1. Find the intersection point of the two lines

The lines are $ax+by+c=0 \quad ...(1)$ $bx+ay+c=0 \quad ...(2)$

Subtract (2) from (1): $ax+by-(bx+ay)=0$ $(a-b)x-(a-b)y=0$ $(a-b)(x-y)=0$

Since $a>b$, we have $a-b\neq 0$. Hence, $x=y$

Put $y=x$ in (1): $ax+bx+c=0$ $(a+b)x+c=0$ $x=-\frac{c}{a+b}$

So the intersection point is $\left(-\frac{c}{a+b},-\frac{c}{a+b}\right)$

2. Use the given distance condition

Distance between $\left(1,1\right)$ and $\left(-\frac{c}{a+b},-\frac{c}{a+b}\right)$ is $d=\sqrt{\left(1+\frac{c}{a+b}\right)^2+\left(1+\frac{c}{a+b}\right)^2}$ $d=\sqrt{2\left(1+\frac{c}{a+b}\right)^2}$ $d=\sqrt{2}\left(1+\frac{c}{a+b}\right)$

Since $a,b,c>0$, the quantity $1+\frac{c}{a+b}>0$.

Given $d<2\sqrt{2}$ so $\sqrt{2}\left(1+\frac{c}{a+b}\right)<2\sqrt{2}$ $1+\frac{c}{a+b}<2$ $\frac{c}{a+b}<1$ $c<a+b$ $a+b-c>0$

Thus Option A is true.

3. Check the other options

• A: $a+b-c>0$ — true.
• B: $a-b+c<0$ — false, because $a-b>0$ and $c>0$, so $a-b+c>0$
• C: Interpreting as $a-b+c\ge 0$ — true, in fact strictly positive. However, this appears to be a typo in the options since both B and C concern the same expression with opposite inequalities, and the stored answer indicates a single-correct MCQ. The conclusion forced by the distance condition is specifically $a+b-c>0$.
• D: $a+b-c<0$ — false.

So the intended single correct option is A.

4. Final answer

$\boxed{A}$