Lấy \(N'\) đối xứng với \(N\) qua \(I\Rightarrow N'=\left(3;\dfrac{5}{3}\right)\)
Phương trình đường thẳng AB: \(\dfrac{x-2}{3-2}=\dfrac{y-\dfrac{4}{3}}{\dfrac{5}{3}-\dfrac{4}{3}}\Leftrightarrow x-3y+2=0\)
Phương trình đường thẳng BD: \(ax+by-3a-3b=0\left(a^2+b^2\ne0\right)\)
\(\Rightarrow AB=\sqrt{BI^2+AI^2}=\sqrt{BI^2+4BI^2}=\sqrt{5}BI\)
\(\Rightarrow cosABD=\dfrac{BI}{AB}=\dfrac{1}{\sqrt{5}}=\dfrac{\left|a-3b\right|}{\sqrt{10.\left(a^2+b^2\right)}}\)
\(\Leftrightarrow2\left(a^2+b^2\right)=\left(a-3b\right)^2\)
\(\Leftrightarrow\left(a-b\right)\left(a+7b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=-7b\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}BD:x+y-6=0\\BD:7x-y-18=0\end{matrix}\right.\)