In the triangle ABC, BC = 2Ab, ∠ ABC = 2 ∠ C, BD = CD Point D is on BC

It is proved that if the bisector BD through B intersects with AC at D, then ∵ ABC = 2 ∠ ACB, BC = 2Ab, ∵ DBC = ∠ ABC, ≌ DBC is isosceles triangle, if de ⊥ BC through D intersects with E, then E is the midpoint of BC at the same time, that is EB = EC = (1 / 2) BC = AB, ∵ DB = dB, ≌ DBC = ∠ DBA, ≌ DBA, ≌ DAB = ∠ DEB = 90 degree, that is ≌ B

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