Equilateral triangle geometric problem I have an Equilateral triangle with unknown side $a$. The next thing I do is to make a random point inside the triangle $P$. The distance $|AP|=3$ cm, $|BP|=4$ cm, $|CP|=5$ cm.

It is the red triangle in the picture. The exercise is to calculate the area of the Equilateral triangle (without using law of cosine and law of sine, just with simple elementary argumentation).
The first I did was to reflect point $A$ along the opposite side $a$, therefore I get $D$. Afterwards I constructed another Equilateral triangle $\triangle PP_1C$.
Now it is possible to say something about the angles, namely that $\angle ABD=120^{\circ}$, $\angle PBP_1=90^{\circ} \implies \angle APB=150^{\circ}$ and $\alpha+\beta=90^{\circ}$
Now I have no more ideas. Could you help me finishing the proof to get $a$ and therefore the area of the $\triangle ABC$. If you have some alternative ideas to get the area without reflecting the point $A$ it would be interesting.

A generic Solution for any lengths $d_1, d_2, d_3$:
The distances $d1, d2$ and $d3$ are from points $C, B$ and $A$, respectively, 
to a fourth point $E$ in the interior of the equilateral $\triangle ABC$.
First, we rotate the figure of $\triangle ABC$ and $E$ about $C$ through $60^\circ$, 
resulting in an equilateral triangle $\triangle BHC$ with $G$ congruent to $\triangle ABC$ with 
$E$. In particular we see that $CE$ is rotated through $60^\circ$ to $CG$, 
which shows that $\triangle CEG$ is an equilateral triangle, so that $EG = CE = CG 
= d1$. This triangle I will now refer to as $EQUI(d1)$: an equilateral 
triangle with side $d1$:

Now we draw some simple conclusions:


*

*The point $E$ divides the original triangle into three triangles, 
which I have made into a red one $\triangle AEC$, a yellow one $\triangle AEB$, and a white 
one $\triangle BEC$.

*The area of the white and red triangles together is the same as 
the area of $\triangle BEG$ and $\triangle CEG$ together.

*BEG is the triangle with sides equal to the given $d_1, d_2, d_3$, 
which I will from now refer to such a triangle as $T$.

*We conclude that the white and red triangles together have an area 
equal to $[T] + [EQUI(d1)]$ (where $[x]$ is the area of $x$).
With similar reasoning we can draw similar conclusions about the 
areas of the white and yellow triangle together, and about the areas 
of the red and yellow triangle together.
Adding the areas of white+red, white+yellow, and red+yellow gives 
$2*[ABC]$ It also gives $3*[T]+[EQUI(d1)]+[EQUI(d2)]+[EQUI(d3)]$
So, if we read in areas, we have:
$[ABC] = 1.5*[T]+ ([EQUI(d1)]+[EQUI(d2)]+[EQUI(d3)])/2.$
We can also find the side length a by using 
Area of equilateral triangle => 
$[ABC] = sqrt(3)/4 * a^2$
$√3/4 * a^2 = 1.5*[T]+ ([EQUI(d1)]+[EQUI(d2)]+[EQUI(d3)])/2.$
Which gives $a = √((4/√3)*(1.5*[T]+ ([EQUI(d1)]+[EQUI(d2)]+[EQUI(d3)])/2))$
