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.
 A: 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.
A: 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.
A: Consider the areas of triangles. Let $[ABC]$ be the area of a triangle $ABC$.
Since $[OAB]=[OAC]+[OBC]$, one has
$$\frac{1}{2}ab\sin120^\circ=\frac{1}{2}ac\sin60^\circ+\frac 12bc\sin60^\circ.$$
This leads
$$ab=ac+bc.$$
Now divide the both sides by $abc$.
A: Here's a geometric argument. 
Construct the equilateral triangles $ODC,OEC$, like in the figure. Then you have
$$ \frac{c}{a}+\frac{c}{b} = \frac{OD}{OA}+\frac{OE}{OB} = \frac{BC}{AB}+\frac{AC}{AB} = 1$$
(note that $DC || OB$ and $CE || OA$)
A: 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
$$  \frac{c}{a} \ + \ \frac{c}{b} \ = \ \frac{2 \ \cdot \ \frac{1}{\sqrt{3}} }{\sqrt{1 + \frac{1}{3}}} \ = \ \frac{\frac{2}{\sqrt{3}}}{\sqrt{\frac{4}{3}}} \ = \ 1 \ \ . $$
A: [I suspected that there was a similarity argument (apart from the fine one Beni Bogosel has provided) using the diagram in the way I'd been looking at it.  After trying a few horrible ones, this one came to me after I gave up -- there's a lesson in this somewhere...]

We start with the diagram with $ \ \overline{OC} \ $ placed on the $ \ y-$ axis and lines $ \ \overline{OA} \ \ \text{and} \ \ \overline{OB} \ $ arranged so that $ \ m(\angle AOC) \ = \ m(\angle BOC) \ = \ \theta \ $ . The length   $ \  OC  \ $ is $ \ c \ $ ; if we label the intersections of $ \ \overline{OA} \ \ \text{and} \ \ \overline{OB} \ $ with the $ \ x-$ axis (passing through $ \ C \ $ ) as points $ \ P \ \ \text{and} \ \ Q \ $ , respectively, then the lengths   $ \  PC  \ \ \text{and} \ \  QC  \ $ are both $ \ c \ \tan \theta \ $ . (We will only need one of these.)
As elsewhere among the posted answers, the oblique line $  \ \overline{AB} \ $ produces the lengths $ \ OA \ \ \text{and} \ \ OB \ $ , which are $ \ a \ \ \text{and} \ \ b \ $ , respectively.  If we add segments $ \ \overline{AA \ '} \ \ \text{and} \ \ \overline{BB \ '} \ $ parallel to $ \ \overline{PQ} \ $ , we have the sets of similar (right) triangles $ \ \Delta OBB \ ' \ \sim \ \Delta OQC \  \sim \ \Delta OAA \ ' \   $ and $ \ \Delta CBB \ ' \ \sim \ \Delta CAA \ ' \ $ , since $ \ \angle ACA \ ' \ \ \text{and} \ \ \angle BCB \ ' $ are "vertical angles".
By similarity (or trigonometric ratio definitions), we have segment lengths $ \ BB \ ' \ = \ b \ \sin \theta \ $ , $ \ OB \ ' \ = \ b \ \cos \theta \ $ , $ \ AA \ ' \ = \ a \ \sin \theta \ $ , and $ \ OA \ ' \ = \ a \ \cos \theta \ $ . Thus, $  \ CB \ ' \ = \ c \ - \ b \ \cos \theta \ $ and $  \ CA \ ' \ = \  a \ \cos \theta \ - \ c \ $ .
Now, by similarity between the two newly added right triangles, we also have 
$$ \frac{CB \ '}{BB \ '} \ = \ \frac{CA \ '}{AA \ '} \ \ \Rightarrow \ \ \frac{c \ - \ b \ \cos \theta}{b \ \sin \theta} \ = \ \frac{a \ \cos \theta \ - \ c}{a \ \sin \theta} $$ 
$$ \Rightarrow \ \ \frac{c }{b } \ - \ \cos \theta \ = \ \cos \theta \ - \ \frac{ c}{a } \ \ \Rightarrow \ \ \frac{ c}{a } \ + \ \frac{c }{b } \  = \ 2 \ \cos \theta  \ \ . $$ 
As before, with $ \ \theta \ = \ \frac{\pi}{3} \ $ , we obtain the relation sought after a simple re-arrangement.
A: This is an outline of my proof. Note that
$\alpha + \beta = 60^{\circ}$,
$\bigtriangleup OAC \sim \, \bigtriangleup BPC$, and
$\bigtriangleup OBC \sim \, \bigtriangleup APC$.
I got this picture by drawing the excircle of $\bigtriangleup OAB$ and extending line $OC$ until it intersected the excircle at $P$. But if you are careful with your construction, you can avoid using the excircle.

