Proving $ \frac{1}{c} = \frac{1}{a} + \frac{1}{b}$ in a geometric context 
Prove or disprove
$$
\frac{1}{c} = \frac{1}{a} + \frac{1}{b}.
$$
I have no idea where to start, but it must be a simple proof.
Trivia. This fact was used for determination of resistance of two parallel resistors in some circumstances long time ago.

In general, if the vertex angle is $2\theta$ and $OC$ is the angle bisector, since
$$\text{Area of triangle }OAB = \text{Area of triangle }OAC + \text{Area of triangle }OBC$$
we have
$$\dfrac{OA \cdot OB \cdot \sin(2\theta)}2 = \dfrac{OA \cdot OC \cdot \sin(\theta)}2 + \dfrac{OC \cdot OB \cdot \sin(\theta)}2$$
$$ab\sin(2\theta) = ac\sin(\theta) + cb\sin(\theta) \implies \dfrac{2\cos(\theta)}c = \dfrac1a + \dfrac1b$$
Taking $\theta=\pi/3$, we obtain what you want.
