Find the area of a triangle given the radius of its incircle and a tangential point A friend gave me recently the following interesting problem and I would like to share a couple of solutions. Any additional contributions are welcome.
A triangle $\vartriangle ABC$ is given and we know the radius $r$ of its incircle $(O, r)$. Let $D$ be the tangential point of the incircle on $AC$ which partitions $AC$ into $AD=\alpha$ and $DC=\beta$. Determine the area of $\vartriangle ABC$ as a function of $\alpha$, $\beta$ and $r$.

 A: Let $s,S,R$ be semiperimeter, area and $B$-exradius of triangle $ABC$ respectively. It is well-known that $rR=\alpha\beta$ and $rs=S=R(s-\alpha-\beta)$ (see below). From the latter we get $s=\frac{R(\alpha+\beta)}{R-r}$ which leads to $$S=rs=\frac{rR(\alpha+\beta)}{R-r}=\frac{r\alpha\beta(\alpha+\beta)}{rR-r^2}=\frac{r\alpha\beta(\alpha+\beta)}{\alpha\beta-r^2}.$$
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For a proof that $rR=\alpha\beta$ denote the $B$-excenter by $J$. Let $X$ be a point of tangency of $B$-excircle with side $AC$. It is well-known that $AX=CD$. Note that triangles $JAX$ and $COD$ are similar as their angles are equal. Thus $$\frac{R}{\beta}=\frac{JX}{AX}=\frac{CD}{OD}=\frac{\alpha}{r}$$ yielding $rR=\alpha\beta$.
Now we'll prove that $S=R(s-\alpha-\beta)$. We'll use a notation $[\mathcal F]$ which denotes area of $\mathcal F$. We have $$S = [ABC] = [ABJ] + [CBJ] - [BCJ] = \frac{AB \cdot R}{2} + \frac{BC \cdot R}{2} - \frac{AC\cdot R}{2} = \frac{AB+BC-AC}{2} \cdot R = \left(\frac{AB+BC+AC}{2} - AC\right) \cdot R = \left(s-(\alpha+\beta)\right) \cdot R = R(s-\alpha-\beta)$$
A: Solution 1: Draw $AO$, $AO$ and $CO$. We know that these are the bisectors of $\angle BAC$, $\angle ABC$ and $\angle BCA$ respectively. 
Let $\angle OAC = \frac{1}{2} \angle BAC =: \phi_A$ and $\angle OBE = \frac{1}{2} \angle ABC =: \phi_B$ and $\angle OCA = \frac{1}{2}\angle BCA =: \phi_C$.
We have that $\angle ODA$ is a right angle (and so is $\angle ODC$). Thus $\tan \phi_A = \frac{r}{\alpha}$ and $\tan \phi_C = \frac{r}{\beta}$. Therefore $\angle AOD = \frac{\pi}{2} - \phi_A$. 
The two right-angled triangles $\vartriangle ADO$ and $\vartriangle AEO$ are equal (indeed, they share a common side $AO$ and $OE=OD=r$). Thus, $\angle EOA = \angle DOA = \frac{\pi}{2} - \phi_A$. 
Similarly, $\angle COD = \frac{\pi}{2}-\phi_C$ and since $\vartriangle COD = \vartriangle COF$ it is $\angle COF = \angle COD =  \frac{\pi}{2}-\phi_C$.
Using the fact that 
$$
\angle AOD + \angle DOC + \angle COF + \angle FOE + \angle EOA = 2\pi,\\
2(\frac{\pi}{2}-\phi_A) + 2(\frac{\pi}{2}-\phi_C) + \angle FOE = 2\pi,\\
\angle FOE = 2(\phi_A + \phi_C).
$$
Because of the equality of $\vartriangle BEO$ and $\vartriangle BFO$ we have $\angle EOB = \frac{\angle FOE}{2} = \phi_A + \phi_C$, so
$$
\tan \angle EOB = \frac{BE}{r},\\
BE = r \tan(\phi_A + \phi_C) = r\frac{\tan\phi_A + \tan\phi_C}{1-\tan\phi_A\tan\phi_C}=\\
=r\frac{\frac{r}{\alpha}+\frac{r}{\beta}}{1-\frac{r}{\alpha}\frac{r}{\beta}}=r\frac{\frac{r(\alpha+\beta)}{\alpha\beta}}{\frac{\alpha\beta-r^2}{\alpha\beta}}=r^2\frac{\alpha+\beta}{\alpha\beta-r^2}
$$
From the aforementioned triangle equalities we have $AE = \alpha$, $CF = \beta$ and $BF = BE$, therefore, the semiperimeter of $\vartriangle ABC$ is
$$
\tau = \alpha + \beta + BE = \alpha + \beta + r^2\frac{\alpha+\beta}{\alpha\beta-r^2}=\frac{\alpha\beta(\alpha+\beta)}{\alpha\beta-r^2},
$$
and the area of the triangle is 
$$S = r\tau = r\frac{\alpha\beta(\alpha+\beta)}{\alpha\beta-r^2}$$
Solution 2: This is a solution without trigonometry and is less elegant. Since $\angle BOE = \phi_A + \phi_C$ there is a point $M$ on $BE$ so that $\angle MOE = \phi_A$ and $\angle MOB = \phi_C$ as in the following figure:

