Area of triangle with incircle tangent to semicircle whose diameter is on one triangle side

The area of the triangle ABC is

\begin{align} Area & = \frac12 r(AB + BC +CA) = \frac12 rAB \frac{\sin A+\sin B +\sin C}{\sin C} \\ &= \frac12 rAB \frac{4\cos \frac A2 \cos \frac B2\cos\frac C2 }{2\sin \frac C2\cos \frac C2} = rAB \frac{\cos \frac A2 \cos \frac B2 }{\cos \frac{A+B}2} = \frac{rAB }{1- \tan\frac A2\tan\frac B2 } \end{align}

Apply the Pythagorean theorem to the right triangle IFO, with $AB = 2R$

$$d^2 = (R-r)^2 - r^2 = R^2 - 2rR$$ $$\tan\frac A2\tan\frac B2= \frac r{R+d} \frac r{R-d}=\frac {r^2}{R^2-d^2} = \frac {r^2}{2rR} = \frac r{AB} $$ Thus $$Area = \frac{r AB }{1- \frac r{AB} } = \frac {rAB^2}{AB -r}$$