Let:

Q lie on AC and N lie outside http://www.cut-the-knot.org/gifs/triangle.gif ABC;
http://www.cut-the-knot.org/gifs/triangle.gif AQN be congruent to http://www.cut-the-knot.org/gifs/triangle.gif BPM;
D be the centre of http://www.cut-the-knot.org/gifs/triangle.gif AQN.

Then:

DF passes through E (the centre of symmetry between triangles AQN and PMB);
http://www.cut-the-knot.org/gifs/triangle.gif BCD is congruent to http://www.cut-the-knot.org/gifs/triangle.gif ACF.

So:

http://www.cut-the-knot.org/gifs/triangle.gif DCF is equilateral and CE is its height.

http://geocities.com/narbab/ctk/equilat.png

Even simpler. Very good.

D be the centre of
http://www.cut-the-knot.org/gifs/triangle.gif AQN.

I realise this isn't very significant, but in the diagram you have D as the centre of PMB, and F as the centre of AQN.

>I realise this isn't very significant, but in the diagram
>you have D as the centre of PMB, and F as the centre of AQN.

Of course. I was so excited with this idea that I posted the very
first version without rereading it.
I even forgot to sign it with my name: Narcyz R. Babnis.