Let ABC be any triangle. Equilateral triangles BCX, ACY, and BAZ are constructed such that none of these triangles overlaps triangle ABC.

Here is the pic but I don't have an answer yet.

Chris you are better at these than I am, why don't you give it a go?

Using your diagram, I can prove that  AX  = BY

Note that we have two triangles, ACX  and BCY

AC    =  CY   =  "b"

And

XC  = BC  =  "a"

And  angle ACX  = angle YCB

So.... by SAS   .... triangle  ACX  is congruent to triangle YCB

So... AX  =  BY

But...I have not figured out how to prove that CZ  is also equal to these

Maybe someone can provide the final step  ???

Thanks Chris,

The bit that follows is exactly the same as Chris said and then i have continued the proof   

I am going to call the angles in the original triangle 

All angles in the 3 other triangles are all 60 degrees because they are equilateral triangles.

Consider

  \(\triangle ACX\;\; and \;\; \triangle YCB \\ AC=b, \;\;CX=a\;\; and \;\; \angle ACX=\angle C+60^\circ\\ YC=b, \;\;CB=a\;\; and \;\; \angle YCB=\angle C+60^\circ\\ \therefore \triangle ACX\cong \triangle YCB\qquad \text{(2 sides and included angle test)} \\ \therefore AX=BY \qquad \text{Corresponding sides in congruent triangles.}\)

\(\text{Now consider} \triangle ABY\;\; and \;\; \triangle AZC \\ AB=AZ=c\\ AY=AC=b\\ \angle YAB=\angle CAZ=\angle A+60^\circ\\ \therefore \triangle ABY\cong \triangle AZC\qquad \text{(2 sides and included angle test)} \\ \therefore BY=ZC \qquad \text{Corresponding sides in congruent triangles.}\\ \)

\(\therefore AX=BY=ZC\\~\\ QED\)

Ah....thanks, Melody.....it's funny how it's sometimes hard to see....and then when someone else points it out.....it seems really easy.....LOL!!!!!

I think that my drawing of this maybe wasn't quite correct.....Melody's made it easy to see.....