5. Given two circles $\Gamma_{1}$ and $\Gamma_{2}$ are externally tangent at point $A$, circle $\Gamma$ is externally tangent to $\Gamma_{1}$ and $\Gamma_{2}$ at points $B$ and $C$ respectively. Extend the chord $B A$ of circle $\Gamma_{1}$ to intersect circle $\Gamma_{2}$ at point $D$, extend the chord $C A$ of circle $\Gamma_{2}$ to intersect circle $\Gamma_{1}$ at point $E$, and extend the chords $E B$ and $D C$ to intersect circle $\Gamma$ at points $F$ and $G$ respectively: If $B C=5, B F=12$, find the length of $B G$.
5. First, prove that quadrilateral $B C G F$ is a rectangle:

In fact, let $K L, B M, C N$ be the common tangents of circles $\Gamma_{1}$ and $\Gamma_{2}$, $\Gamma_{1}$ and $\Gamma$, $\Gamma_{2}$ and $\Gamma$ (Figure 2). According to the inscribed angle and the angle between a tangent and a chord, we have
$$
\angle A B E=\angle E A K=\angle C A L=\angle A D C \text {. }
$$

Then $B E / / C D$
$$
\begin{array}{l}
\Rightarrow \angle C B E+\angle B C D=180^{\circ} . \\
\text { Also } \angle B C N=\angle C B M, \\
\angle A C N=\angle C A L=\angle A B E, \\
\angle A B M=\angle B A L=\angle A C D,
\end{array}
$$
$$
\begin{array}{l}
\text { Therefore } \angle E B C=\angle D C B=90^{\circ} \\
\Rightarrow \angle B F G=\angle B C D=90^{\circ} \text {. }
\end{array}
$$

Thus, $B G=\sqrt{5^{2}+12^{2}}=13$.