Dividing a Triangle into Two Parts of Equal Area with Constraints Given a $\Delta ABC$ and a line labelled '$n$' that passes through the triangle but it's not parallel to any of the sides, how do we construct a line parallel to '$n$' that will divide $\Delta ABC$ into two parts of equal area.
 A: Up to vertex relabelling, we can assume that there is a point $P$ on the $BC$-side such that $AP$ is parallel to $n$ and $BP\leq PC$. The ratio between the areas of $APC$ and $APB$ is equal to $\frac{PC}{PB}$, so it is $\geq 1$. Let $t$ be the ratio between the area of $CAP$ and half the area of $ABC$: $t\geq 1$ holds. If $Q$ lies between $P$ and $C$, the parallel to $n$ through $Q$ cuts $AC$ in $R$ and $\frac{CP}{CQ}=\sqrt{t}$, the area of $CQR$ is half the area of $ABC$.
In summary, if $M$ is the midpoint of $BC$ and $Q$ lies between $P$ and $M$ such that $CQ^2=CM\cdot CP$, the line through $Q$ parallel to $AP$ splits $ABC$ in a triangle and a convex quadrilateral having equal areas.
A: You can construct the line to cut the point A, B, or C. This would hopefully get your result. Just a guest though.
A: If the triangle is a primitive Pythagorean triple with
$\space B\space$ as the even side, we begin by constructing
$\space n\space$ as the line from the $\space AC\space$ vertex to the midpoint of $\space B.\quad$ This allows us to see that side
$\space A\space$ is the altitude and that
$\space \dfrac{B}{2}\space$ is the base to be used in calculating area for both new triangles.
Now that we have this image, we can generalize this to making
$\space n\space$ a line from any vertex to the midpoint of the opposite side of any triangle.
