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.
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You should check this nice answer: http://math.stackexchange.com/q/235626/4058 – Javier Álvarez Nov 16 '12 at 12:05
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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.
Jack D'Aurizio
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You can construct the line to cut the point A, B, or C. This would hopefully get your result. Just a guest though.
Baari Gaffar
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That won't work unless the line $n$ happens already to be parallel to a bisector of the triangle through one of its vertices, which typically is not the case. – coffeemath Nov 16 '12 at 05:45
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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.
poetasis
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