\(\dfrac{a+b+c}{a+b-c}=\dfrac{a-b+c}{a-b-c}\)
\(\Leftrightarrow\left(a+b+c\right)\left(a-b-c\right)=\left(a-b+c\right)\left(a+b-c\right)\)\(\Leftrightarrow a^2-ab-ac+ab-b^2-bc+ac-bc-c^2=a^2-ab+ac+ab-b^2+bc-ac+bc-c^2\)
\(\Leftrightarrow4bc=0\) \(\Leftrightarrow bc=0\)
\(\Rightarrow D=0\)