Problem

Source: RMO KV 2024 Q3

Tags: geometry, perpendicular bisector



Let $ABC$ be an equilateral triangle. Suppose $D$ is the point on $BC$ such that $BD+DC = 1:3$. Let the perpendicular bisector of $AD$ intersect $AB,AC$ at $E,F$ respectively. Prove that $49 \cdot [BDE] = 25 \cdot [CDF]$, where $[XYZ]$ denotes the area of the triangle $XYZ$.