4. Let $A B C D$ be a rectangle and $E$ be a point on segment $A D$. We are given that quadrilateral $B C D E$ has an inscribed circle $\omega_{1}$ that is tangent to $B E$ at $T$. If the incircle $\omega_{2}$ of $A B E$ is also tangent to $B E$ at $T$, then find the ratio of the radius of $\omega_{1}$ to the radius of $\omega_{2}$.
Answer:
$$
\frac{3+\sqrt{5}}{2}
$$

Solution: Let $\omega_{1}$ be tangent to $A D, B C$ at $R, S$ and $\omega_{2}$ be tangent to $A D, A B$ at $X, Y$. Let $A X=A Y=r, E X=E T=E R=a, B Y=B T=B S=b$. Then noting that $R S \| C D$, we see that $A B S R$ is a rectangle, so $r+2 a=b$. Therefore $A E=a+r, A B=b+r=2(a+r)$, and so $B E=(a+r) \sqrt{5}$. On the other hand, $B E=b+a=r+3 a$. This implies that $a=\frac{1+\sqrt{5}}{2} r$. The desired ratio is then $\frac{R S}{2 A Y}=\frac{A B}{2 r}=\frac{a+r}{r}=\frac{3+\sqrt{5}}{2}$.