Proof: From the given, we know that $QA$ and $QB$ are two secants of the incircle $\odot O$ of $\angle EQF$. Let $AC$ and $BD$ intersect at point $R$.
Then, by Property 3, we know that $P, F, R, E$ are collinear.
Therefore, $P, E, F$ are collinear.
To introduce Example 8, let's discuss the figure represented by Property 2(2).

As shown in Figure 7, $\odot O$ is tangent to the sides of $\angle T'N'S'$ at points $T'$ and $S'$. Then, $M$ is the midpoint of the chord $IJ$ cut by the secant through point $N'$ on $\odot O$ if and only if $N', T', M, S'$ are concyclic.

Handling this figure with inversion, we can obtain a new proposition: $\square$

Let the power of point $M$ with respect to $\odot O$ be $k$, and perform the inversion $I(M, k)$. Then $\odot O$ is a self-inverse circle. Let the inverse of point $X'$ be $X$. The inverse of line $N'S'$ is a circle $\odot O_1$ passing through points $M$ and $N$ and internally tangent to $\odot O$ at point $S$. The inverse of line $N'T'$ is a circle $\odot O_2$ passing through points $M$ and $N$ and internally tangent to $\odot O$ at point $T$. The line $MN'$ remains unchanged. When $M$ does not coincide with the center $O$, $\odot O_1$ and $\odot O_2$ are not equal. And $N', T', M, S'$ are concyclic if and only if $S, N, T$ are collinear.
Thus, the inverted proposition is the following example.