a+b-c/a+b-c + 2c/a+b-c = a-b-c/a-b-c + 2c/a-b-c
suy ra 2c/a+b-c = 2c/a-b-c
Dấu = xảy ra khi c=0
\(\dfrac{a+b+c}{a+b-c}=\dfrac{a-b+c}{a-b-c}\)
\(\Leftrightarrow\left(a+b+c\right)\left(a-b-c\right)=\left(a-b+c\right)\left(a+b-c\right)\)
\(\Leftrightarrow a^2-\left(b+c\right)^2=a^2-\left(b-c\right)^2\)
\(\Leftrightarrow\left(b+c\right)^2-\left(b-c\right)^2=0\)
\(\Leftrightarrow\left(b+c-b+c\right)\left(b+c+b-c\right)=0\)
\(\Leftrightarrow4bc=0\)
Do b\(\ne\) 0\(\Rightarrow c=0\)
Vậy c=0 thì thỏa tỉ lệ thức (đcpcm)