Compare Medians against Side Comparison

Solution is obvious naming M as the point where the three medians meet. The three triangles AMB, BMC, CMA have same area therefore the smallest side base c AB in our case has to have the largest median CF.

The picture below will help to visualize the proof, just comparing areas of colored paralellograms.

I will apply the Apollonius' Theorem .

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By applying Apollonius' Theorem , we can have─

$c^2 + b^2 = 2 ( d^2 + (\frac{a}{2})^2)$

$\implies c^2 + b^2 = 2 ( d^2 + \frac{a^2}{4})$

$\implies c^2+ b^2 = 2 \times \frac{4d^2+a^2}{4}$

$\implies c^2 + b^2 = \frac{4d^2+a^2}{2}$

$\implies 4d^2= 2(b^2+c^2)-a^2$

$\implies 4d^2 = 2(a^2+b^2+c^2) -3a^2 \ldots \ldots \ldots (I)$ .

Similarly,

$4e^2 = 2(a^2+b^2+c^2) -3b^2 \ldots \ldots \ldots (II)$ .

$4f^2 = 2(a^2+b^2+c^2) -3c^2 \ldots \ldots \ldots (III)$ .

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As $a > b > c$ , we can conclude from $(I), (II)$ and $(III)$ that

$4d^2 < 4e^2 < 4f^2$

$\implies d < e < f$ .

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Yes, The larger side supports the smaller median in a triangle . So, $\boxed{f}$ is the answer.