To calculate side of the Equilateral triangle The figure is an equilateral triangle. 3 line segments , which meet at a(any) point in the triangle , are of the length 5cm, 4cm, and 3 cm as shown in the figure. Find the side of the equilateral triangle.

Theorem: Let $ABC$ be an equilateral triangle, with a point $P$ inside it such that $PC^2=PA^2+PB^2$, then $\angle APB=150^{\circ}$.

So to do that, we cleverly construct a point $D$ such that $APD=60^{\circ}$ and $AP=PD$. Hence, $APD$ is equilateral and $AP=AD$.
Now, as $60^{\circ}=\angle CAB=\angle DAP$, so, $\angle CAP = \angle BAD$, which is just $\angle PAB$ subtracted from them. We know that $AB=AC$.
So, by SAS criteria $PAC \cong BAD$. Hence, $PC=BD$. This means that $BD^2=PD^2+BP^2$. Thus by converse of Pythagoras Theorem, $\angle BPD = 90^{\circ}$, which means $\angle APB=150^{\circ}$.
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Corollary: If $ABC$ be an equilateral triangle, with a point $P$ inside it such that $PC=5$, $PB=4$ and $PA=3$ then $AB=\sqrt{25+12\sqrt3}$.
Note that $3^2+4^2=5^2$, so $\angle APB=150^{\circ}$. Hence, by the cosine law, $AB^2=3^2+4^2-2\cdot12\cdot \cos 150^{\circ}=25+12\sqrt3$. So: $$AB=\sqrt{25+12\sqrt3}$$
