10. $[7]$ Let $A B C$ be a triangle with $B C=2007, C A=2008, A B=2009$. Let $\omega$ be an excircle of $A B C$ that touches the line segment $B C$ at $D$, and touches extensions of lines $A C$ and $A B$ at $E$ and $F$, respectively (so that $C$ lies on segment $A E$ and $B$ lies on segment $A F$ ). Let $O$ be the center of $\omega$. Let $\ell$ be the line through $O$ perpendicular to $A D$. Let $\ell$ meet line $E F$ at $G$. Compute the length $D G$.
Answer: 2014024 Let line $A D$ meet $\omega$ again at $H$. Since $A F$ and $A E$ are tangents to $\omega$ and $A D H$ is a secant, we see that $D E H F$ is a harmonic quadrilateral. This implies that the pole of $A D$ with respect to $\omega$ lies on $E F$. Since $\ell \perp A D$, the pole of $A D$ lies on $\ell$. It follows that the pole of $A D$ is $G$.

Thus, $G$ must lie on the tangent to $\omega$ at $D$, so $C, D, B, G$ are collinear. Furthermore, since the pencil of lines $(A E, A F ; A D, A G)$ is harmonic, by intersecting it with the line $B C$, we see that $(C, B ; D, G)$ is harmonic as well. This means that
$$
\frac{B D}{D C} \cdot \frac{C G}{G B}=-1 .
$$
(where the lengths are directed.) The semiperimeter of $A B C$ is $s=\frac{1}{2}(2007+2008+2009)=3012$. So $B D=s-2009=1003$ and $C D=s-2008=1004$. Let $x=D G$, then the above equations gives
$$
\frac{1003}{1004} \cdot \frac{x+1004}{x-1003}=1 \text {. }
$$

Solving gives $x=2014024$.
Remark: If you are interested to learn about projective geometry, check out the last chapter of Geometry Revisited by Coxeter and Greitzer or Geometric Transformations III by Yaglom.