Draw seg AP ⊥ seg PE and seg EQ ⊥ seg QC.

Side AP || side EQ and AC is their transversal.
∴ ∠PAE ≅ ∠QEC …(i) [Corresponding angles]
In ∆ APE and ∆ EQC,
∠PAE ≅ ∠QEC …[From (i)]
∠APE ≅ ∠EQC … [Each angle is of measure 90°]
side PE ≅ side QC …. [Perpendicular distance between parallel lines]
∴ ∆ APE ≅ ∆ EQC … [By AAS test]
∴ AE = EC … [Corresponding sides of congruent triangles]
∴ E is the midpoint of AC.
∴ seg BE is the median. Similarly, seg CF is the median.
Since, the medians of a triangle are concurrent.
∴ G is the centroid of ∆ABC.