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


$\triangle ABC = \frac{1}{2} AB \times h$
Let's get onto finding $h$ and then we are done.
We know $R = \frac{AB}{2} \,( R$ being the radius of the semicircle).
$AB = h \cot A + h \cot B \implies h = \frac{AB}{\cot A + \cot B}$
Now, $\cot \frac{A}{2} = \frac{AF}{IF} = \frac{R+x}{r}$
Similarly, $\cot \frac{B}{2} = \frac{R - x}{r}$
Using the identity $\, \cot 2 \alpha  = \frac{1}{2} (\cot \alpha - \tan \alpha), \,$ we have,
$2 \cot A = \frac{(R+x)^2 - r^2}{r(R+x)} \,, \, 2 \cot B = \frac{(R-x)^2 - r^2}{r(R-x)}$
$ 2(\cot A + \cot B) = \frac{(R^2-x^2)(R+x) + (R^2-x^2)(R -x) - 2 r^2R}{r(R^2-x^2)}$
Also, $R^2 - x^2 = R^2 - ((R-r)^2 - r^2) = 2Rr$
So, $ 2(\cot A + \cot B) = \frac{4R - 2r}{2r} = \frac{AB - r}{r}$
And hence $\frac{h}{2} = \frac{AB \, r}{AB - r} \, , \,$ Area $\triangle ABC = \frac{AB^2 \, r}{AB - r}$
