<s>[INST] Find the area of triangle, given an angle and the length of the segments cut by the projection of the incenter on the opposite side. 
In a triangle $ABC$, one of the angles (say $\widehat{C}$) equals $60^\circ$.
  Given that the incircle touches the opposite side ($AB$) in a point that splits it in two segments having length $a$ and $b$, what is the area of $ABC$?

 [/INST] 
Area $S_{ABC}$ in terms of $|AB|$ and altitude $h_C$ at the point $C$: 
\begin{align}
h_C&=(b+r\sqrt{3})\sin(60)
=\frac{\sqrt{3}}{2}(b+r\sqrt{3})
\\
S&=
\frac{\sqrt{3}}{4}
(a+r\sqrt{3})
(b+r\sqrt{3})
\\
&=
\frac{3}{4}\left(
\sqrt{3}r^2+(a+b)r
\right)
+\frac{\sqrt{3}}{4}ab
\quad(1)
\end{align}
Also, $S_{ABC}=S_{OAB}+S_{OBC}+S_{OCA}$:
\begin{align}
S_{ABC}&=
\frac{r}{2}
(2a+2b+2r\sqrt3)
\\
&=
\sqrt{3}r^2 + (a+b)r
\quad(2)
\end{align}
$\frac{4}{3} (1)-(2):$
\begin{align}
\frac{1}{3}S_{ABC}
&=
\frac{\sqrt3}{3}ab,
\end{align}
thus, $S_{ABC}=ab\sqrt{3}$. 
Edit:
This approach also provides a generalization for 
$\angle BAC =\alpha$. By replacing $r\sqrt{3}$
with $r/\tan(\alpha/2)$, similar steps result in
\begin{align}
S_{ABC}(\alpha)&=\frac{ab}{\tan{\alpha/2}}
=
\frac{\sin{\alpha}}{1-\cos{\alpha}}ab.
\end{align}
Particularly, $S(90°)=ab$.
</s>