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.

You can solve this without any trig if you consider the properties of a equilateral triangle, and the fact that you've created six right triangles in which you know the length of the hypotenuse and the relationship:
$a+b = c+d = e+f$
$a^2 + g^2 = |AP|^2$
$b^2 + g^2 = |BP|^2$
$c^2 + h^2 = |BP|^2$
$d^2 + h^2 = |CP|^2$
$e^2 + i^2 = |CP|^2$
$f^2 + i^2 = |AP|^2$
Then:
$a+b = c+d = e+f$
$a^2 + g^2 = 9$
$b^2 + g^2 = 16$
$c^2 + h^2 = 16$
$d^2 + h^2 = 25$
$e^2 + i^2 = 25$
$f^2 + i^2 = 9$
Then: 
$a+b = c+d = e+f = s$
$b^2 - a^2 = 7$
$d^2 - c^2 = 9$
$e^2 - f^2 = 16$
Then:
$b^2 - (s-b)^2 - 7 = s^2 + 2sb - 7 = 0$
$d^2 - (s-d)^2 - 9 = s^2 + 2sd - 9 = 0$
$e^2 - (s-e)^2 - 16 = s^2 + 2se - 16 = 0$
