<s>[INST] If $a^2+b^2=5c^2$ where $a,b,c$ are the sides of a triangle, prove that the area of triangle is $c^2\tan C$ 
If $a^2+b^2=5c^2$ where $a,b,c$ are the sides of a triangle, prove that the area of triangle is $c^2tan C$

Let median through $C$ be $CF$. $AF=FB=\frac{c}{2}$
$$CF=\frac{1}{2}\sqrt{2(a^2+b^2)-c^2}=\frac{3c}{2}$$
$CG=c$ where $G$ is the centroid and $GF=\frac{c}{2}$
$$\frac{3}{4}(a^2+b^2+c^2)=M_a^2+M_b^2+M_c^2$$
$$\frac{9c^2}{2}=M_a^2+M_b^2+\frac{9c^2}{4}$$
$$c^2=(\frac{2}{3}M_a^2)+(\frac{2}{3}M_b^2)$$
$$BC^2=AG^2+BG^2$$
So medians through $A$ and $B$ are perpendicular.
I don't know if this information is useful for finding area.
Is $CF$ perpendicular to $AB$? 
 [/INST] Using Heron's formula,
$$\begin{align}
 16 S_{\triangle ABC}^2 &= (a+b+c)(a+b-c)(b+c-a)(c+a-b) \\
&= ((a+b)^2 - c^2)(c^2 - (a-b)^2) \\
&= (a^2 + 2ab + b^2 - c^2)(c^2 + 2ab - a^2 - b^2) \\
&= (2ab + 4c^2)(2ab - 4c^2) \\
&= 4a^2b^2 - 16c^4
\end{align}$$
On the other hand,
$$ c^2 = a^2 + b^2 - 2ab\cos{C} = 5c^2 - 2ab\cos{C}$$
thus
$$ ab = \frac{2c^2}{\cos{C}} = 2c^2 \sec{C}.$$
$$ 16 S_{\triangle ABC}^2 = 4a^2b^2 - 16c^4 = 4\cdot 4c^4 \sec^2{C} - 16c^4 = 16c^2 \tan^2{C} $$
So
$$ S_{\triangle ABC} = c\tan{C}.$$
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