36 Given three fixed points $A, B, C$ and a moving point $P$, if the areas of $\triangle P A B$ and $\triangle P B C$ are equal, find the equation of the locus of point $P$.
36 First, let points $A$, $B$, $C$ be non-collinear. As shown in Figure 25.
Let $E$ be the midpoint of $AC$, and for any point $P$ on line $BE$,
$$
\frac{\text{the distance from } P \text{ to } AB}{\text{the distance from } E \text{ to } AB} = \frac{PB}{EB} = \frac{\text{the distance from } P \text{ to } BC}{\text{the distance from } E \text{ to } BC}
$$

Therefore,
$$
\frac{S_{\triangle PAB}}{S_{\triangle EAB}} = \frac{S_{\triangle PBC}}{S_{\triangle EBC}}
$$

Thus, $S_{\triangle PAB} = S_{\triangle PBC}$.
Now, draw a line $BF$ parallel to $AC$ through $B$, for any point $Q$ on $BF$,
$$
S_{\triangle QAB} = S_{\triangle QBC} = S_{\triangle CQB} = S_{\triangle QBC}.
$$

Therefore, the points on lines $BE$ and $BF$ all meet the requirement.
Conversely, suppose $S_{\triangle PAB} = S_{\triangle PBC}$. There are two cases:
(i) $P$ is inside $\angle ABC$ or its vertical angle.

In this case, (2) holds, so

Thus, $P$ lies on line $BE$.
(ii) $P$ is inside the exterior angle of $\angle ABC$.

From $S_{\triangle APB} = S_{\triangle CPB}$, we know $BP \parallel AC$, i.e., $P$ lies on line $BE$.
Therefore, $BE$ and $BF$ are the required loci.
When points $A$, $B$, $C$ are collinear, there are two cases: (1) When $B$ is the midpoint of segment $AC$, $P$ is any point in the plane (except on line $AC$). (2) When $B$ is not the midpoint of segment $AC$, this problem has no solution.