Six, point $P$ is inside
$\square A B C D$.
$P A, P B, P C, P D$
divide the quadrilateral $ABCD$ into four
triangles, their areas are $a, a r, a r^{2}, a r^{3}(a>0, r>0)$. Determine the position of point $P$. Please explain your reasoning.
Six, Hint: Using $S \triangle P^{A} A_{R}+S \triangle P C$ $=S \triangle \mathrm{PBC}+S \triangle \mathrm{PDA}$, we can get $r=1$, so point $P$ is the intersection of $A C$ and $B D$.

