4. If $\triangle A B C$ satisfies $\frac{\angle A}{3}=\frac{\angle B}{4}=\frac{\angle C}{5}$, then the value of $\frac{\overrightarrow{A B} \cdot \overrightarrow{A C}}{|B C|^{2}}$ is
According to the problem, $A=45^{\circ}, B=60^{\circ}, C=75^{\circ}$. Therefore,
$$
\frac{\overrightarrow{A B} \cdot \overrightarrow{A C}}{|B C|^{2}}=\frac{b c \cos A}{a^{2}}=\frac{\sin B \sin C \cos A}{\sin ^{2} A}=\frac{\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{6}+\sqrt{2}}{4} \cdot \frac{\sqrt{2}}{2}}{\frac{1}{2}}=\frac{3+\sqrt{3}}{4} .
$$