Given point $K$ in a triangle, is it possible to construct a line through $K$ that bisects the area of the triangle?

Question

$K$ is an arbitary fixed point inside $\triangle ABC$. Is it possible to construct a line that pass through point $K$ such that area of $\triangle AJH =\triangle ABC/2$

Answer 1

Yes, it can be constructed with ruler and compass. We use cartesian coordinates and give an algebraic proof. If, for example, $K$ is very close to $A,$ then $J$ will be on $BC$ and $H$ might be on $AB.$ We assume that things have been rotated and reflected as necessary so that the given diagram is valid.

Let $A=(0,0), C=(c,0), H=(t,0), B=(r,s)$ and $K=(u,v).$ We shall show that $t$ and therefore $H$ is constructible.

We deduce that $J=k(r,s),$ where $$k={tv\over rv-su+st}$$ To verify this, we must show that $${ks\over kr-t}={v\over u-t}$$ This is equivalent to $$s(u-t)=(r-{t\over k})v$$ and to $$-su+st+rv={tv\over k},$$ which agrees with our definition of $k.$

The area condition means that $2kst=sc,$ so that $c=2tk.$ and therefore, $$(rv-su+st)c=2t^2v.$$ Since $t$ satisfies a quadratic equation with constructible coefficients, the proof is complete.

Answer 2

Converting and augmenting a comment ...

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There certainly is such a line. @achillehui's comment to the question suggests how to prove existence. If you're asking whether such a line is straightedge-and-compass constructible ...

Note that, as discussed in this answer of mine to a separate question, area-bisecting lines are precisely the tangents to certain hyperbolas that have the triangle's side-lines as asymptotes. Here's that answer's picture of an equilateral case:

Note. Only the arcs of the hyperbolas bounding the curvilinear triangle have area-bisecting tangents; these are the tangents that meet the triangle on its "other" sides. That is, a tangent that cuts the interior of the triangle side associated with its hyperbola is not an angle bisector. Luckily, as the image shows, area-bisecting tangents sweep across the entire interior of the triangle; so, any point $K$ is guaranteed to lie on an area-bisecting tangent of at least one of these hyperbolas.

When the coordinates of the vertices are $(1,0)$, $(-1/2, \sqrt{3}/2)$, $(-1/2,-\sqrt{3}/2)$, the equation of one hyperbola is $$8(x-1)^2 - 24 y^2 = 9$$ from which we see that all necessary metric elements of the hyperbola (say, major axis and eccentricity) are straightedge-and-compass constructible, as they require, at worst, extraction of square roots. Likewise, the tangents from a given point $K$ to any of these hyperbolas must also be straightedge-and-compass constructible.

But we can be a little more explicit ...

Leaving the reader to construct the foci and transverse axis, I'll cite "Construction of tangents from a point outside the hyperbola" at Nabla.hr for the final steps. Paraphrasing:

The circle about $K$ through focus $F$, and the circle about other focus $F'$ with radius equal to the transverse axis, meet at $P$ and $Q$ (not shown). The tangents from $K$ are the perpendicular bisectors of $\overline{FP}$ and $\overline{FQ}$.

Referring back to the animation, observe that if $K$ is "inside" one hyperbola, then the associated area-bisecting tangent would touch one of the other hyperbolas. So, $K$ is always "outside" the hyperbola it needs. $\square$