Cho $\triangle ABC$ cân với góc B =120 độ. Gọi D là giao điểm của đường thẳng BC với tiếp tuyến tại A của (O) ngoại tiếp $\triangle ABC$.
  a/Cm $\triangle ADC$ vuông suy ra tỉ số $\frac{DB}{BC}$
   b/ Cm $\frac{1}{AD} +\frac{1}{AC}= \frac{\sqrt{3}}{AB} $
  c/ Đường thẳng qua D và qua tâm O cắt AB, AC tại E,F. Gọi M,N là trung điểm AB,AC. Cm các đường thẳng AO,MF,NE đồng quy
Ta có : $\Delta ADB \sim \Delta ADC$ (dễ cm) $\Rightarrow DB\times DC=DA^2\Rightarrow \frac{DB}{DC}=\frac{DA^2}{DC^2}$ :)
Ta có : $\Delta ABC$ cân và $\widehat{ABC}=120^o\Rightarrow \widehat{C}=\widehat{A}=30^o$ theo tính chất tiếp tuyến thì $\widehat{DAB}=\widehat{C}=30^o,\widehat{DBA}=180^o-\widehat{ABC}=60^o\Rightarrow \widehat{D}=180^0-60^o-30^o=90^o\Rightarrow \Delta ADC$  vuông :)
Ta có : $\widehat{DBA}=60^o\Rightarrow \sin 60^o=\frac{AB}{AD}=\frac{\sqrt{3}}{2}$ tương tự do tam giác đồng dạng nên $\frac{AB}{AC}=\frac{\sqrt{3}}{2}\Rightarrow \frac{1}{AD}+\frac{1}{AC}=\frac{\sqrt{3}}{AB}$

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