the area of the triangle $ ABC $ is constant so we must have the angle $ ACB $ must be a right angle. so we have to construct the circle of diameter $ AB $. if we construct two congruent circles with centers in $ A,B $ which have at least one common point. if I'm not wrong we can construct a perpendicular line on $ AB $ so if we constructed it then using the straight edge we can construct the parallel line through the point of intersection of the circles and we'll obtain the midpoint of $ AB $ and we're done.

the straightedge is a ruler... so you can draw a line through two points given,don't you?

Instasolvable? Answer: Construct the circle with diameter $AB$. If it meets $l$ somewhere, that point satisfies the condition. Else, take the intersection of the perpendicular bisector of $AB$ with $l$. Proof sketch: "Double count" the area of $\Delta ABC$ to get that we want $\angle C$ to be closest to $90^\circ$, from here it's pretty obvious.