As shown in the figure, ∠ B = ∠ C = 90 °, point m is the midpoint of BC, am bisects ∠ DAB

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But it seems that we can also think of a triangle AFM as isosceles triangle AF = MF, AB parallel to MF parallel to CD and BM = cm, so DF = AF, so MF = FD, so triangle MFD is isosceles triangle, then FMD = angle FDM, and then parallel to CD, so inner angles are equal Angle FMD = angle CDM, so it is called FDM = angle CDM, so the ADC is divided equally by DM

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