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過A作AO⊥BC,交BC于點(diǎn)O,以BC所在的直線為x軸,AO所在的直線為y軸建立平面直角坐標(biāo)系,
設(shè)A(0,a),B(b,0),C(c,0),D(d,0),
∵|AB|2=|AD|2+|BD|?|DC|,
∴a2+b2=a2+d2+(d-b)(c-d),即d2-b2+(d-b)(c-d)=0,
∴(d+b)(d-b)+(d-b)(c-d)=0,即(d-b)(b+c)=0,
∵D與B不重合,∴d≠b,即d-b≠0,
∴b+c=0,即b=-c,
∴B與C關(guān)于y軸對稱,
∴AB=AC,
則△ABC為等腰三角形.
故選C