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 $AB$?
Best Answer
The law of cosines says:
$a^2 + b^2 - 2ab\cos(C) = c^2$
One also has the following area formula for a triangle:
$\text{Area}(ABC) = \frac{1}{2}ab\sin(C)$
Can you figure it out from here?