29. Side $B C$ of triangle $A B C$ is equal to $a$, the radius of the inscribed circle is $r$. Find the area of the triangle if the inscribed circle touches the circle constructed on $B C$ as its diameter.
29. Let $O$ be the center of the inscribed circle, $M$ the midpoint of $BC$, and $K, L, M$ the points of tangency of the inscribed circle with the sides $AC$, $AB$, and $BC$ of the triangle, respectively. Denote $|AK|=|AL|=x$, $|CK|=$ $=|CN|=y$, $|BL|=|BN|=z$, and $y+z=a$. By the condition, $|OM|=$ $=\frac{a}{2}-r$. Therefore, $|NM|=\sqrt{|OM|^2-|ON|^2}=\sqrt{\frac{a^2}{4}-ar}$, and one of the segments $y$ or $z$ equals $\frac{a}{2}-\sqrt{\frac{a^2}{4}-ar}$, while the other equals $\frac{a}{2}+\sqrt{\frac{a^2}{4}-ar}$, and $yz=ar$. Equating the expressions for the area of the triangle using Heron's formula and $S=pr$:

$$
\sqrt{(x+y+z)xyz}=(x+y+z)r \Rightarrow xar=(x+a)r^2 \Rightarrow x=\frac{ar}{a-r}
$$

Thus, $S=\left(\frac{ar}{a-r}+a\right)r=\frac{a^2r}{a-r}$.