Áp dụng định lý hàm sin cho tam giác BCD:
\(\frac{CD}{sin30^0}=\frac{BD}{sinC}=\frac{BD}{\frac{AB}{AC}}\Rightarrow AC.BD=2\)
Ta có: \(cosA=\frac{AB}{AC}=\frac{1}{AC}\)
Mà \(\frac{BD}{sinA}=\frac{AD}{sin60^0}\Rightarrow\frac{BD^2}{sin^2A}=\frac{AD^2}{sin^260}\Rightarrow\frac{4}{AC^2.\left(1-cos^2A\right)}=\frac{4\left(AC-1\right)^2}{3}\)
\(\Leftrightarrow\frac{1}{AC^2\left(1-\frac{1}{AC^2}\right)}=\frac{\left(AC-1\right)^2}{3}\) \(\Leftrightarrow\frac{1}{AC^2-1}=\frac{\left(AC-1\right)^2}{3}\)
\(\Leftrightarrow\left(AC^2-1\right)\left(AC-1\right)^2=3\)
\(\Leftrightarrow AC^4-2AC^3+2AC-4=0\)
\(\Leftrightarrow AC^3\left(AC-2\right)+2\left(AC-2\right)=0\)
\(\Leftrightarrow\left(AC^3+2\right)\left(AC-2\right)=0\Rightarrow AC=2\)