# 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.

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 $$Area = \frac{r AB }{1- \frac r{AB} } = \frac {rAB^2}{AB -r}$$

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, $$\Delta=rs=c\left(s-c\right)=c\cdot\left(\frac{cr}{c-r}\right)=\boxed {\frac{c^2r}{c-r}}$$

$$\triangle ABC = \frac{1}{2} AB \times h$$

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 $$\triangle ABC = \frac{AB^2 \, r}{AB - r}$$

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. $$\square$$