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 $r$.


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 $$S = [ABC] = [ABJ] + [CBJ] - [BCJ] = \frac{AB \cdot R}{2} + \frac{BC \cdot R}{2} - \frac{AC\cdot R}{2} = \frac{AB+BC-AC}{2} \cdot R = \left(\frac{AB+BC+AC}{2} - AC\right) \cdot R = \left(s-(\alpha+\beta)\right) \cdot R = R(s-\alpha-\beta)$$
