Determine all points $P$ on $\triangle ABC$ so that $|\triangle PAB| = |\triangle PBC|= |\triangle PAC|$. 
Determine all points $P$ on $\triangle ABC$ so that $|\triangle PAB| = |\triangle PBC| = |\triangle PAC|$.  Here, $|\triangle XYZ|$ denotes the area of $\triangle XYZ$.

I've tried drawing it up, and $|\triangle PAB|$ = $|\triangle PAC|$ has to mean that the distance from the points $B$ and $C$ to where $PA$ is, are equal. Which in turn has to mean that also $AP$ = $AB$ and $AC$ right? Does that mean that $P$ is always in a place so that $AP$=$BP$=$CP$? 
 A: Here's a coordinate proof of the generalized question, where $P$ is not necessarily confined to the interior of $\triangle ABC$. Let's take
$$A = (0,0) \qquad B = (h,2b) \qquad C = (h,2c)$$
so that
$$\overleftrightarrow{AB}:\; 2b x - h y = 0 \qquad\overleftrightarrow{AC}:\;2c x-hy=0 \qquad \overleftrightarrow{BC}:\;x=h$$
Now, we consider how to make two of the desired areas equal. For $P=(x,y)$,
$$\begin{align}
|\triangle APC| = |\triangle ABP| &\iff \frac12\,|AC|\,\left(\text{dist from $P$ to $\overleftrightarrow{AC}$}\right) = \frac12\,|AB|\,\left(\text{dist from $P$ to $\overleftrightarrow{AB}$}\right) \\[6pt]
&\iff \sqrt{4b^2+h^2}\cdot\frac{\left|\;2bx-hy\;\right|}{\sqrt{4b^2+h^2}} = \sqrt{4c^2+h^2}\cdot\frac{\left|\;2cx-hy\;\right|}{\sqrt{4c^2+h^2}} \\[6pt]
&\iff \left|\;2bx-hy\;\right|=\left|\;2cx-hy\;\right| \\[6pt]
&\iff 2bx-hy = \pm\left(2cx-hy\right) \\[6pt]
&\iff x = 0 \quad\text{or}\quad hy=(b+c)x
\end{align}$$
Thus, to make two triangle areas match, $P$ must lie on the line through $A$ parallel to $\overline{BC}$, or else the (median) line through $A$ and the midpoint of $\overline{BC}$.
To make all three triangle areas match, $P$ must also lie on either the parallel or the median through $B$, and on either the parallel or the median through $C$. This happens in four ways, as shown in the image below: where all three medians meet (the centroid, $K$), or the three ways in which one median meets two parallels (points $A^\prime$, $B^\prime$, $C^\prime$, each of which completes a parallelogram with the vertices of $\triangle ABC$). $\square$

A: It says that $P$ is the center of gravity of $\Delta ABC$.
Indeed, let $AP\cap BC=\{A'\}.$
Thus, by the given $$d(C,AP)=d(B,AP),$$ which says $$BA'=CA',$$ which gives $AA'$ is a median of the triangle.
A: Here is a proof of Blue's claim.  Without the restriction that $P$ lie inside $\triangle ABC$, there are exactly four possible locations of $P$.  We first need the following lemma.

Lemma:  Let $X,Y,Z$ be given points on a plane.  For a geometric shape $S$, write $|S|$ for its area.  Then, the locus of the point $T$ such that $|\triangle TXY|=|\triangle TXZ|$ is the union $L_1\cup L_2$ of two straight lines $L_1$ and $L_2$: the first one $L_1$ is the line joining $X$ to the midpoint of the line segment $YZ$, and the second one $L_2$ is the line parallel to $YZ$ passing through $X$.

First, it is easy to see that any $T$ on $L_1\cup L_2$ satisfies the required condition.  We will show that every $T$ that satisfies the condition that $|\triangle TXY|=|\triangle TXZ|$ must be on $L_1$ or $L_2$.  We start by drawing the line $TX$.  There are two possibilities: $Y$ and $Z$ are on the opposite side of $TX$, or $Y$ and $Z$ are on the same side of $TX$.
In the case that $Y$ and $Z$ are on the opposite side of $TX$, let $M$ be the intersection of the line $TX$ and the segment $YZ$.  Draw perpendiculars from $Y$ and $Z$ to meet the line $TX$ and $Y'$ and $Z'$, respectively.  Then, $|\triangle TXY|=|\triangle TXZ|$ translates to $YY'=ZZ'$.  So, we have right triangles $\triangle YY'M$ and $\triangle ZZ'M$ with $YY'=ZZ'$, $\angle YY'M=90^\circ=\angle ZZ'M$, and $\angle YMY'=\angle ZMZ'$.  Thus, $\triangle YY'M\cong \triangle ZZ'M$, so $YM=ZM$, and $M$ is the midpoint of $YZ$.  Hence, $T\in L_1$.
Now, we suppose that $Y$ and $Z$ are on the same side of $TX$.  Draw perpendiculars from $Y$ and $Z$ to meet the line $TX$ and $Y'$ and $Z'$, respectively.  Then, $|\triangle TXY|=|\triangle TXZ|$ translates to $YY'=ZZ'$.  Therefore, $\square YY'Z'Z$ is a rectangle, so $YZ\parallel TX$, making $T\in L_2$.

Now, for our triangle $\triangle ABC$, let $L_1^V$ for $V\in\{A,B,C\}$ be the line joining $V$ to the midpoint of its opposite side, and $L_2^V$ the line passing through $V$ parallel to its opposite side.  Note that there are only four possible intersections of the form $L_i^A\cap L_j^B\cap L_k^C$, where $i,j,k\in\{1,2\}$.  When $i=j=k=1$, this produces the centroid $G$.  When exactly one among $i,j,k$ is $1$, and the remaining indices are $2$, this gives you the points $Q_A$, $Q_B$, $Q_C$ such that $\square ABQ_AC$, $\square BCQ_B A$, and $\square CA Q_CB$ are parallelograms.  Thus, all possible points $P$ are $P\in\{G,Q_A,Q_B,Q_C\}$.
A: Let's say $|\triangle PBC|=|\triangle ABP|=|\triangle ACP|=S/2$, then the corresponding heights (distance of P to AB, BC, AC, respectively) $h_1=S/a, h_2=S/b, h_3=S/c$
For AB, points of distance $h_1$ to it consists of 2 lines on either side of it (parallel to it); name them $l_1,l_2$. Similarly, there are 2 lines whose points has h_2 distance to BC; name them $l_3,l_4$. These 4 lines have 4 distinct intersection for given $S>0$. If we can somehow adjust value of $S$ so that one of these intersection point Q is of distance $h_3$ to AC, we have find a solution to original problem. This proved that we can have at most 4 such points.
