Three medians meet (centroid, 2:1)

Dependencies: Triangle midsegment theorem (the midsegment is parallel to the third side and half its length), AA similarity, Vertical angles are equal, Parallel ⇒ alternate angles equal (alternate interior angles ⇒ parallel); together with Ruler axiom, which guarantees the uniqueness of the point dividing a segment in a given ratio.

Statement

Let $\triangle ABC$ be an arbitrary triangle. Denote the midpoints of its sides by

Then the three medians $AM_a$, $BM_b$, $CM_c$ meet at a single point $G$, called the centroid of the triangle. Moreover, each median is divided by $G$ in the ratio $2:1$ (the segment from the vertex to $G$ is exactly twice the segment from $G$ to the midpoint of the opposite side):