Given that the perpendicular bisectors of the sides of AXYZ meet at a single point S, which is the circumcenter of the triangle, I can use the properties of the circumcenter to find the lengths XS, YZ, and VY.
To start, I can use the fact that the circumcenter of a triangle is equidistant from its vertices. Therefore, I can write:
SV = SY (since S is equidistant from V and Y)
SZ = SX (since S is equidistant from Z and X)
SV = VX (since S is equidistant from V and X)
SU = UY (since S is equidistant from U and Y)
From the given information, we know that VS = 66, YU = 86, and ZS = 110. I can use this information to solve for the unknown lengths.
To find XS, I can use the fact that SZ = SX. We know that ZS = 110, so:
XS = ZS = 110
To find YZ, I can first find VY by using the fact that SV = VX. I know that VS = 66, so:
VY = VS - SY = 66 - YU/2
Now I can use the fact that SU = UY to solve for YZ:
YZ = YU - 2UY/2 = YU - UY = 86 - (66 - VY) = 20 + VY
So, YZ = 20 + VY
Therefore, I still need to find the value of VY. We can use the fact that SV = SY and the equation we found for VY above to solve for it:
SV = SY
66 = VY + 86/2
66 = VY + 43
VY = 23
Therefore, we have:
XS = 110
YZ = 20 + VY = 43
VY = 23
So, XS = 110, YZ = 43, and VY = 23.