33. $[\mathbf{1 7}]$ Let $A B C$ be a triangle with $A B=5, B C=8$, and $C A=7$. Let $\Gamma$ be a circle internally tangent to the circumcircle of $A B C$ at $A$ which is also tangent to segment $B C$. $\Gamma$ intersects $A B$ and $A C$ at points $D$ and $E$, respectively. Determine the length of segment $D E$.
Answer: $\frac{40}{9}$
First, note that a homothety $h$ centered at $A$ takes $\Gamma$ to the circumcircle of $A B C, D$ to $B$ and $E$ to $C$, since the two circles are tangent. As a result, we have $D E \| B C$. Now, let $P$ be the center of $\Gamma$ and $O$ be the circumcenter of $A B C$ : by the homothety $h$, we have $D E / B C=A P / A O$.
Let $\Gamma$ be tangent to $B C$ at $X$, and let ray $\overrightarrow{A X}$ meet the circumcircle of $A B C$ at $Y$. Note that $Y$ is the image of $X$ under $h$. Furthermore, $h$ takes $B C$ to the tangent line $l$ to the circumcircle of $A B C$ at $Y$, and since $B C \| l$, we must have that $Y$ is the midpoint of arc $\overparen{B C}$. Therefore, $A X$ bisects $\angle B A C$.
Now, let $Z$ be the foot of the altitude from $A$ to $B C$, and let $M$ be the midpoint of $B C$, so that $O M \perp B C$. Note that $A P / A O=Z X / Z M$. Now, letting $B C=a=8, C A=b=7$, and $A B=c=5$, we compute
$$
B Z=c \cos B=\frac{c^{2}+a^{2}-b^{2}}{2 a}=\frac{5}{2}
$$
by the Law of Cosines,
$$
B X=\frac{a c}{b+c}=\frac{10}{3}
$$
by the Angle Bisector Theorem, and
$$
B M=4 .
$$

To finish,
$$
D E=\frac{(A P)(B C)}{A O}=\frac{(Z X)(B C)}{Z M}=\frac{(5 / 6)(8)}{(3 / 2)}=\frac{40}{9}
$$