Example 3. In the square $ABCD$, there is a point $P$. If $PA, PB, PC, PD$ divide $\triangle ABCD$ into four triangles with areas in geometric progression, try to determine: How is the position of point $P$ determined?
Analysis: Draw $EF \perp AB$ through $P$, with $E, F$ as the feet of the perpendiculars. It is easy to prove that $S_{\triangle PAB} + S_{\triangle PCD} = \frac{1}{2} \cdot EF \cdot AB = \frac{1}{2} S_{DABCD}$. Therefore, 
$S_{\triangle PAB} + S_{\triangle PCD} = S_{\triangle PBC} + S_{\triangle PDA}$.
Let the areas of the four triangles be $a, ar, ar^2, ar^3$ ($a > 0, r > 0$). Using the implicit equal sum relationship, the possible equations are: (1) $a + ar = ar^2 + ar^3$;
(2) $a + ar^2 = ar + ar^3$;
(3) $a + ar^3 = ar + ar^2$.
From (1) and (2), it is easy to get $r = 1$. From (3), by rearranging and factoring, it is also not difficult to get $r = 1$. Thus, $a = ar = ar^2 = ar^3$, or $S_{\triangle PAB} = S_{\triangle PBC} = S_{\triangle PCD} = S_{\triangle PDA}$. At this point, $P$ must be the intersection point of the diagonals of $\square ABCD$, which is the only one.