Triangle $\triangle ABC$ with incenter $I$ and inradius $r$ has the following property: the incircle is tangent to the semicircle with diameter $AB$ and is within the semicircle. Find the area of $\triangle ABC$ as a function of the side $AB$ and the inradius $r$.
So far I've noticed that $\angle AEF=\angle BEF=\frac{\pi}{4}$ but I don't think that that's useful in any way.
Best Answer
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 $$Area = \frac{r AB }{1- \frac r{AB} } = \frac {rAB^2}{AB -r}$$