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.

I'm not sure what the protocol is for adding a separate answer when it uses a different approach.  If there is an issue with doing this, I can move this to my existing entry.

Here is a method using analytic geometry. I have changed the labeling of two points, since I wish to place the origin of the coordinate system at a different point from $ \ O \ $ in the original diagram.  For convenience, I will place the apex of $ \ \Delta ACB \ $ at $ \ (0, c) \ $ , with line $ \ OC \ $ on the $ \ y-$ axis, and lines $ \ AC \ $ and $ \ BC \ $ making equal angles to the $ \ y-$ axis.  The equations for the lines along which the sides of the triangle lie are  $ \ AC : \ y \ = \ c \ + \ mx \ , \ BC : \ y \ = \ c \ - \ mx \ , \ AB : \ y \ = \ kx \ , $ with $ \ k \ < \ m \ $ . (This condition proves to be important.)
We find the locations of the intersection points $ \ A \ $ and $ \ B \ $ from
$$ k \ x_A \ = \ c \ + \ m \ x_A \ \ \Rightarrow \ \ x_A \ = \ \frac{c}{k - m} \ \ , \ \ y_A \ = \ c \ + \  \frac{mc}{k - m} \ \ , $$
$$ k \ x_B \ = \ c \ - \ m \ x_B \ \ \Rightarrow \ \ x_B \ = \ \frac{c}{k + m} \ \ , \ \ y_A \ = \ c \ - \  \frac{mc}{k + m} \ \ . $$
[We will see shortly that it is preferable to express the $ \ y-$ coordinates in this manner.]
The squares of the lengths of sides $ \ AC \ $ and $ \ BC \ $ are then obtained 
$$ a^2 \ = \ \left( \ \frac{c}{k - m} \ - \ 0 \ \right)^2 \ + \ \left( \ c \ - \ \left[ \ c \ + \frac{mc}{k - m} \ \right] \ \right)^2 \ = \ \ c^2 \ \left[ \ \frac{1 + m^2}{(k - m)^2} \ \right] \ \ ,  $$
$$ b^2 \ = \ \left( \ \frac{c}{k + m} \ - \ 0 \ \right)^2 \ + \ \left( \ c \ - \ \left[ \ c \ - \frac{mc}{k + m} \ \right] \ \right)^2 \ = \ \ c^2 \ \left[ \ \frac{1 + m^2}{(k + m)^2} \ \right] \ \ .  $$
We can write directly that $ \ \frac{c}{b} \ = \ \frac{k + m}{\sqrt{1 + m^2}} \ $ .  However, since we are constructing ratios of positive lengths, we must write $ \ \frac{c}{a} \ = \ \frac{\sqrt{(k - m)^2}}{\sqrt{1 + m^2}} \ = \ \frac{|k - m|}{\sqrt{1 + m^2}} \ = \ \frac{m - k}{\sqrt{1 + m^2}} \ $ .  (I found this  to be a little trap for the unwary... I'll also note at this point that forming these ratios now links this approach to those using trigonometry in some fashion.)
Summing the two ratios produces
$$  \frac{c}{a} \ + \ \frac{c}{b} \ = \ \frac{m - k}{\sqrt{1 + m^2}} \ + \  \frac{k + m}{\sqrt{1 + m^2}} \ = \ \frac{2m }{\sqrt{1 + m^2}} \ \ . $$
For the special case of $ \ m(\angle ACO) \ = \ m(\angle BCO) \ = \ \frac{\pi}{3} \ $ , we have the slope $ \ m \ = \ \frac{1}{\sqrt{3}} \ $ .  So our general relation reduces to
$$  \frac{c}{a} \ + \ \frac{c}{b} \ = \ \frac{2 \ \cdot \ \frac{1}{\sqrt{3}} }{\sqrt{1 + \frac{1}{3}}} \ = \ \frac{\frac{2}{\sqrt{3}}}{\sqrt{\frac{4}{3}}} \ = \ 1 \ \ . $$
