78. Find the perimeter of triangle $ABC$, if $|BC|=$ $=a$ and the segment of the line tangent to the inscribed circle and parallel to $BC$, contained within the triangle, is equal to $b$.
78. Let $KM$ be a segment parallel to $BC$, and $N$ and $L$ be the points of tangency of the inscribed circle with sides $AC$ and $BC$. As is known (see problem 18, section I), $|AN|=|AL|=p-a$, where $p$ is the semiperimeter of $\triangle ABC$. On the other hand, $|AN|=|AL|$ is the semiperimeter of $\triangle AKM$, similar to $\triangle ABC$. Therefore, $\frac{p-a}{p}=\frac{b}{a}$, $p=\frac{a^2}{a-b}$.

Answer: $\frac{2a^2}{a-b}$.