+98 Let \(ABC\) be any triangle. Equilateral triangles \(BCX\), \(ACY\), and \(BAZ\) are constructed such that none of these triangles overlaps triangle \(ABC\).
(a) Draw a triangle \(ABC\) and then sketch the remainder of the figure. It will help if \(\triangle{ABC}\) is not isosceles (or equilateral).
(b) Show that, regardless of choice of \(\triangle{ABC}\), we always have \(AX=BY=CZ\).
+118608 We need someone better than me at geometry proofs here 
Any takers ??
Thios is the diagram.
The aim is to show that the 3 red lines are equal in length.
+128407
This isn't as difficult as it first appears....although I have to admit, it threw me for a bit...!!!
Note that we have triangles ZAC and BAY
And two sides of ZAC - namely, b and c - are equal to two sides of BAY
And angle ZAB + angle CAB = angle YAC + angle CAB
Thus....by SAS......triangle ZAC is congruent to triangle BAY....therefore ZC = BY
Likewise...we have two other triangles, BCY and XCA
And two sides of BCY - namely, a and b - are equal to to sides of XCA
And angle XCB + angle ACB = angle YCA + angle ACB
Thus.....by SAS......triangle BCY is congruent to triangle XCA
Thus BY = XA = AX
But BY was also proved equal to ZC = CZ
Therefore.... AX = BY = CZ
P.S. - Melody should get some points for providing the diagram....without that, I wouldn't have understood the layout....!!!!
