6 Given non-zero vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$ satisfy $\left(\frac{\overrightarrow{A B}}{|\overrightarrow{A B}|}+\frac{\overrightarrow{A C}}{|\overrightarrow{A C}|}\right) \cdot \overrightarrow{B C}=0$ and $\frac{\overrightarrow{A B}}{|\overrightarrow{A B}|} \cdot \frac{\overrightarrow{A C}}{|\overrightarrow{A C}|}=\frac{1}{2}$, then $\triangle A B C$ is ( ).
(A) a triangle with all sides of different lengths
(B) a right triangle
(C) an isosceles but not equilateral triangle
(D) an equilateral triangle
6) D Hint: Since the line on which $\frac{\overrightarrow{A B}}{|\overrightarrow{A B}|}+\frac{\overrightarrow{A C}}{|\overrightarrow{A C}|}$ lies passes through the incenter of $\triangle A B C$, it follows from $\left(\frac{\overrightarrow{A B}}{|\overrightarrow{A B}|}+\frac{\overrightarrow{A C}}{|\overrightarrow{A C}|}\right) \cdot \overrightarrow{B C}=0$ that $|\overrightarrow{A B}|=|\overrightarrow{A C}|$. Also, $\frac{\overrightarrow{A B}}{|\overrightarrow{A B}|} \cdot \frac{\overrightarrow{A C}}{|\overrightarrow{A C}|}=\frac{1}{2}$, so $\angle A=\frac{\pi}{3}$, which means $\triangle A B C$ is an equilateral triangle. Therefore, the answer is D.