Need geometry proofs for equilateral triangle. The side length of square $ABCD$ is $x$, $P$ is a dot in square $ABCD$, $AP=DP$, $\angle APD=150$ degrees, please prove that $\triangle BCP$ is an equilateral triangle. 

I already proved that  $\triangle APB \cong \triangle DPC$, and is there any other methods?
 A: Here are some possibilities to show that $\Delta PBC$ is equilateral.
Because of the obvious symmetry w.r.t. the line $p$ through $P$ which is parallel
to $AB$ and $CD$, an perpendicular on the other two sides, the triangle $\Delta PBC$
is isosceles, $PB=PC$.
We need only one angle in this triangle. Let us start.

(1)
Reverse engineering solution:
We show that $P$ has the stated property by introducing the point $Q$ inside the (interior of the) square $\square ABCD$ with $\Delta QBC$ equilateral,
then compute the angles in $\Delta AQD$, obtain the fact that $P=Q$. This "converse" approach leads to a quick end, since we "start with more"
and need to show a simple thing. (Starting with the "simple thing" is not the same game.)
Let $Q$ be as above. 

Then $BA=BC=BQ$, so $\Delta ABQ$ isosceles, so the angles of this triangle in $A,Q$ are equal to
$(180^\circ -\widehat{ABP})/2=(180^\circ - 30^\circ)/2=75^\circ$. This implies $\widehat{QAD}=90^\circ -75^\circ =15^\circ$.
By symmety we also have $QDA=15^\circ$, so $\widehat{AQD}=150^\circ$. This shows $Q=P$.
$\square$

(2)
Trigonometric proof: We may and do assume that the side of the square is equal to $2$.
Let us draw the height from $P$ on the side $BC$ of $\Delta PBC$, it it supported on the
already mentioned symmetry line $p$. It intersects $BC$ in a point $M$, say, and $DA$ in $N$.
We know $MN=AB=2$, and
$$
PN=AN\cdot\tan 15^\circ=\tan 15^\circ\ .
$$
So we only need an explicit formula for the above value, let us denote it by $y:=\tan 15^\circ$ for short. We obtain $y$ from
$$
\frac 1{\sqrt3}=\tan 30^\circ=\tan(2\cdot 15^\circ)=\frac{2y}{1-y^2}\ ,
$$
which leads to the equation $0=y^2 + 2\sqrt 3 y-1$, with solutions $y=-\sqrt 2\pm \sqrt {3+1}=-\sqrt 3\pm 2$.
We accept only the positive solution for $PN$, so $PN=2-\sqrt 3$, so $PM=MN-PN=2-(2-\sqrt 3)=\sqrt 3$.
This gives $\tan \widehat{PBM}=PM:BM=\sqrt 3:1$, so
$$
\widehat{PBM}=60^\circ\ .
$$
$\square$

(3)
Direct proof: We accept the challenge and give (in contrast to (1)) a "direct proof", i.e. construct points starting with $P$
and in direct relation to the data given for $P$. We construct the points $Q,R,S$  as we have also constructed $P$ inside of the square
$\square ABCD$, so that the triangles
$\Delta QAB$,
$\Delta RBC$,
$\Delta SCD$
are isosceles in $Q,R,S$ and have in these vertices the angle of $150^\circ$.
Equivalently, we can construct $Q,R,S$ by rotations of $P$ around the center of symmetry of the square
corresponding to angles of $90^\circ$, $180^\circ$, $270^\circ$.
We then obtain the following figure.

In particular
$$
PA=AQ=QB=BR=RC=CS=SD=DP\ .
$$
The acute angles in the constructed triangles
$\Delta PDA$,
$\Delta QAB$,
$\Delta RBC$,
$\Delta SCD$
have each $(180^\circ -150^\circ)/2=15^\circ$,
so for instance the angle $\widehat{PAQ}$ is $90^\circ-15^\circ-15^\circ=60^\circ$.
In particular, $\Delta PAQ$ is equilateral. (Similarly the other three triangles
are also equilateral.) We can extend the above equalities with
$$
PA=AQ=QB=BR=RC=CS=SD=DP=PQ=QR=RS=SP\ .
$$
It is easy now to compute the angles in $PQRS$, so that we quickly show it is a square.
Now let us consider the triangles $\Delta QAB$ and $\Delta QPB$. They are isosceles, the "short" sides
are equal, $QA=QB=QP$, and the angles in $Q$ have the same measure
$\widehat{AQB}=150^\circ=60^\circ+90^\circ=\widehat{BQR}+\widehat{RQP}=\widehat{BQP}$.
We obtain their equality, so in particular
$$
AB=PB\ .
$$
This leads to $PB=BC=CP$, so $\Delta PBC$ equilateral.
$\square$
A: Adapted from a previous answer of mine:

A: 
Here is a geometric proof:
Extend AP to meet CD at Q and DT$\perp$AQ. Then, AQ = 2PD due to isosceles triangle PDQ and DT = $\frac {1}2$PD due to 30-60 right triangle PDT.  The area of the right triange ADQ is 
$$\frac12 AD\cdot DQ = \frac12 AQ \cdot DT \implies \frac {DQ}{DP} = \frac{DP}{DC}$$
Due to the shared angle $\angle$PDQ, the triangles PQD and PDC are similar, which yields $\angle$PCB = 90 - $\angle$PCD = 90 - $\angle$DPQ = 60. Thus, the triangle PBC is equilateral.
