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 $ABC$?

 A: Let the triangle be $\triangle ABC$. Let $\angle A=60^{\circ}$. Let the incircle touch the triangle on the points $A_1,B_1,C_1$ (opposite to $A,B,C$, respectively). Let $A_1B=a,\,A_1C=b$. Then $C_1B=a,\, B_1C=b$. Let $AC_1=AB_1=x$. Then by the Law of cosines:
$$(a+b)^2=(x+a)^2+(x+b)^2-2(x+a)(x+b)\cos 60^{\circ}$$
Solve this quadratic equation. The area is $\frac{1}{2}(x+a)(x+b)\sin 60^{\circ}$.
A: An alternative approach: let we just find the inradius $r$. Assuming that $\widehat{C}=60^\circ$, $I_C$ is the projection of the incenter on the $AB$-side, $AI_C=a,BI_C=b$, we have:
$$\widehat{A}=2\arctan\frac{r}{a},\qquad \widehat{B}=2\arctan\frac{r}{b}\tag{1}$$
hence:
$$ \frac{\pi}{3} = \arctan\frac{r}{a}+\arctan\frac{r}{b}\tag{2} $$
must hold, from which:
$$ \sqrt{3} = \frac{\frac{r}{a}+\frac{r}{b}}{1-\frac{r^2}{ab}}\tag{3}$$
follows. Since:
$$ \Delta = \frac{1}{2}(a+\sqrt{3}\,r)(b+\sqrt{3}\,r)\sin 60^\circ =\frac{\sqrt{3}}{4}ab\left(1+\sqrt{3}\left(\frac{r}{a}+\frac{r}{b}\right)+3\frac{r^2}{ab}\right)\tag{4}$$
by exploiting $(3)$ we have:

$$ \Delta = \color{red}{\sqrt{3}\,ab},\tag{5}$$

pretty nice, don't you think?
A: 
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, $S(90°)=ab$.
