How to construct a normal to side $AB$ of acute triangle so that it halves it area?
So we have $${c\cdot v\over 2} = 2{x\cdot v' \over 2} \implies {v\over v'} = {2x\over c}$$
From similar triangles $BEF$ and $BDC$ we have $${v\over v'} = {d\over x} $$
And thus we have $cd =2x^2$ which remainds me on the power of the point $B$ on some circle through $A$ and $D$. So I draw an arbitray circle through $A$ and $D$ and now we have to find a point $I$ on this circle so that $J$ which is also on a circle that halves $IB$. And this is dead end. Any idea how to solve this one? It should be not difficult.
Your idea is fine: construct any circle passing through $A$ and $D$, then construct the tangent $BT$ from $B$ to the circle, so that $BT=\sqrt{cd}=\sqrt2 x$. To find $x$, you can then construct the side of a square whose diagonal is $BT$.