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.
 A: 
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
$$Area =  \frac{r AB }{1- \frac r{AB} } = \frac {rAB^2}{AB -r}$$
A: 
$\triangle ABC = \frac{1}{2} AB \times h$
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 $\triangle ABC = \frac{AB^2 \, r}{AB - r}$
A: 
Let the midpoint of $AB$ be $O$. Draw $OE$ and $IF$. Label the sides $BC$, $CA$ and $AB$ as $a$, $b$ and $c$ respectively. Let the semiperimeter of $\triangle ABC$ be $s$. Points $O$, $I$ and $E$ will be collinear.
$OI=OE-IE=\frac{c}{2}-r$
$OF=OB-FB=\frac{c}{2}-\left(s-b\right)$
Applying Pythagorean theorem on $\triangle OIF$ gives,
$OF^2 + IF^2= OI^2$
$\Rightarrow  \{\frac{c}{2}-\left(s-b\right)\}^2 + r^2=(\frac{c}{2}-r)^2$
On simplification, this leads to, $r=\frac{\left(s-b\right)\left(s-a\right)}{c}$
But, $r= \frac{\Delta}{s}=\sqrt{\frac{\left(s-a\right)\left(s-b\right)\left(s-c\right)}{s}}$
$\Rightarrow  r^2=\frac{\left(s-a\right)\left(s-b\right)\left(s-c\right)}{s}$
$\Rightarrow  r^2=\frac{cr\left(s-c\right)}{s}$
$\Rightarrow    r=\frac{c\left(s-c\right)}{s}$
$\Rightarrow    rs=c\left(s-c\right)$
Hence, $\Delta=rs=c\left(s-c\right)$
$rs=c\left(s-c\right)$ $\Rightarrow  s=\frac{c^2}{c-r}$ $\Rightarrow  \left(s-c\right)=\frac{cr}{c-r}$
Thus, $\Delta=rs=c\left(s-c\right)=c\cdot\left(\frac{cr}{c-r}\right)=\boxed {\frac{c^2r}{c-r}}$
A: 

Lemma.
$$|\triangle ABC|=\frac12ch=\frac{c^2r}{c-c'} \qquad\qquad
\left(\;\leftarrow\quad\frac{h}{c}=\frac{h-2r}{c'}\;\right)\tag1$$

Therefore, to get the formula derived in other answers, "all we have to do" is show that the particular circumstances of the original problem imply that $c'=r$. We can do this by focusing on the tangential trapezoid $\square ABB'A'$ and introducing a concentric semicircle.

From the three shaded right triangles in the figure (two determined by angle-chasing, one via Thales' Theorem), we observe three instances of the geometric mean construction, whence
$$r^2 = pp' = qq' = (p-r)(q-r) \tag{2}$$
Then we have
$$(p-r)(q-r)=r^2\quad\to\quad pq=r(p+q) \quad\to\quad \frac1r=\frac1p+\frac1q =\frac{p'+q'}{r^2} \tag3$$
Thus, $c' := p'+q' = r$, as desired. $\square$
