Let $\theta=\angle B A C$, and $\cos \theta=\frac{7}{9}$ implies $\cos \frac{\theta}{2}=\sqrt{\frac{1+\frac{7}{9}}{2}}=\frac{2 \sqrt{2}}{3} ; \sin \frac{\theta}{2}=\frac{1}{3} ; B C=$ $\sqrt{4+25-2(2)(5) \frac{7}{9}}=\frac{11}{3}$. Let $O_{1}$ be the excircle of $\triangle A B C$ tangent to lines $A B$ and $A C$, and let $r_{1}$ be its radius; let $O_{1}$ be tangent to line $A B$ at point $P_{1}$. Then $A P_{1}=\frac{A B+B C+C A}{2}$ and $\frac{r_{1}}{A P_{1}}=\tan \frac{\theta}{2}=$ $\frac{1}{2 \sqrt{2}} \Longrightarrow r_{1}=\frac{16}{3 \cdot 2 \sqrt{2}}$. Let $O_{n}$ be a circle tangent to $O_{n-1}$ and the lines $A B$ and $A C$, and let $r_{n}$ be its radius; let $O_{n}$ be tangent to line $A B$ at point $P_{n}$. Then $\frac{O_{n} P_{n}}{A O_{n}}=\sin \frac{\theta}{2}=\frac{1}{3}$; since $\triangle A P_{n} O_{n} \sim$ $\triangle A P_{n-1} O_{n-1}$ and $O_{n} O_{n-1}=r_{n}+r_{n-1}$, we have $\frac{1}{3}=\frac{O_{n} P_{n}}{A O_{n}}=\frac{r_{n}}{A O_{n-1}+O_{n-1} O_{n}}=\frac{r_{n}}{3 r_{n-1}+r_{n}+r_{n-1}} \Longrightarrow$ $r_{n}=2 r_{n-1}=2^{n-1} \frac{16}{3 \cdot 2 \sqrt{2}}$. We want the highest $n$ such that $O_{n}$ is contained inside $\triangle A D E$. Let the incircle of $\triangle A D E$ be tangent to $A D$ at $X$; then the inradius of $\triangle A D E$ is $\frac{A X}{\tan \frac{\theta}{2}}=\frac{\frac{500+200-\frac{1100}{3}}{2}}{2 \sqrt{2}}=\frac{500}{3 \cdot 2 \sqrt{2}}$. We want the highest $n$ such that $r_{n} \leq \frac{500}{3 \cdot 2 \sqrt{2}}$; thus $2^{n-1} \cdot 16 \leq 500 \Longrightarrow n=5$.
Convex quadrilateral $B C D E$ lies in the plane. Lines $E B$ and $D C$ intersect at $A$, with $A B=2$, $A C=5, A D=200, A E=500$, and $\cos \angle B A C=\frac{7}{9}$. What is the largest number of nonoverlapping circles that can lie in quadrilateral $B C D E$ such that all of them are tangent to both lines $B E$ and $C D$ ?