78. Two circles touch each other internally at point A. A tangent to the smaller circle intersects the larger circle at points $B$ and $C$. Find the geometric locus of the centers of the circles inscribed in triangles $A B C$.
78. Let the radii of the given circles be denoted by \( R \) and \( r (R \geqslant r) \), the point of tangency of the chord \( BC \) with the smaller circle be \( D \); let \( K \) and \( L \) be the points of intersection of the chords \( AC \) and \( AB \) with the smaller circle, and finally, let \( O \) be the center of the circle inscribed in \( \triangle ABC \). Since the angular measures of the arcs \( AK \) and \( AC \) are the same, then \( |AK| = rx, |AC| = Rx \); from this we get \( |DC|^2 = |AC| \cdot |CK| = (R-r)Rx^2 \). Similarly, \( |AB| = Ry, \quad |DB|^2 = (R-r)Ry^2 \), hence \( \frac{|CD|}{|DB|} = \frac{x}{y} = \frac{|AC|}{|AB|} \), i.e., \( AD \) is the bisector of angle \( BAC \). Further, we have: \( \frac{|AO|}{|OD|} = \frac{|AC|}{|CD|} = \frac{Rx}{\sqrt{(R-r)R}x} = \sqrt{\frac{R}{R-r}} \). Thus, the required geometric place at the same point \( A \) with radius \( \rho = r \frac{|AO|}{|AD|} = \frac{r \sqrt{R}}{\sqrt{R} + \sqrt{R-r}} \).