So I was playing around GeoGebra and found this thing out, I don't know if this problem has a name or something.
Triangle ABC is inscribed inside a circle, from point D which is located inside the circle, we draw 3 perpendicular lines to each side of the triangle, what is the maximum area of the triangle whose vertices are the intersections of the perpendicular lines and the sides of the triangle? (maximum area of triangle EFG, the red triangle in the picture)
Using Geogebra I found out that this area is always maximal when point D is located at the center of the circle, or in other words, when the perpendiculars divide the sides into 2 equal segments.
If someone could provide a proof/explain why, I would be grateful.
See the diagram below:
$+1$ for re-discovering this neat property of pedal triangles: the area is proportional to the power of the point with respect to the circumcircle, in other words it only depends on the distance from the point to the circumcenter of the original triangle:
$$ S_{EFG}=\frac{R^2-OD^2}{4 R^2} \,S_{ABC} $$
Mathworld quotes on this Johnson, R. A. Modern Geometry: An Elementary Treatise on the Geometry of the Triangle and the Circle. Boston, MA: Houghton Mifflin, 1929.
A proof can be found for example on cut-the-knot.