6. As shown in Figure 1, in $\triangle A B C$, $A D$, $B E$, and $C F$ intersect at point $P$, $\frac{A F}{B F}=\frac{A E}{C E}$, and $A D \perp B C$. A line parallel to $A B$ is drawn through point $C$, intersecting the extension of $B E$ at point $G$. If $P E=1$ and $E G=2$, then the value of $B P$ is ( ).
(A) 3
(B) 2
(C) $\sqrt{3}$
(D) $\sqrt{2}$
6. C.

By Ceva's Theorem,
$$
\frac{B D}{D C}=\frac{E A}{C E} \cdot \frac{F B}{A F}=1
$$
$\Rightarrow D$ is the midpoint of side $B C$
$$
\Rightarrow A B=A C \text {. }
$$

It is easy to see that, $\angle A B P=\angle A C P, B P=C P$.
By $\angle G=\angle A B P=\angle A C P$
$\Rightarrow \triangle P C E \sim \triangle P G C$
$\Rightarrow B P^{2}=P C^{2}=P E \cdot P G=3$
$\Rightarrow B P=\sqrt{3}$.