Solution. Let the radius of $k_{1}$ be $r$, and of $k_{2}$ be $t$, and let the common tangent of $k_{1}$ and $k_{2}$ at point $A$ intersect line $B C$ at $G$. Note that $\overline{A G}=\overline{B G}=\overline{C G}$, as tangent segments. We also have

$$
\measuredangle A P B=180^{\circ}-\measuredangle A G B=\measuredangle A G C
$$

Thus, the isosceles triangles $\triangle A P B$ and $\triangle A G C$ are similar. Since we know that $\overline{A C}=\frac{\overline{A B}}{2}$, it follows that $r=\overline{A P}=2 \overline{A G}$, hence $\overline{B C}=r$. Let the line parallel to $B C$ through $R$ intersect $P B$ at point $F$. Then, by the Pythagorean theorem for the right triangle $\triangle P R F$, we have

$$
(r+t)^{2}=(r-t)^{2}+\overline{R F}^{2} \text {, i.e. } \overline{R F}=2 \sqrt{r t}
$$

From this, we have $\overline{B C}=2 \sqrt{r t}$. Now we have $r=2 \sqrt{r t}$. This gives us $r=4 t$. From the similarity of the right triangles $\triangle R E C$ and $\triangle P R F$, we have

$$
\frac{\overline{P R}}{\overline{P F}}=\frac{\overline{R E}}{\overline{R C}}
$$

i.e.,

$$
\frac{r+t}{r-t}=\frac{t+\overline{D E}}{t}
$$

from which we get $\frac{5 t}{3 t}=1+\frac{\overline{D E}}{t}, \overline{D E}=\frac{2}{3} t$. Now we have: $\frac{\overline{B C}}{\overline{D E}}=\frac{r}{\frac{2}{3} t}=\frac{12 t}{2 t}=6$.
2. Given are two circles $k_{1}$ and $k_{2}$ with centers $P$ and $R$ respectively, which touch each other externally at point $A$. Let $p$ be their common tangent, which does not pass through $A$, and touches $k_{1}$ and $k_{2}$ at points $B$ and $C$, respectively. The line $P R$ intersects the line $B C$ at point $E$, and the circle $k_{2}$ at points $A$ and $D$. If $\overline{A C}=\frac{\overline{A B}}{2}$, find the ratio $\frac{\overline{B C}}{\overline{D E}}$.