Parallel through the centroid and ratio of areas

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In triangle $A\hat{B}C$, $G$ is the centroid. Draw through $G$ the line parallel to $AB$, meeting side $AC$ at $K$ and side $BC$ at $L$. Prove that the area of triangle $K\hat{C}L$ is $\dfrac49$ of the area of $A\hat{B}C$.

Solution

Let $h$ be the distance of $C$ from the base $AB$ (the height relative to $AB$). The centroid $G$ divides each median in the ratio $2:1$ from the vertex; in particular $G$ is at distance $\dfrac{h}{3}$ from base $AB$. Hence $G$ is at distance from $C$ $h-\frac{h}{3}=\frac{2}{3}h.$ The line $KL$ passes through $G$ and is parallel to $AB$: triangle $K\hat{C}L$ is similar to $A\hat{B}C$ (common angle $\hat C$, and $KL\parallel AB$ gives congruent corresponding angles). The similarity ratio is the ratio of the heights from $C$: $k=\frac{\tfrac23 h}{h}=\frac23.$ Areas scale as the square of the similarity ratio: $\frac{[KCL]}{[ABC]}=k^2=\left(\frac23\right)^2=\boxed{\frac49.}\qquad\blacksquare$