不妨设\(\angle DAE=\theta\),则\(\angle DAF=\arctan\left(\dfrac{3}{4}\right)-\theta\),故
\[\tan\angle DAE=\tan\angle DAG=\frac{DE}{AD}=\frac{DG}{AG}\Longrightarrow\tan\theta=\dfrac{p}{\sqrt{x^2+4}}=\dfrac{x}{2}\]
\[\tan\angle DAF=\frac{DF}{AD}\Longrightarrow\tan\left[\arctan\left(\dfrac{3}{4}\right)-\theta\right]=\dfrac{q}{\sqrt{x^2+4}}\\\frac{\tan\left(\arctan\dfrac{3}{4}\right)-\tan\theta}{1+\tan\left(\arctan\dfrac{3}{4}\right)\tan\theta}=\dfrac{q}{\sqrt{x^2+4}}\Longrightarrow\dfrac{q}{\sqrt{x^2+4}}=\frac{\dfrac{3}{4}-\dfrac{x}{2}}{1+\dfrac{3}{4}\cdot\dfrac{x}{2}}=\frac{6-4x}{8+3x}\]
从而得到:\[\begin{cases}
\dfrac{p}{\sqrt{x^2+4}}=\dfrac{x}{2}\\
\dfrac{q}{\sqrt{x^2+4}}=\dfrac{6-4x}{8+3x}
\end{cases}\Longrightarrow
\begin{cases}
p=\dfrac{x}{2}\sqrt{x^2+4}\\
q=\left(\dfrac{6-4x}{8+3x}\right)\sqrt{x^2+4}
\end{cases}\]
\[|EF|=|DE|+|DF|=p+q=\left(\dfrac{x}{2}+\dfrac{6-4x}{8+3x}\right)\sqrt{x^2+4}=\left[\dfrac{x}{2}+\dfrac{50}{3(3x+8)}-\dfrac{4}{3}\right]\sqrt{x^2+4}
\] |