Prove that $d$ passes through a fixed point.

Well, when there is no idea then coordinate system comes at handy. And actualy it is an easy problem with c.s.

Let $B=(2b,0)$, $C= (2c,0)$, $A= (0,2a)$ and $A'= (2t,0)$, for some fixed $a,b,c$ and variable $t$. The midpoint of $A'B$ is $N = (b+t,0)$. Since $B'$ is on a line $$AC:\;\;\;{x\over 2b}+{y\over 2c}=1$$ we have $$B' = (b+t,{a(b-t)\over b}) $$ and analougly we get $$C' = (c+t,{a(c-t)\over c}) $$

Now the slope of segment $B'C'$ is $$k= {at\over bc}$$ so the slope of $d$ is $$k' = -{1\over k} = -{bc\over at}$$

So the line $d$ has equation $$ y= {bc\over at}x +{2bc\over a}$$

which means that this line goes always through the point $F=(0,{2bc\over a})$.