Problem
In any given Triangle ABC, construct a perpendicular XY upon line BC using only a compass and a straight edge such that Area of Quadrilateral ABYX is equal to Area of Triangle XYC.
Diagram
What I've done so far
Construct the median AM from apex A on line BC such that BM = MC.
Now, Area[ABM] = Area[ACM].
Suppose the line XY is drawn. Then let it cross the median at O. Therefore, by sine-area rule on Triangles AOX and MOY, 1/2 * AO * OX * angle[AOX] = 1/2 * MO * OY * angle[MOX]
=> AO * OX = MO * OY.
I'm stuck here. Can anybody please help?
Let's draw an altitude $AH$ on $BC$ and denote $b=BH$, $c=HC$, $h=AH$ and $y=YC$. We can also see, that $x=XY=yh/c$ (from similarity of $\triangle XYC$ and $\triangle AHC$).
The area of $XYC$: $$A_1=\frac{xy}2=\frac{y^2h}{2c}.$$
The area of $ABC$: $$ A_2=\frac{h(b+c)}2. $$
You want $A_1=A_2/2$, thus $$y^2 = \frac{c(b+c)}2,\quad\text{or}\quad y=\sqrt{c\cdot\frac{b+c}2}$$
Can you try to construct the latter?