It is easy to see that the two triangles $\vartriangle MEO$ and $\vartriangle OEA$ are similar (they have all their angles equal to one another), so, 
$$
\frac{EM}{EO}=\frac{EO}{EA},\\
EM = \frac{r^2}{\alpha}.
$$
We now need to determine $BM$. What is the same, $\vartriangle BMO$ and $\vartriangle BOC$ are similar (all angles equal to one another). Note that (Pythagorean theorem on $\vartriangle BEO$)
$$BO=\sqrt{BE^2 + r^2} = \sqrt{(BM+ME)^2 + r^2},$$  
and (Pythagorean theorem on $\vartriangle OCD$)
$$OC = \sqrt{r^2 + \beta^2}.$$ 
We also have 
$$
MO = \sqrt{r^2 + ME^2} = \sqrt{r^2 + \frac{r^4}{\alpha^2}} = r\sqrt{1+(r/a)^2}
$$
Using the fact that $\vartriangle BMO$ and $\vartriangle BOC$ are similar (all angles equal to one another), we arrive at
$$
\frac{BM}{BO}=\frac{MO}{OC},\\
\frac{BM}{\sqrt{(BM+\frac{r^2}{\alpha^2})^2 + r^2}}=\frac{r\sqrt{1+\frac{r^2}{\alpha^2}}}{\sqrt{r^2 + \beta^2}},\\
\frac{BM^2}{(BM+\frac{r^2}{\alpha^2})^2 + r^2}=\frac{r^2 (1+\frac{r^2}{\alpha^2})}{r^2 + \beta^2},
$$
which is a quadratic equation one can solve to derive a formula for $BM$ (the non-positive solution should be discarded) in terms of $r$, $\alpha$ and $\beta$. Having determined $BM$ and $ME$ we can easily determine the semiperimeter of $\vartriangle ABC$, $\tau = \alpha + \beta + BM + ME$ and the area of $\vartriangle ABC$ is again $S = r\tau$.
A: If $I$ is the incenter of $ABC$ and $I_A,I_B,I_C$ are the projections of the incenter on the sides $BC,AC,AB$ (sorry, but to call the incenter $O$ gives me some itch), we have:
$$ AI_B=AI_C=\frac{b+c}{2}\,\quad BI_A=BI_C=\frac{a+c}{2},\quad  CI_A=CI_B=\frac{a+b}{2} $$
and $2\Delta = r(a+b+c)$, or $2\Delta = r(AI_B+BI_C+CI_A) $, so we just need to find $BI_C$ in terms of $r,AI_B,CI_B$. Pretty easy:
$$ BI_C = \frac{r}{\tan\frac{\widehat{B}}{2}} = r\tan\frac{\widehat{A}+\widehat{C}}{2}=r\cdot\frac{\tan\frac{\widehat{A}}{2}+\tan\frac{\widehat{C}}{2}}{1-\tan\frac{\widehat{A}}{2}\tan\frac{\widehat{C}}{2}}$$
hence:
$$ BI_C = r^2\cdot\frac{AI_B+CI_B}{AI_B\cdot CI_B-r^2} $$
and:
$$ 2\Delta = \frac{(AI_B+CI_B)\,AI_B\,CI_B\,r}{AI_B\,CI_B-r^2}.$$
A: I'll use the more-traditional notation with tangent points $D$, $E$, $F$ opposite respective vertices $A$, $B$, $C$. I'll also take $\alpha$, $\beta$, $\gamma$ (the first two given, the last unknown) to be the lengths of the tangent segments from vertices $A$, $B$, $C$. Finally, I'll reduce some notational clutter by writing $A_2$, $B_2$, $C_2$ for half-angles $A/2$, $B/2$, $C/2$. (So, $A_2 + B_2 + C_2 = 90^\circ$.)
Since $\alpha + \beta + \gamma$ is the semi-perimeter, we know the area of the triangle is given by
$$|\triangle ABC| = r (\alpha + \beta + \gamma) \tag{$\star$}$$
Our task is to replace $\gamma$. Clearly, 
$$r = \alpha \tan A_2 = \beta \tan B_2 = \gamma \tan C_2$$
But, $$\tan C_2 = \tan(90^\circ - A_2 - B_2) = \cot(A_2 + B_2) = \frac{\cot A_2 \cot B_2 - 1}{\cot A_2 + \cot B_2} = \frac{\frac{\alpha}{r}\frac{\beta}{r}-1}{\frac{\alpha}{r}+\frac{\beta}{r}} = \frac{\alpha\beta-r^2}{r(\alpha+\beta)}$$
Therefore,
$$\gamma = \frac{r}{\tan C_2} = \frac{r^2(\alpha+\beta)}{\alpha\beta-r^2}$$
so that, by $(\star)$,
$$|\triangle ABC| = r\left(\alpha + \beta + \frac{r^2(\alpha+\beta)}{\alpha\beta-r^2}\right)= \frac{r\,\alpha\beta\,(\alpha+\beta)}{\alpha \beta - r^2}$$

Observation: Recall that the circle with diameter $\overline{AB}$ meets the perpendicular at $F$ in points $P$ and $Q$ such that $|\overline{AP}|^2 = |\overline{AQ}|^2 = |\overline{AF}||\overline{BF}| = \alpha \beta$. This says exactly that the "power" of point $F$ with respect to that circle is $\alpha\beta$. Moreover, since that perpendicular passes through $I$ (which, we'll say, lies between $F$ and $P$), we have 
$$|\overline{IP}||\overline{IQ}| = \left(|\overline{AP}| - |\overline{AI}|\right)\left(|\overline{AQ}| + |\overline{AI}|\right) = \left(\sqrt{\alpha\beta} - r\right)\left(\sqrt{\alpha\beta} + r\right) = \alpha \beta - r^2$$
Thus, the denominator of the area formula is the power of $I$ with respect to that circle. This seems significant ... or coincidental. (It may also be worth noting that that circle contains the feet of the altitudes from $A$ and $B$.) 
