81. Given a circle of radius $R$ with center $O$. Two other circles touch the given one internally and intersect at points $A$ and $B$. Find the sum of the radii of the two latter circles, given that $\widehat{O A B}=90^{\circ}$.
81. Let (Fig. 17) $O_{1}$ and $O_{2}$ be the centers of intersecting circles. Denote their radii by $x$ and $y$, $|O A|=a$. Since triangles $A O O_{1}$ and $A O O_{2}$, as follows from the condition, are equal in area, expressing their areas using Heron's formula, considering that $\left|O_{1} A\right|=x$, $\left|0 O_{1}\right|=R-x$,
$\left|O_{2} A\right|=y$, $0 O_{2}=R-y$, we find

$$
\begin{aligned}
& \sqrt{\frac{(R+a)}{2} \cdot \frac{(R-a)}{2} \cdot \frac{(R+a-2 x)}{2} \cdot \frac{(a+2 x-R)}{2}}= \\
& \quad=\sqrt{\frac{(R+a)}{2} \cdot \frac{(R-a)}{2} \cdot \frac{(R+a-2 y)}{2} \cdot \frac{(a+2 y-R)}{2}} \Rightarrow \\
& \Rightarrow a^{2}-(R-2 x)^{2}=a^{2}-(R-2 y)^{2}
\end{aligned}
$$

from which, since $x \neq y$, we get $x+y=R$.