Divide triangle into two equivalent parts by line perpendicular to the base. Originally the problem wants me to divide into 3 equivalent(equal areas) parts but I thought that I should do 2 first. Tried 2 and could not after many attempts. DF below is movable so it is not really the correct solution.

 A: (I'm too lazy to draw a figure. Therefore I use coordinates to fix ideas. Nevertheless the solution is not in terms of analytic geometry.)
Assume $B=(0,0)$, $F=(x,0)$, $C=(c,0)$, and $A=(p,q)$ with ${c\over2}\leq p\leq c$. We then want
$${1\over2}\  x\cdot{q\over p}x={1\over2}\ {c q\over2}\ ,$$
or
$$x^2={1\over2}\>cp\ .$$
Since $x$ is independent of $q$ we may replace $A$ by $A=(p,q')$ such that the resulting triangle has a right angle at $A'$ (draw a Thales circle over $BC$ to find $A'$). Letting $|BA'|=:s$ we have $cp=s^2$, by "Euclid's theorem". We therefore have to divide $s$ by $\sqrt{2}$ in order t obtain $x$. Constructively this can be done by drawing a square with side $s$ and taking half the diagonal.
Use similar ideas to divide the given triangle into three parts of equal area.
A: The Math

\begin{align}
\text {Area of }\triangle \text{ ABH} &= \frac{mbh}{2} \\[1.5ex]
\text {Area of }\triangle \text{ AXY} &= \frac{mx^2}{2} \\[1.5ex]
\text{For } \triangle \text { AXY to be half the area of } \triangle { ABH:} \\[1.5ex]
\frac{mx^2}{2} &= \frac{1}{2} \times \frac{mbh}{2} \\[1.5ex]
\text{giving } x \text{ (independent of gradient } m \text{)}: \\[1.5ex]
x^2 &= \frac{bh}{2}\\ 
x &= \sqrt{\frac{bh}{2}}
\end{align}
The Construction

1. Draw a line $\perp$ to AC through A, and a line $\parallel$ to AC through B. Mark D, the point of intersection.

2. Mark E, the midpoint of AC. Then with centre A and radius AE, draw a circle. Mark F, the point of intersection of the circle and the perpendicular.
\begin{align}
\text{DF} &= \text{AD} + \text{AF} = h \text{ (height)} + \frac{b \text{ (base)}}{2}
\end{align}



*Mark G, the midpoint of DF. Then with centre G and radius DG, draw a circle cutting AC in H.
AH = $\sqrt{\frac{bh}{2}}$, which is the required length (see Note 1 below).





*Draw the line $\perp$ to AC through H, cutting AB in I.



*$\triangle$ AHI is half the area of $\triangle$ ABC


Notes:


*

*In step 3 above,
\begin{align}
\text {AD } & = h \\
\text {AF } & = \frac{b}{2} \\[1.5ex]
\text {DG } & = \frac{h + \frac{b}{2}}{2} = \frac{h}{2} + \frac{b}{4} \\[1ex]
\text {So, AG} &= h - (\frac{h}{2} + \frac{b}{4}) = \frac{h}{2} - \frac{b}{4}  \\[1ex]  
\text {And, GH} & = DG = \frac{h + \frac{b}{2}}{2} = \frac{h}{2} + \frac{b}{4} \\[1ex]
\text{Now, let AH } &= x \\[1ex]
\text{Then by Pythagoras' Theorem (} \triangle \text{AGH):} \\
x^2 & = (\frac{h}{2} + \frac{b}{4})^2 - (\frac{h}{2} + \frac{b}{4})^2 \\[1ex]
      & = \frac{bh}{2} \\
\text{And, } x &= \sqrt{\frac{bh}{2}}
\end{align}
