Geometry Problem 1612: Perpendicular Medians and Squares of Sides

Diagram showing triangle ABC with perpendicular medians AM and CN intersecting at centroid G

Given: $\triangle ABC$, medians $AM$ and $CN$ such that $AM \perp CN$. Let $G$ be the centroid.
To Prove: $AB^2 - AC^2 = 3 GA^2$

Problem Statement: Theorem: Metric Relations in a Triangle with Orthogonal Medians

In a triangle $ABC$, let $AM$ and $CN$ be the medians that intersect perpendicularly at the centroid $G$. Prove that the difference between the squares of sides $AB$ and $AC$ is equal to three times the square of the segment $GA$:

$$AB^2 - AC^2 = 3GA^2$$

Understanding the Problem

The orthogonality of $AM$ and $CN$ implies that $\angle AGC = 90^\circ$. Because $G$ is the centroid, it partitions the medians into specific ratios that allow for direct application of the Pythagorean Theorem.