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.

Here's a trigonometric approach (which is in some way equivalent to other answers here, since one could start from one of these relations to prove the others).  We will start with a general angle  $ \ O \ $ with $ \ m(\angle O ) \ = \ \theta \ $ , as user17762 does, and bisect it. Upon declaring $ \ m(\angle OCA ) \ = \ \phi \ $ , the other angle measures shown in the diagram follow directly.
 
From the Law of Sines, we have in $ \ \Delta OAC \ $ , 
$$ \frac{\sin \phi}{a} \ \ = \ \ \frac{\sin(180º \ - \ \theta \ - \ \phi)}{c} \ \ = \ \ \frac{\sin( \theta \ + \ \phi)}{c} \ \ , $$
upon applying the relation that the sine of an angle equals the sine of the supplementary angle.  In $ \ \Delta OBC \ $ , we find
$$ \frac{\sin (180º \ - \ \phi)}{b} \ \ = \ \ \frac{\sin  \phi}{b} \ \ = \ \ \frac{\sin( \phi \ - \ \theta)}{c} \ \ , $$
again using the "supplementary sine" relation.
We may then add these two equations together and apply the "angle-addition/difference" formulas for sine to obtain
$$ \frac{\sin \phi}{a} \ + \ \frac{\sin \phi}{b} \ \ = \ \ \frac{\sin( \theta \ + \ \phi)}{c} \ + \ \frac{\sin( \phi \ - \ \theta)}{c}   $$
$$  = \ \ \frac{\sin  \theta \ \cos \phi \ + \ \cos \theta \ \sin \phi}{c} \ + \ \frac{\sin  \phi \ \cos \theta \ - \ \cos \phi \ \sin \theta}{c} \ \ = \ \ \frac{2 \ \sin  \phi \ \cos \theta }{c} \ \ . $$
Dividing through by $ \ \sin \phi \ $ produces
$$ \frac{1}{a} \ + \ \frac{1}{b} \ \ = \ \ 2 \ \cos \theta \ \cdot \ \frac{1}{c} \ \ ,  $$
which differs from the desired equation only by a multiplicative constant dependent upon the choice of $ \ m(\angle O ) \ $ . (This means that employing the diagram as a hodographic device as VividD describes in the Trivial note, any angle will do after dividing out this constant factor.)  For $ \ \theta \ = \ \frac{\pi}{3} \ $ , we have exactly the relation sought.
