Cevians AD and BE meet at point X inside triangle ABC. CX extended meets AB at F. If [AEX] = [CEX] = [CDX] = [BDX] = [BXF] = [AXF], (<--- areas)
must X be the centroid of triangle ABC?
The answer is yes; line segments AD, BE, and CF would have to be medians, which would by definition make X the centeroid, if the following pairs of triangles have the same area
:\(\Delta AXF\ and\ \Delta FXB\)
\(\Delta AXE\ and\ \Delta EXC\)
\(\Delta BXD\ and\ \Delta DXC\)
Let's look at the reasoning for the last pair; the other two would be similar.
The line segment \(X{H}_{3}\)is perpendicular to BC, so it must be perpendicular to BD and DC both; So it would serve as a heigt for triangles BXD and DXC. If the two triangles have the same height and the same area, it follows that they have the same base; that is, the line segments BD and DC must have the same length. In other words AD is a median.
The answer is yes; line segments AD, BE, and CF would have to be medians, which would by definition make X the centeroid, if the following pairs of triangles have the same area
:\(\Delta AXF\ and\ \Delta FXB\)
\(\Delta AXE\ and\ \Delta EXC\)
\(\Delta BXD\ and\ \Delta DXC\)
Let's look at the reasoning for the last pair; the other two would be similar.
The line segment \(X{H}_{3}\)is perpendicular to BC, so it must be perpendicular to BD and DC both; So it would serve as a heigt for triangles BXD and DXC. If the two triangles have the same height and the same area, it follows that they have the same base; that is, the line segments BD and DC must have the same length. In other words AD is a median.
