$G$ is the centroid of $\triangle ABC$. A line $a$ though $G$ is constructed such that $a$ is parallel to $AB$ and intersects $AC$ and $BC$ at $M$ and $P$, respectively. I should find the area of $\triangle ABC$ if the area of $\triangle MPC$ is $24$.
I was able to show that $\triangle ABC \sim MPC$. Therefore, $\dfrac{S_{\triangle ABC}}{S_{\triangle MPC}}=\dfrac{AB^2}{MP^2}=k^2$. How to approach the problem further?
Hint. Since the centroid $G$ is equal to $\frac{A+B+C}{3}$, it follows that $|CC_1|=3|GC_1|$. Moreover, from your work, $$\dfrac{S_{\triangle ABC}}{S_{\triangle MPC}}=\dfrac{|CC_1|^2}{|CG_1|^2}$$