Geometry Contruction - Bisect Areas of 2 Triangles

Source: Asked to me by Sankeerth Rao (EE IITB 4th year Student)

Problem:
Given any two triangles in a plane construct a line which bisects both their areas.

Background:
In fact the existence of such a line is true in a very general setting - for any two polygons in a plane there exists a line which bisects both their areas. In fact its true for any two Jordan measurable sets in a plane. Further generalized version is called the Ham Sandwich Theorem and is proved using Borsuk Ulam Theorem.

Solution:

Highlight the part between the * symbols for the answer.
* Key fact: in every direction there is a line that bisects a given triangle's area, and it moves continuously as the direction rotates. (For direction theta, sweep a line of that slope across the triangle; the area on one side goes continuously from 0 to the full area, so some position halves it - and the halving position varies continuously with theta.)

Now parameterize by angle theta in [0, pi): let L(theta) be the bisecting line of triangle 1 in direction theta, and let f(theta) = (area of triangle 2 on the left side of L(theta)) - (half the area of triangle 2). f is continuous, and rotating by pi swaps the two sides, so f(theta + pi) = -f(theta). By the intermediate value theorem some theta* has f(theta*) = 0: the line L(theta*) bisects both triangles.

For an explicit construction: a bisecting line of a triangle must cut two of its sides; requiring it to halve the area gives a quadratic condition on the cut point (the product of the two side-fractions is fixed), so the cut points - and hence the line - are constructible with straightedge and compass, and solving the same condition simultaneously for both triangles reduces to intersecting conics.

The same continuity argument works for any two polygons (or any two measurable sets) - the two-dimensional case of the Ham Sandwich Theorem.

Solution by Betzalel from the comments. *