∠ D, ② AC = DB, ③ AB = DC, where _ _ is the one where △ ABC △ DCB cannot be determined (only fill in the serial number).

Solution: ∫ Known ∠ ABC = ∠ DCB, and BC = CB.

∴ If ① ∠ A = ∠ D is added, △ ABC △ DCB can be determined by AAS;

If 2AC = DB is added, it belongs to the angular order, and it is impossible to judge △ ABC △ DCB;

If 3AB = DC is added, which belongs to the angular order, it can be judged as △ ABC △ DCB.

So the answer is: ②.