6. As shown in Figure $3, P, D, A$ and $P, B, C$ are collinear respectively, $AB$ intersects $CD$ at $Q$, and $PQ$ bisects $\angle A P C$. Let $P A=$ $a, P B=b, P C=c, P D=$ $d$. Then the relationship between $a, b, c, d$ is ( ).
(A) $a b=c d$
(B) $a+b=c+d$
(C) $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}+\frac{1}{d}$
(D) $\frac{a c}{b d}=\frac{a+c}{b+d}$
6.C.

As shown in Figure 8, draw $Q E / / P C$ intersecting $P A$ at $E$.
Let $Q E=x$, it is easy to see that
$$
P E=x, A E=a-x .
$$

By $\triangle A Q E \backsim \triangle A B P$, we get
$$
\frac{x}{b}=\frac{a-x}{a}=1-\frac{x}{a} .
$$

Rearranging gives $\frac{x}{a}+\frac{x}{b}=1$, which means
$$
\frac{1}{a}+\frac{1}{b}=\frac{1}{x} \text {. }
$$

Similarly, by $\triangle D Q E \backsim \triangle D C P$, we get
$$
\frac{1}{c}+\frac{1}{d}=\frac{1}{x} \text {. }
$$

Therefore, $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}+\frac{1}{d}$.